How to detect integer overflow in C [duplicate]How do I detect unsigned integer multiply overflow?Detecting signed overflow in C/C++How do I check if A+B exceed long long? (both A and B is long long)Test for overflow in integer additionHow do you set, clear, and toggle a single bit?How do I detect unsigned integer multiply overflow?Improve INSERT-per-second performance of SQLite?Catch and compute overflow during multiplication of two large integersWhen to address integer overflow in CDetecting signed overflow in C/C++How to find (all) integer overflows in a C program?Speed comparison with Project Euler: C vs Python vs Erlang vs HaskellHow to check if overflow occured?automatic overflow detection in C++?
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How to detect integer overflow in C [duplicate]
How do I detect unsigned integer multiply overflow?Detecting signed overflow in C/C++How do I check if A+B exceed long long? (both A and B is long long)Test for overflow in integer additionHow do you set, clear, and toggle a single bit?How do I detect unsigned integer multiply overflow?Improve INSERT-per-second performance of SQLite?Catch and compute overflow during multiplication of two large integersWhen to address integer overflow in CDetecting signed overflow in C/C++How to find (all) integer overflows in a C program?Speed comparison with Project Euler: C vs Python vs Erlang vs HaskellHow to check if overflow occured?automatic overflow detection in C++?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;
This question already has an answer here:
Detecting signed overflow in C/C++
12 answers
How do I detect unsigned integer multiply overflow?
31 answers
We know CPython promotes integers to long integers (which allow arbitrary-precision arithmetic) silently when the number gets bigger.
How can we detect overflow of int
and long long
in pure C?
c overflow
marked as duplicate by sleske, ead, phuclv, Cody Gray♦ 2 days ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Detecting signed overflow in C/C++
12 answers
How do I detect unsigned integer multiply overflow?
31 answers
We know CPython promotes integers to long integers (which allow arbitrary-precision arithmetic) silently when the number gets bigger.
How can we detect overflow of int
and long long
in pure C?
c overflow
marked as duplicate by sleske, ead, phuclv, Cody Gray♦ 2 days ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
It's very tricky since you just can't add two numbers and check if the value is above some threshold (because signed integer arithmetic overflow and such). A simple solution might be to check ifx
(the value you want to check) is above a specific threshold, or if adding one goes above a threshold. If it does and the other number you want to add is larger than one, then you have an overflow situation.
– Some programmer dude
2 days ago
1
Nitpick, but, it was CPython 2.7 that did this. CPython 3 doesn't "promote" anything, even internally there is just one type.
– Antti Haapala
2 days ago
1
there are a lot of duplicates depending on what you want to do with the values (add/sub/mul/div/...?) How to check if A+B exceed long long? (both A and B is long long), Detecting signed overflow in C/C++, Test for overflow in integer addition
– phuclv
2 days ago
and add 1 more codereview.stackexchange.com/questions/37177/…
– NoChance
2 days ago
add a comment |
This question already has an answer here:
Detecting signed overflow in C/C++
12 answers
How do I detect unsigned integer multiply overflow?
31 answers
We know CPython promotes integers to long integers (which allow arbitrary-precision arithmetic) silently when the number gets bigger.
How can we detect overflow of int
and long long
in pure C?
c overflow
This question already has an answer here:
Detecting signed overflow in C/C++
12 answers
How do I detect unsigned integer multiply overflow?
31 answers
We know CPython promotes integers to long integers (which allow arbitrary-precision arithmetic) silently when the number gets bigger.
How can we detect overflow of int
and long long
in pure C?
This question already has an answer here:
Detecting signed overflow in C/C++
12 answers
How do I detect unsigned integer multiply overflow?
31 answers
c overflow
c overflow
edited 2 days ago
Peter Mortensen
13.9k1987113
13.9k1987113
asked 2 days ago
DeanDean
1054
1054
marked as duplicate by sleske, ead, phuclv, Cody Gray♦ 2 days ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by sleske, ead, phuclv, Cody Gray♦ 2 days ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
It's very tricky since you just can't add two numbers and check if the value is above some threshold (because signed integer arithmetic overflow and such). A simple solution might be to check ifx
(the value you want to check) is above a specific threshold, or if adding one goes above a threshold. If it does and the other number you want to add is larger than one, then you have an overflow situation.
– Some programmer dude
2 days ago
1
Nitpick, but, it was CPython 2.7 that did this. CPython 3 doesn't "promote" anything, even internally there is just one type.
– Antti Haapala
2 days ago
1
there are a lot of duplicates depending on what you want to do with the values (add/sub/mul/div/...?) How to check if A+B exceed long long? (both A and B is long long), Detecting signed overflow in C/C++, Test for overflow in integer addition
– phuclv
2 days ago
and add 1 more codereview.stackexchange.com/questions/37177/…
– NoChance
2 days ago
add a comment |
3
It's very tricky since you just can't add two numbers and check if the value is above some threshold (because signed integer arithmetic overflow and such). A simple solution might be to check ifx
(the value you want to check) is above a specific threshold, or if adding one goes above a threshold. If it does and the other number you want to add is larger than one, then you have an overflow situation.
– Some programmer dude
2 days ago
1
Nitpick, but, it was CPython 2.7 that did this. CPython 3 doesn't "promote" anything, even internally there is just one type.
– Antti Haapala
2 days ago
1
there are a lot of duplicates depending on what you want to do with the values (add/sub/mul/div/...?) How to check if A+B exceed long long? (both A and B is long long), Detecting signed overflow in C/C++, Test for overflow in integer addition
– phuclv
2 days ago
and add 1 more codereview.stackexchange.com/questions/37177/…
– NoChance
2 days ago
3
3
It's very tricky since you just can't add two numbers and check if the value is above some threshold (because signed integer arithmetic overflow and such). A simple solution might be to check if
x
(the value you want to check) is above a specific threshold, or if adding one goes above a threshold. If it does and the other number you want to add is larger than one, then you have an overflow situation.– Some programmer dude
2 days ago
It's very tricky since you just can't add two numbers and check if the value is above some threshold (because signed integer arithmetic overflow and such). A simple solution might be to check if
x
(the value you want to check) is above a specific threshold, or if adding one goes above a threshold. If it does and the other number you want to add is larger than one, then you have an overflow situation.– Some programmer dude
2 days ago
1
1
Nitpick, but, it was CPython 2.7 that did this. CPython 3 doesn't "promote" anything, even internally there is just one type.
– Antti Haapala
2 days ago
Nitpick, but, it was CPython 2.7 that did this. CPython 3 doesn't "promote" anything, even internally there is just one type.
– Antti Haapala
2 days ago
1
1
there are a lot of duplicates depending on what you want to do with the values (add/sub/mul/div/...?) How to check if A+B exceed long long? (both A and B is long long), Detecting signed overflow in C/C++, Test for overflow in integer addition
– phuclv
2 days ago
there are a lot of duplicates depending on what you want to do with the values (add/sub/mul/div/...?) How to check if A+B exceed long long? (both A and B is long long), Detecting signed overflow in C/C++, Test for overflow in integer addition
– phuclv
2 days ago
and add 1 more codereview.stackexchange.com/questions/37177/…
– NoChance
2 days ago
and add 1 more codereview.stackexchange.com/questions/37177/…
– NoChance
2 days ago
add a comment |
3 Answers
3
active
oldest
votes
You cannot detect signed int
overflow. You have to write your code to avoid it.
Signed int overflow is Undefined Behaviour and if it is present in your program, the program is invalid and the compiler is not required to generate any specific behaviour.
3
You can check you input values before doing a calculation to prevent overflow.
– A.R.C.
2 days ago
7
I think, it would be nice and informative to explain why signed int overflow undefined, whereas unsigned apperantly isn't..
– hetepeperfan
2 days ago
6
@hetepeperfan It's because that's what the language standard says.
– Sneftel
2 days ago
6
@sneftel thats an authoritative argument lacking an authoritative source, despise it is probably correct. On top of that, standards make more sense to people, once they start to understand the language, which is perhaps a reason they visit stackoverflow in the first place.
– hetepeperfan
2 days ago
5
@hetepeperfan the reason for why the standard is written as it is, is for the most part outside the scope of Stack Overflow.
– Antti Haapala
2 days ago
|
show 3 more comments
You can predict signed int overflow
but attempting to detect it after the summation is too late. You have to test for possible overflow before you do a signed addition.
It's not possible to avoid undefined behaviour by testing for it after the summation. If the addition overflows then there is already undefined behaviour.
If it were me, I'd do something like this:
#include <limits.h>
int safe_add(int a, int b)
if (a >= 0)
if (b > (INT_MAX - a))
/* handle overflow */
else
if (b < (INT_MIN - a))
/* handle underflow */
return a + b;
Refer this paper for more information. You can also find why unsigned integer overflow is not undefined behaviour and what could be portability issues in the same paper.
EDIT:
GCC and other compilers have some provisions to detect the overflow. For example, GCC
has following built-in functions allow performing simple arithmetic operations together with checking whether the operations overflowed.
bool __builtin_add_overflow (type1 a, type2 b, type3 *res)
bool __builtin_sadd_overflow (int a, int b, int *res)
bool __builtin_saddl_overflow (long int a, long int b, long int *res)
bool __builtin_saddll_overflow (long long int a, long long int b, long long int *res)
bool __builtin_uadd_overflow (unsigned int a, unsigned int b, unsigned int *res)
bool __builtin_uaddl_overflow (unsigned long int a, unsigned long int b, unsigned long int *res)
bool __builtin_uaddll_overflow (unsigned long long int a, unsigned long long int b, unsigned long long int *res)
Visit this link.
EDIT:
Regarding the question asked by someone
I think, it would be nice and informative to explain why signed int overflow undefined, whereas unsigned apperantly isn't..
The answer depends upon the implementation of the compiler. Most C implementations (compilers) just used whatever overflow behaviour was easiest to implement with the integer representation it used.
In practice, the representations for signed values may differ (according to the implementation): one's complement
, two's complement
, sign-magnitude
. For an unsigned type there is no reason for the standard to allow variation because there is only one obvious binary representation
(the standard only allows binary representation).
Why extra parentheses? Also you could save one test on average withif (a >= 0) test overflow else test underflow return a + b;
– chqrlie
2 days ago
@chqrlie that is not sufficient because there is no possibility of overflow whena == 0
.
– Antti Haapala
2 days ago
It is not necessary to test overflow ifa == 0
but testinga
just once saves one comparison ifa < 0
, which is half the cases.
– chqrlie
2 days ago
8
Also, both are technically called overflow. Underflow means that the value is too small in magnitude to be representable in a floating point variable.
– Antti Haapala
2 days ago
3
@AnttiHaapala it does not ignore the casea == 0
where there is no possible overflow, it just handles it differently.
– chqrlie
2 days ago
|
show 3 more comments
Signed operands must be tested before the addition is performed. Here is a safe addition function with 2 comparisons in all cases:
#include <limits.h>
int safe_add(int a, int b)
if (a >= 0)
if (b > INT_MAX - a)
/* handle overflow */
else
return a + b;
else
if (b < INT_MIN - a)
/* handle negative overflow */
else
return a + b;
If the type long long
is known to have a larger range than type int
, you could use this approach, which might prove faster:
#include <limits.h>
int safe_add(int a, int b)
long long res = (long long)a + b;
if (res > INT_MAX
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
You cannot detect signed int
overflow. You have to write your code to avoid it.
Signed int overflow is Undefined Behaviour and if it is present in your program, the program is invalid and the compiler is not required to generate any specific behaviour.
3
You can check you input values before doing a calculation to prevent overflow.
– A.R.C.
2 days ago
7
I think, it would be nice and informative to explain why signed int overflow undefined, whereas unsigned apperantly isn't..
– hetepeperfan
2 days ago
6
@hetepeperfan It's because that's what the language standard says.
– Sneftel
2 days ago
6
@sneftel thats an authoritative argument lacking an authoritative source, despise it is probably correct. On top of that, standards make more sense to people, once they start to understand the language, which is perhaps a reason they visit stackoverflow in the first place.
– hetepeperfan
2 days ago
5
@hetepeperfan the reason for why the standard is written as it is, is for the most part outside the scope of Stack Overflow.
– Antti Haapala
2 days ago
|
show 3 more comments
You cannot detect signed int
overflow. You have to write your code to avoid it.
Signed int overflow is Undefined Behaviour and if it is present in your program, the program is invalid and the compiler is not required to generate any specific behaviour.
3
You can check you input values before doing a calculation to prevent overflow.
– A.R.C.
2 days ago
7
I think, it would be nice and informative to explain why signed int overflow undefined, whereas unsigned apperantly isn't..
– hetepeperfan
2 days ago
6
@hetepeperfan It's because that's what the language standard says.
– Sneftel
2 days ago
6
@sneftel thats an authoritative argument lacking an authoritative source, despise it is probably correct. On top of that, standards make more sense to people, once they start to understand the language, which is perhaps a reason they visit stackoverflow in the first place.
– hetepeperfan
2 days ago
5
@hetepeperfan the reason for why the standard is written as it is, is for the most part outside the scope of Stack Overflow.
– Antti Haapala
2 days ago
|
show 3 more comments
You cannot detect signed int
overflow. You have to write your code to avoid it.
Signed int overflow is Undefined Behaviour and if it is present in your program, the program is invalid and the compiler is not required to generate any specific behaviour.
You cannot detect signed int
overflow. You have to write your code to avoid it.
Signed int overflow is Undefined Behaviour and if it is present in your program, the program is invalid and the compiler is not required to generate any specific behaviour.
answered 2 days ago
Jesper JuhlJesper Juhl
17.4k32647
17.4k32647
3
You can check you input values before doing a calculation to prevent overflow.
– A.R.C.
2 days ago
7
I think, it would be nice and informative to explain why signed int overflow undefined, whereas unsigned apperantly isn't..
– hetepeperfan
2 days ago
6
@hetepeperfan It's because that's what the language standard says.
– Sneftel
2 days ago
6
@sneftel thats an authoritative argument lacking an authoritative source, despise it is probably correct. On top of that, standards make more sense to people, once they start to understand the language, which is perhaps a reason they visit stackoverflow in the first place.
– hetepeperfan
2 days ago
5
@hetepeperfan the reason for why the standard is written as it is, is for the most part outside the scope of Stack Overflow.
– Antti Haapala
2 days ago
|
show 3 more comments
3
You can check you input values before doing a calculation to prevent overflow.
– A.R.C.
2 days ago
7
I think, it would be nice and informative to explain why signed int overflow undefined, whereas unsigned apperantly isn't..
– hetepeperfan
2 days ago
6
@hetepeperfan It's because that's what the language standard says.
– Sneftel
2 days ago
6
@sneftel thats an authoritative argument lacking an authoritative source, despise it is probably correct. On top of that, standards make more sense to people, once they start to understand the language, which is perhaps a reason they visit stackoverflow in the first place.
– hetepeperfan
2 days ago
5
@hetepeperfan the reason for why the standard is written as it is, is for the most part outside the scope of Stack Overflow.
– Antti Haapala
2 days ago
3
3
You can check you input values before doing a calculation to prevent overflow.
– A.R.C.
2 days ago
You can check you input values before doing a calculation to prevent overflow.
– A.R.C.
2 days ago
7
7
I think, it would be nice and informative to explain why signed int overflow undefined, whereas unsigned apperantly isn't..
– hetepeperfan
2 days ago
I think, it would be nice and informative to explain why signed int overflow undefined, whereas unsigned apperantly isn't..
– hetepeperfan
2 days ago
6
6
@hetepeperfan It's because that's what the language standard says.
– Sneftel
2 days ago
@hetepeperfan It's because that's what the language standard says.
– Sneftel
2 days ago
6
6
@sneftel thats an authoritative argument lacking an authoritative source, despise it is probably correct. On top of that, standards make more sense to people, once they start to understand the language, which is perhaps a reason they visit stackoverflow in the first place.
– hetepeperfan
2 days ago
@sneftel thats an authoritative argument lacking an authoritative source, despise it is probably correct. On top of that, standards make more sense to people, once they start to understand the language, which is perhaps a reason they visit stackoverflow in the first place.
– hetepeperfan
2 days ago
5
5
@hetepeperfan the reason for why the standard is written as it is, is for the most part outside the scope of Stack Overflow.
– Antti Haapala
2 days ago
@hetepeperfan the reason for why the standard is written as it is, is for the most part outside the scope of Stack Overflow.
– Antti Haapala
2 days ago
|
show 3 more comments
You can predict signed int overflow
but attempting to detect it after the summation is too late. You have to test for possible overflow before you do a signed addition.
It's not possible to avoid undefined behaviour by testing for it after the summation. If the addition overflows then there is already undefined behaviour.
If it were me, I'd do something like this:
#include <limits.h>
int safe_add(int a, int b)
if (a >= 0)
if (b > (INT_MAX - a))
/* handle overflow */
else
if (b < (INT_MIN - a))
/* handle underflow */
return a + b;
Refer this paper for more information. You can also find why unsigned integer overflow is not undefined behaviour and what could be portability issues in the same paper.
EDIT:
GCC and other compilers have some provisions to detect the overflow. For example, GCC
has following built-in functions allow performing simple arithmetic operations together with checking whether the operations overflowed.
bool __builtin_add_overflow (type1 a, type2 b, type3 *res)
bool __builtin_sadd_overflow (int a, int b, int *res)
bool __builtin_saddl_overflow (long int a, long int b, long int *res)
bool __builtin_saddll_overflow (long long int a, long long int b, long long int *res)
bool __builtin_uadd_overflow (unsigned int a, unsigned int b, unsigned int *res)
bool __builtin_uaddl_overflow (unsigned long int a, unsigned long int b, unsigned long int *res)
bool __builtin_uaddll_overflow (unsigned long long int a, unsigned long long int b, unsigned long long int *res)
Visit this link.
EDIT:
Regarding the question asked by someone
I think, it would be nice and informative to explain why signed int overflow undefined, whereas unsigned apperantly isn't..
The answer depends upon the implementation of the compiler. Most C implementations (compilers) just used whatever overflow behaviour was easiest to implement with the integer representation it used.
In practice, the representations for signed values may differ (according to the implementation): one's complement
, two's complement
, sign-magnitude
. For an unsigned type there is no reason for the standard to allow variation because there is only one obvious binary representation
(the standard only allows binary representation).
Why extra parentheses? Also you could save one test on average withif (a >= 0) test overflow else test underflow return a + b;
– chqrlie
2 days ago
@chqrlie that is not sufficient because there is no possibility of overflow whena == 0
.
– Antti Haapala
2 days ago
It is not necessary to test overflow ifa == 0
but testinga
just once saves one comparison ifa < 0
, which is half the cases.
– chqrlie
2 days ago
8
Also, both are technically called overflow. Underflow means that the value is too small in magnitude to be representable in a floating point variable.
– Antti Haapala
2 days ago
3
@AnttiHaapala it does not ignore the casea == 0
where there is no possible overflow, it just handles it differently.
– chqrlie
2 days ago
|
show 3 more comments
You can predict signed int overflow
but attempting to detect it after the summation is too late. You have to test for possible overflow before you do a signed addition.
It's not possible to avoid undefined behaviour by testing for it after the summation. If the addition overflows then there is already undefined behaviour.
If it were me, I'd do something like this:
#include <limits.h>
int safe_add(int a, int b)
if (a >= 0)
if (b > (INT_MAX - a))
/* handle overflow */
else
if (b < (INT_MIN - a))
/* handle underflow */
return a + b;
Refer this paper for more information. You can also find why unsigned integer overflow is not undefined behaviour and what could be portability issues in the same paper.
EDIT:
GCC and other compilers have some provisions to detect the overflow. For example, GCC
has following built-in functions allow performing simple arithmetic operations together with checking whether the operations overflowed.
bool __builtin_add_overflow (type1 a, type2 b, type3 *res)
bool __builtin_sadd_overflow (int a, int b, int *res)
bool __builtin_saddl_overflow (long int a, long int b, long int *res)
bool __builtin_saddll_overflow (long long int a, long long int b, long long int *res)
bool __builtin_uadd_overflow (unsigned int a, unsigned int b, unsigned int *res)
bool __builtin_uaddl_overflow (unsigned long int a, unsigned long int b, unsigned long int *res)
bool __builtin_uaddll_overflow (unsigned long long int a, unsigned long long int b, unsigned long long int *res)
Visit this link.
EDIT:
Regarding the question asked by someone
I think, it would be nice and informative to explain why signed int overflow undefined, whereas unsigned apperantly isn't..
The answer depends upon the implementation of the compiler. Most C implementations (compilers) just used whatever overflow behaviour was easiest to implement with the integer representation it used.
In practice, the representations for signed values may differ (according to the implementation): one's complement
, two's complement
, sign-magnitude
. For an unsigned type there is no reason for the standard to allow variation because there is only one obvious binary representation
(the standard only allows binary representation).
Why extra parentheses? Also you could save one test on average withif (a >= 0) test overflow else test underflow return a + b;
– chqrlie
2 days ago
@chqrlie that is not sufficient because there is no possibility of overflow whena == 0
.
– Antti Haapala
2 days ago
It is not necessary to test overflow ifa == 0
but testinga
just once saves one comparison ifa < 0
, which is half the cases.
– chqrlie
2 days ago
8
Also, both are technically called overflow. Underflow means that the value is too small in magnitude to be representable in a floating point variable.
– Antti Haapala
2 days ago
3
@AnttiHaapala it does not ignore the casea == 0
where there is no possible overflow, it just handles it differently.
– chqrlie
2 days ago
|
show 3 more comments
You can predict signed int overflow
but attempting to detect it after the summation is too late. You have to test for possible overflow before you do a signed addition.
It's not possible to avoid undefined behaviour by testing for it after the summation. If the addition overflows then there is already undefined behaviour.
If it were me, I'd do something like this:
#include <limits.h>
int safe_add(int a, int b)
if (a >= 0)
if (b > (INT_MAX - a))
/* handle overflow */
else
if (b < (INT_MIN - a))
/* handle underflow */
return a + b;
Refer this paper for more information. You can also find why unsigned integer overflow is not undefined behaviour and what could be portability issues in the same paper.
EDIT:
GCC and other compilers have some provisions to detect the overflow. For example, GCC
has following built-in functions allow performing simple arithmetic operations together with checking whether the operations overflowed.
bool __builtin_add_overflow (type1 a, type2 b, type3 *res)
bool __builtin_sadd_overflow (int a, int b, int *res)
bool __builtin_saddl_overflow (long int a, long int b, long int *res)
bool __builtin_saddll_overflow (long long int a, long long int b, long long int *res)
bool __builtin_uadd_overflow (unsigned int a, unsigned int b, unsigned int *res)
bool __builtin_uaddl_overflow (unsigned long int a, unsigned long int b, unsigned long int *res)
bool __builtin_uaddll_overflow (unsigned long long int a, unsigned long long int b, unsigned long long int *res)
Visit this link.
EDIT:
Regarding the question asked by someone
I think, it would be nice and informative to explain why signed int overflow undefined, whereas unsigned apperantly isn't..
The answer depends upon the implementation of the compiler. Most C implementations (compilers) just used whatever overflow behaviour was easiest to implement with the integer representation it used.
In practice, the representations for signed values may differ (according to the implementation): one's complement
, two's complement
, sign-magnitude
. For an unsigned type there is no reason for the standard to allow variation because there is only one obvious binary representation
(the standard only allows binary representation).
You can predict signed int overflow
but attempting to detect it after the summation is too late. You have to test for possible overflow before you do a signed addition.
It's not possible to avoid undefined behaviour by testing for it after the summation. If the addition overflows then there is already undefined behaviour.
If it were me, I'd do something like this:
#include <limits.h>
int safe_add(int a, int b)
if (a >= 0)
if (b > (INT_MAX - a))
/* handle overflow */
else
if (b < (INT_MIN - a))
/* handle underflow */
return a + b;
Refer this paper for more information. You can also find why unsigned integer overflow is not undefined behaviour and what could be portability issues in the same paper.
EDIT:
GCC and other compilers have some provisions to detect the overflow. For example, GCC
has following built-in functions allow performing simple arithmetic operations together with checking whether the operations overflowed.
bool __builtin_add_overflow (type1 a, type2 b, type3 *res)
bool __builtin_sadd_overflow (int a, int b, int *res)
bool __builtin_saddl_overflow (long int a, long int b, long int *res)
bool __builtin_saddll_overflow (long long int a, long long int b, long long int *res)
bool __builtin_uadd_overflow (unsigned int a, unsigned int b, unsigned int *res)
bool __builtin_uaddl_overflow (unsigned long int a, unsigned long int b, unsigned long int *res)
bool __builtin_uaddll_overflow (unsigned long long int a, unsigned long long int b, unsigned long long int *res)
Visit this link.
EDIT:
Regarding the question asked by someone
I think, it would be nice and informative to explain why signed int overflow undefined, whereas unsigned apperantly isn't..
The answer depends upon the implementation of the compiler. Most C implementations (compilers) just used whatever overflow behaviour was easiest to implement with the integer representation it used.
In practice, the representations for signed values may differ (according to the implementation): one's complement
, two's complement
, sign-magnitude
. For an unsigned type there is no reason for the standard to allow variation because there is only one obvious binary representation
(the standard only allows binary representation).
edited yesterday
answered 2 days ago
abhiaroraabhiarora
2,49931533
2,49931533
Why extra parentheses? Also you could save one test on average withif (a >= 0) test overflow else test underflow return a + b;
– chqrlie
2 days ago
@chqrlie that is not sufficient because there is no possibility of overflow whena == 0
.
– Antti Haapala
2 days ago
It is not necessary to test overflow ifa == 0
but testinga
just once saves one comparison ifa < 0
, which is half the cases.
– chqrlie
2 days ago
8
Also, both are technically called overflow. Underflow means that the value is too small in magnitude to be representable in a floating point variable.
– Antti Haapala
2 days ago
3
@AnttiHaapala it does not ignore the casea == 0
where there is no possible overflow, it just handles it differently.
– chqrlie
2 days ago
|
show 3 more comments
Why extra parentheses? Also you could save one test on average withif (a >= 0) test overflow else test underflow return a + b;
– chqrlie
2 days ago
@chqrlie that is not sufficient because there is no possibility of overflow whena == 0
.
– Antti Haapala
2 days ago
It is not necessary to test overflow ifa == 0
but testinga
just once saves one comparison ifa < 0
, which is half the cases.
– chqrlie
2 days ago
8
Also, both are technically called overflow. Underflow means that the value is too small in magnitude to be representable in a floating point variable.
– Antti Haapala
2 days ago
3
@AnttiHaapala it does not ignore the casea == 0
where there is no possible overflow, it just handles it differently.
– chqrlie
2 days ago
Why extra parentheses? Also you could save one test on average with
if (a >= 0) test overflow else test underflow return a + b;
– chqrlie
2 days ago
Why extra parentheses? Also you could save one test on average with
if (a >= 0) test overflow else test underflow return a + b;
– chqrlie
2 days ago
@chqrlie that is not sufficient because there is no possibility of overflow when
a == 0
.– Antti Haapala
2 days ago
@chqrlie that is not sufficient because there is no possibility of overflow when
a == 0
.– Antti Haapala
2 days ago
It is not necessary to test overflow if
a == 0
but testing a
just once saves one comparison if a < 0
, which is half the cases.– chqrlie
2 days ago
It is not necessary to test overflow if
a == 0
but testing a
just once saves one comparison if a < 0
, which is half the cases.– chqrlie
2 days ago
8
8
Also, both are technically called overflow. Underflow means that the value is too small in magnitude to be representable in a floating point variable.
– Antti Haapala
2 days ago
Also, both are technically called overflow. Underflow means that the value is too small in magnitude to be representable in a floating point variable.
– Antti Haapala
2 days ago
3
3
@AnttiHaapala it does not ignore the case
a == 0
where there is no possible overflow, it just handles it differently.– chqrlie
2 days ago
@AnttiHaapala it does not ignore the case
a == 0
where there is no possible overflow, it just handles it differently.– chqrlie
2 days ago
|
show 3 more comments
Signed operands must be tested before the addition is performed. Here is a safe addition function with 2 comparisons in all cases:
#include <limits.h>
int safe_add(int a, int b)
if (a >= 0)
if (b > INT_MAX - a)
/* handle overflow */
else
return a + b;
else
if (b < INT_MIN - a)
/* handle negative overflow */
else
return a + b;
If the type long long
is known to have a larger range than type int
, you could use this approach, which might prove faster:
#include <limits.h>
int safe_add(int a, int b)
long long res = (long long)a + b;
if (res > INT_MAX
add a comment |
Signed operands must be tested before the addition is performed. Here is a safe addition function with 2 comparisons in all cases:
#include <limits.h>
int safe_add(int a, int b)
if (a >= 0)
if (b > INT_MAX - a)
/* handle overflow */
else
return a + b;
else
if (b < INT_MIN - a)
/* handle negative overflow */
else
return a + b;
If the type long long
is known to have a larger range than type int
, you could use this approach, which might prove faster:
#include <limits.h>
int safe_add(int a, int b)
long long res = (long long)a + b;
if (res > INT_MAX
add a comment |
Signed operands must be tested before the addition is performed. Here is a safe addition function with 2 comparisons in all cases:
#include <limits.h>
int safe_add(int a, int b)
if (a >= 0)
if (b > INT_MAX - a)
/* handle overflow */
else
return a + b;
else
if (b < INT_MIN - a)
/* handle negative overflow */
else
return a + b;
If the type long long
is known to have a larger range than type int
, you could use this approach, which might prove faster:
#include <limits.h>
int safe_add(int a, int b)
long long res = (long long)a + b;
if (res > INT_MAX
Signed operands must be tested before the addition is performed. Here is a safe addition function with 2 comparisons in all cases:
#include <limits.h>
int safe_add(int a, int b)
if (a >= 0)
if (b > INT_MAX - a)
/* handle overflow */
else
return a + b;
else
if (b < INT_MIN - a)
/* handle negative overflow */
else
return a + b;
If the type long long
is known to have a larger range than type int
, you could use this approach, which might prove faster:
#include <limits.h>
int safe_add(int a, int b)
long long res = (long long)a + b;
if (res > INT_MAX
edited 2 days ago
answered 2 days ago
chqrliechqrlie
62.4k848105
62.4k848105
add a comment |
add a comment |
3
It's very tricky since you just can't add two numbers and check if the value is above some threshold (because signed integer arithmetic overflow and such). A simple solution might be to check if
x
(the value you want to check) is above a specific threshold, or if adding one goes above a threshold. If it does and the other number you want to add is larger than one, then you have an overflow situation.– Some programmer dude
2 days ago
1
Nitpick, but, it was CPython 2.7 that did this. CPython 3 doesn't "promote" anything, even internally there is just one type.
– Antti Haapala
2 days ago
1
there are a lot of duplicates depending on what you want to do with the values (add/sub/mul/div/...?) How to check if A+B exceed long long? (both A and B is long long), Detecting signed overflow in C/C++, Test for overflow in integer addition
– phuclv
2 days ago
and add 1 more codereview.stackexchange.com/questions/37177/…
– NoChance
2 days ago