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declare as function pointer and initialize in the same line
The Next CEO of Stack OverflowWhat are the differences between a pointer variable and a reference variable in C++?What is a smart pointer and when should I use one?Returning unique_ptr from functionsWhy is reading lines from stdin much slower in C++ than Python?Image Processing: Algorithm Improvement for 'Coca-Cola Can' RecognitionWhy should I use a pointer rather than the object itself?Replacing a 32-bit loop counter with 64-bit introduces crazy performance deviationsFunction pointer, which can point to every thing?Has a std::byte pointer the same aliasing implications as char*?declare and define function pointer variable in one line
In C++ how do we do the following
// fundamental language construct
type name = value;
// for example
int x = y;
with function pointers?
typedef (char)(*FP)(unsigned);
// AFAIK not possible in C++
FP x = y ;
I can use lambdas
FP x = []( unsigned k) -> char return char(k);
But I do not know how to do this without lambda. Any ideas? This is not an answer. we know this works:
void whatever ()
typedef void (*FP) (void);
FP x = whatever ;
The question is if one can do this in one line in C++. Like one can do it in one line in C++ with every other type kind.
c++
edited yesterday
Guillaume Racicot
15.9k53770
asked yesterday
emma brainemma brain
1083
New contributor
emma brain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
In C++ how do we do the following
// fundamental language construct
type name = value;
// for example
int x = y;
with function pointers?
typedef (char)(*FP)(unsigned);
// AFAIK not possible in C++
FP x = y ;
I can use lambdas
FP x = []( unsigned k) -> char return char(k);
But I do not know how to do this without lambda. Any ideas? This is not an answer. we know this works:
void whatever ()
typedef void (*FP) (void);
FP x = whatever ;
The question is if one can do this in one line in C++. Like one can do it in one line in C++ with every other type kind.
c++
edited yesterday
Guillaume Racicot
15.9k53770
asked yesterday
emma brainemma brain
1083
New contributor
emma brain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
You could always stick withauto x = &the_function;'.
– François Andrieux
yesterday
1
The name of a function pointer variable appears between the return type and the arguments It won't look liketype name = value;.
– François Andrieux
yesterday
You're missing the&beforewhatever.FP x = &whatever ;
– dave
yesterday
@dave: That's the same for functions, you don't need the ampersand. @emma: Why should that not be possible in C++? This should work fine (except for the(char), which should becharin yourtypedef)
– andreee
yesterday
add a comment |
In C++ how do we do the following
// fundamental language construct
type name = value;
// for example
int x = y;
with function pointers?
typedef (char)(*FP)(unsigned);
// AFAIK not possible in C++
FP x = y ;
I can use lambdas
FP x = []( unsigned k) -> char return char(k);
But I do not know how to do this without lambda. Any ideas? This is not an answer. we know this works:
void whatever ()
typedef void (*FP) (void);
FP x = whatever ;
The question is if one can do this in one line in C++. Like one can do it in one line in C++ with every other type kind.
c++
edited yesterday
Guillaume Racicot
15.9k53770
asked yesterday
emma brainemma brain
1083
New contributor
emma brain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In C++ how do we do the following
// fundamental language construct
type name = value;
// for example
int x = y;
with function pointers?
typedef (char)(*FP)(unsigned);
// AFAIK not possible in C++
FP x = y ;
I can use lambdas
FP x = []( unsigned k) -> char return char(k);
But I do not know how to do this without lambda. Any ideas? This is not an answer. we know this works:
void whatever ()
typedef void (*FP) (void);
FP x = whatever ;
The question is if one can do this in one line in C++. Like one can do it in one line in C++ with every other type kind.
c++
c++
edited yesterday
Guillaume Racicot
15.9k53770
asked yesterday
emma brainemma brain
1083
New contributor
emma brain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Guillaume Racicot
15.9k53770
asked yesterday
emma brainemma brain
1083
New contributor
emma brain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Guillaume Racicot
15.9k53770
edited yesterday
Guillaume Racicot
15.9k53770
edited yesterday
Guillaume Racicot
15.9k53770
15.9k53770
asked yesterday
emma brainemma brain
1083
New contributor
emma brain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
emma brainemma brain
1083
asked yesterday
emma brainemma brain
1083
1083
New contributor
emma brain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
emma brain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
emma brain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
You could always stick withauto x = &the_function;'.
– François Andrieux
yesterday
1
The name of a function pointer variable appears between the return type and the arguments It won't look liketype name = value;.
– François Andrieux
yesterday
You're missing the&beforewhatever.FP x = &whatever ;
– dave
yesterday
@dave: That's the same for functions, you don't need the ampersand. @emma: Why should that not be possible in C++? This should work fine (except for the(char), which should becharin yourtypedef)
– andreee
yesterday
add a comment |
3
You could always stick withauto x = &the_function;'.
– François Andrieux
yesterday
1
The name of a function pointer variable appears between the return type and the arguments It won't look liketype name = value;.
– François Andrieux
yesterday
You're missing the&beforewhatever.FP x = &whatever ;
– dave
yesterday
@dave: That's the same for functions, you don't need the ampersand. @emma: Why should that not be possible in C++? This should work fine (except for the(char), which should becharin yourtypedef)
– andreee
yesterday
3
3
You could always stick with
auto x = &the_function;'.– François Andrieux
yesterday
You could always stick with
auto x = &the_function;'.– François Andrieux
yesterday
1
1
The name of a function pointer variable appears between the return type and the arguments It won't look like
type name = value;.– François Andrieux
yesterday
The name of a function pointer variable appears between the return type and the arguments It won't look like
type name = value;.– François Andrieux
yesterday
You're missing the
& before whatever. FP x = &whatever ;– dave
yesterday
You're missing the
& before whatever. FP x = &whatever ;– dave
yesterday
@dave: That's the same for functions, you don't need the ampersand. @emma: Why should that not be possible in C++? This should work fine (except for the
(char), which should be char in your typedef)– andreee
yesterday
@dave: That's the same for functions, you don't need the ampersand. @emma: Why should that not be possible in C++? This should work fine (except for the
(char), which should be char in your typedef)– andreee
yesterday
add a comment |
2 Answers
2
active
oldest
votes
Whenever you can write a typedef, you can also write a variable declaration with no typedef, with almost identical syntax.
Example:
// typedef
typedef char(*FP)(unsigned);
FP x = y ;
// no typedef
char(*x)(unsigned) = y;
Remove the typedef keyword, and you have a variable declaration. Slap an initialisation on it if you want.
answered yesterday
n.m.n.m.
73.7k885172
Works just as well as in C, and can also be used for array types:char (*img)[width][3] = malloc(height*sizeof(*img));While this syntax is pretty useless in C++ (no array types with runtime sizes), it works well in C and sports the same counterintuitive syntax.
– cmaster
yesterday
add a comment |
You can use auto:
auto fptr = &f;
It skips the need of a typedef and conserve a nice syntax.
answered yesterday
Guillaume RacicotGuillaume Racicot
15.9k53770
I like this answer a bit more than the accepted one. It is more C++11-ish even though it is not clear that OP has or wants a C++11 solution. But hey, C++11 is not the "all new exotic" standard any more.
– TobiMcNamobi
17 hours ago
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Whenever you can write a typedef, you can also write a variable declaration with no typedef, with almost identical syntax.
Example:
// typedef
typedef char(*FP)(unsigned);
FP x = y ;
// no typedef
char(*x)(unsigned) = y;
Remove the typedef keyword, and you have a variable declaration. Slap an initialisation on it if you want.
answered yesterday
n.m.n.m.
73.7k885172
Works just as well as in C, and can also be used for array types:char (*img)[width][3] = malloc(height*sizeof(*img));While this syntax is pretty useless in C++ (no array types with runtime sizes), it works well in C and sports the same counterintuitive syntax.
– cmaster
yesterday
add a comment |
Whenever you can write a typedef, you can also write a variable declaration with no typedef, with almost identical syntax.
Example:
// typedef
typedef char(*FP)(unsigned);
FP x = y ;
// no typedef
char(*x)(unsigned) = y;
Remove the typedef keyword, and you have a variable declaration. Slap an initialisation on it if you want.
answered yesterday
n.m.n.m.
73.7k885172
Works just as well as in C, and can also be used for array types:char (*img)[width][3] = malloc(height*sizeof(*img));While this syntax is pretty useless in C++ (no array types with runtime sizes), it works well in C and sports the same counterintuitive syntax.
– cmaster
yesterday
add a comment |
Whenever you can write a typedef, you can also write a variable declaration with no typedef, with almost identical syntax.
Example:
// typedef
typedef char(*FP)(unsigned);
FP x = y ;
// no typedef
char(*x)(unsigned) = y;
Remove the typedef keyword, and you have a variable declaration. Slap an initialisation on it if you want.
answered yesterday
n.m.n.m.
73.7k885172
Whenever you can write a typedef, you can also write a variable declaration with no typedef, with almost identical syntax.
Example:
// typedef
typedef char(*FP)(unsigned);
FP x = y ;
// no typedef
char(*x)(unsigned) = y;
Remove the typedef keyword, and you have a variable declaration. Slap an initialisation on it if you want.
answered yesterday
n.m.n.m.
73.7k885172
answered yesterday
n.m.n.m.
73.7k885172
answered yesterday
n.m.n.m.
73.7k885172
answered yesterday
n.m.n.m.
73.7k885172
73.7k885172
Works just as well as in C, and can also be used for array types:char (*img)[width][3] = malloc(height*sizeof(*img));While this syntax is pretty useless in C++ (no array types with runtime sizes), it works well in C and sports the same counterintuitive syntax.
– cmaster
yesterday
add a comment |
Works just as well as in C, and can also be used for array types:char (*img)[width][3] = malloc(height*sizeof(*img));While this syntax is pretty useless in C++ (no array types with runtime sizes), it works well in C and sports the same counterintuitive syntax.
– cmaster
yesterday
Works just as well as in C, and can also be used for array types:
char (*img)[width][3] = malloc(height*sizeof(*img)); While this syntax is pretty useless in C++ (no array types with runtime sizes), it works well in C and sports the same counterintuitive syntax.– cmaster
yesterday
Works just as well as in C, and can also be used for array types:
char (*img)[width][3] = malloc(height*sizeof(*img)); While this syntax is pretty useless in C++ (no array types with runtime sizes), it works well in C and sports the same counterintuitive syntax.– cmaster
yesterday
add a comment |
You can use auto:
auto fptr = &f;
It skips the need of a typedef and conserve a nice syntax.
answered yesterday
Guillaume RacicotGuillaume Racicot
15.9k53770
I like this answer a bit more than the accepted one. It is more C++11-ish even though it is not clear that OP has or wants a C++11 solution. But hey, C++11 is not the "all new exotic" standard any more.
– TobiMcNamobi
17 hours ago
add a comment |
You can use auto:
auto fptr = &f;
It skips the need of a typedef and conserve a nice syntax.
answered yesterday
Guillaume RacicotGuillaume Racicot
15.9k53770
I like this answer a bit more than the accepted one. It is more C++11-ish even though it is not clear that OP has or wants a C++11 solution. But hey, C++11 is not the "all new exotic" standard any more.
– TobiMcNamobi
17 hours ago
add a comment |
You can use auto:
auto fptr = &f;
It skips the need of a typedef and conserve a nice syntax.
answered yesterday
Guillaume RacicotGuillaume Racicot
15.9k53770
You can use auto:
auto fptr = &f;
It skips the need of a typedef and conserve a nice syntax.
answered yesterday
Guillaume RacicotGuillaume Racicot
15.9k53770
answered yesterday
Guillaume RacicotGuillaume Racicot
15.9k53770
answered yesterday
Guillaume RacicotGuillaume Racicot
15.9k53770
answered yesterday
Guillaume RacicotGuillaume Racicot
15.9k53770
15.9k53770
I like this answer a bit more than the accepted one. It is more C++11-ish even though it is not clear that OP has or wants a C++11 solution. But hey, C++11 is not the "all new exotic" standard any more.
– TobiMcNamobi
17 hours ago
add a comment |
I like this answer a bit more than the accepted one. It is more C++11-ish even though it is not clear that OP has or wants a C++11 solution. But hey, C++11 is not the "all new exotic" standard any more.
– TobiMcNamobi
17 hours ago
I like this answer a bit more than the accepted one. It is more C++11-ish even though it is not clear that OP has or wants a C++11 solution. But hey, C++11 is not the "all new exotic" standard any more.
– TobiMcNamobi
17 hours ago
I like this answer a bit more than the accepted one. It is more C++11-ish even though it is not clear that OP has or wants a C++11 solution. But hey, C++11 is not the "all new exotic" standard any more.
– TobiMcNamobi
17 hours ago
add a comment |
emma brain is a new contributor. Be nice, and check out our Code of Conduct.
emma brain is a new contributor. Be nice, and check out our Code of Conduct.
emma brain is a new contributor. Be nice, and check out our Code of Conduct.
emma brain is a new contributor. Be nice, and check out our Code of Conduct.
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3
You could always stick with
auto x = &the_function;'.– François Andrieux
yesterday
1
The name of a function pointer variable appears between the return type and the arguments It won't look like
type name = value;.– François Andrieux
yesterday
You're missing the
&beforewhatever.FP x = &whatever ;– dave
yesterday
@dave: That's the same for functions, you don't need the ampersand. @emma: Why should that not be possible in C++? This should work fine (except for the
(char), which should becharin yourtypedef)– andreee
yesterday