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declared variable inside void setup is forgotten in void loop
Coding DebuggingSpeed Control L298 motorsPassing pointers between setup and loopServotimer2, detach “stop”Arduino code to control 4 led's from 4 buttonsAdding music to loopServo doesn't work in 'for' loop running under 'if' loopInterrupts Problem with Flow sensorWhy a servo doesn`t move to angles properlyHow do I make a Servo Stop if it Hits Resistance?
if i declare a var in void setup() and try to do something with it in void loop(), it just says that the variable is undeclared. here is the code:
#include "Servo.h"
void setup()
Servo servo1;
int x_key = A0;
servo1.attach(2);
pinMode(A0, INPUT);
pinMode(2, OUTPUT);
int x_position = 90;
void loop()
if (analogRead(A0) < 512)
x_position++;
servo1.write(x_position);
arduino-uno programming c++ motor servo
edited May 12 at 19:29
Duncan C
2,3492720
asked May 12 at 19:27
ThisIsAronThisIsAron
61
|
show 1 more comment
if i declare a var in void setup() and try to do something with it in void loop(), it just says that the variable is undeclared. here is the code:
#include "Servo.h"
void setup()
Servo servo1;
int x_key = A0;
servo1.attach(2);
pinMode(A0, INPUT);
pinMode(2, OUTPUT);
int x_position = 90;
void loop()
if (analogRead(A0) < 512)
x_position++;
servo1.write(x_position);
arduino-uno programming c++ motor servo
edited May 12 at 19:29
Duncan C
2,3492720
asked May 12 at 19:27
ThisIsAronThisIsAron
61
You need to select your code and tap the(code formatting) button. I did it for you.
– Duncan C
May 12 at 19:29
3
Of course it's undeclared. The variable is in setup. It's not in loop.
– Majenko♦
May 12 at 19:30
1
Because the variable is only valid, where it is declared, to use the official term: it's scope. When the program moves out of the scope of this variable, it will get thrown away and be no longer available. You can declare it in global scope, meaning outside of any function
– chrisl
May 12 at 19:32
This is more a question about C/C++, not about Arduino
– chrisl
May 12 at 19:33
@chrisl true, but the OP doesn't know enough to know where to ask the question. We all have to start somewhere. I just wish people would do some self-study before running to the internet looking for somebody else to answer their questions for them.
– Duncan C
May 12 at 19:35
|
show 1 more comment
if i declare a var in void setup() and try to do something with it in void loop(), it just says that the variable is undeclared. here is the code:
#include "Servo.h"
void setup()
Servo servo1;
int x_key = A0;
servo1.attach(2);
pinMode(A0, INPUT);
pinMode(2, OUTPUT);
int x_position = 90;
void loop()
if (analogRead(A0) < 512)
x_position++;
servo1.write(x_position);
arduino-uno programming c++ motor servo
edited May 12 at 19:29
Duncan C
2,3492720
asked May 12 at 19:27
ThisIsAronThisIsAron
61
if i declare a var in void setup() and try to do something with it in void loop(), it just says that the variable is undeclared. here is the code:
#include "Servo.h"
void setup()
Servo servo1;
int x_key = A0;
servo1.attach(2);
pinMode(A0, INPUT);
pinMode(2, OUTPUT);
int x_position = 90;
void loop()
if (analogRead(A0) < 512)
x_position++;
servo1.write(x_position);
arduino-uno programming c++ motor servo
arduino-uno programming c++ motor servo
edited May 12 at 19:29
Duncan C
2,3492720
asked May 12 at 19:27
ThisIsAronThisIsAron
61
edited May 12 at 19:29
Duncan C
2,3492720
asked May 12 at 19:27
ThisIsAronThisIsAron
61
edited May 12 at 19:29
Duncan C
2,3492720
edited May 12 at 19:29
Duncan C
2,3492720
edited May 12 at 19:29
Duncan C
2,3492720
2,3492720
asked May 12 at 19:27
ThisIsAronThisIsAron
61
asked May 12 at 19:27
ThisIsAronThisIsAron
61
asked May 12 at 19:27
ThisIsAronThisIsAron
61
61
You need to select your code and tap the(code formatting) button. I did it for you.
– Duncan C
May 12 at 19:29
3
Of course it's undeclared. The variable is in setup. It's not in loop.
– Majenko♦
May 12 at 19:30
1
Because the variable is only valid, where it is declared, to use the official term: it's scope. When the program moves out of the scope of this variable, it will get thrown away and be no longer available. You can declare it in global scope, meaning outside of any function
– chrisl
May 12 at 19:32
This is more a question about C/C++, not about Arduino
– chrisl
May 12 at 19:33
@chrisl true, but the OP doesn't know enough to know where to ask the question. We all have to start somewhere. I just wish people would do some self-study before running to the internet looking for somebody else to answer their questions for them.
– Duncan C
May 12 at 19:35
|
show 1 more comment
You need to select your code and tap the(code formatting) button. I did it for you.
– Duncan C
May 12 at 19:29
3
Of course it's undeclared. The variable is in setup. It's not in loop.
– Majenko♦
May 12 at 19:30
1
Because the variable is only valid, where it is declared, to use the official term: it's scope. When the program moves out of the scope of this variable, it will get thrown away and be no longer available. You can declare it in global scope, meaning outside of any function
– chrisl
May 12 at 19:32
This is more a question about C/C++, not about Arduino
– chrisl
May 12 at 19:33
@chrisl true, but the OP doesn't know enough to know where to ask the question. We all have to start somewhere. I just wish people would do some self-study before running to the internet looking for somebody else to answer their questions for them.
– Duncan C
May 12 at 19:35
You need to select your code and tap the
– Duncan C
May 12 at 19:29
You need to select your code and tap the
– Duncan C
May 12 at 19:29
3
3
Of course it's undeclared. The variable is in setup. It's not in loop.
– Majenko♦
May 12 at 19:30
Of course it's undeclared. The variable is in setup. It's not in loop.
– Majenko♦
May 12 at 19:30
1
1
Because the variable is only valid, where it is declared, to use the official term: it's scope. When the program moves out of the scope of this variable, it will get thrown away and be no longer available. You can declare it in global scope, meaning outside of any function
– chrisl
May 12 at 19:32
Because the variable is only valid, where it is declared, to use the official term: it's scope. When the program moves out of the scope of this variable, it will get thrown away and be no longer available. You can declare it in global scope, meaning outside of any function
– chrisl
May 12 at 19:32
This is more a question about C/C++, not about Arduino
– chrisl
May 12 at 19:33
This is more a question about C/C++, not about Arduino
– chrisl
May 12 at 19:33
@chrisl true, but the OP doesn't know enough to know where to ask the question. We all have to start somewhere. I just wish people would do some self-study before running to the internet looking for somebody else to answer their questions for them.
– Duncan C
May 12 at 19:35
@chrisl true, but the OP doesn't know enough to know where to ask the question. We all have to start somewhere. I just wish people would do some self-study before running to the internet looking for somebody else to answer their questions for them.
– Duncan C
May 12 at 19:35
|
show 1 more comment
1 Answer
1
active
oldest
votes
Yes, that is how C and C++ (and most other C-like languages) work. Variables have "scope". Any variable define inside a pair of curly braces (between a and a ) is only visible inside those braces.
If you want to reference a variable in both setup() and loop(), you have to make it a global variable, defined at the top of your code.
#include "Servo.h"
Servo servo1;
int x_position = 90; //Define your global var(s) here
void setup()
int x_key = A0; //This var only exists inside the setup function
servo1.attach(2);
pinMode(A0, INPUT);
pinMode(2, OUTPUT);
void loop()
if (analogRead(A0) < 512)
x_position++;
servo1.write(x_position);
edited May 13 at 8:52
Sim Son
41118
answered May 12 at 19:31
Duncan CDuncan C
2,3492720
2
Surely servo1 should be global as well?
– copper.hat
May 13 at 1:02
Yup, that too. I was in a bit of a hurry when I made that post.
– Duncan C
May 15 at 0:14
add a comment |
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1 Answer
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oldest
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1 Answer
1
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oldest
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active
oldest
votes
Yes, that is how C and C++ (and most other C-like languages) work. Variables have "scope". Any variable define inside a pair of curly braces (between a and a ) is only visible inside those braces.
If you want to reference a variable in both setup() and loop(), you have to make it a global variable, defined at the top of your code.
#include "Servo.h"
Servo servo1;
int x_position = 90; //Define your global var(s) here
void setup()
int x_key = A0; //This var only exists inside the setup function
servo1.attach(2);
pinMode(A0, INPUT);
pinMode(2, OUTPUT);
void loop()
if (analogRead(A0) < 512)
x_position++;
servo1.write(x_position);
edited May 13 at 8:52
Sim Son
41118
answered May 12 at 19:31
Duncan CDuncan C
2,3492720
2
Surely servo1 should be global as well?
– copper.hat
May 13 at 1:02
Yup, that too. I was in a bit of a hurry when I made that post.
– Duncan C
May 15 at 0:14
add a comment |
Yes, that is how C and C++ (and most other C-like languages) work. Variables have "scope". Any variable define inside a pair of curly braces (between a and a ) is only visible inside those braces.
If you want to reference a variable in both setup() and loop(), you have to make it a global variable, defined at the top of your code.
#include "Servo.h"
Servo servo1;
int x_position = 90; //Define your global var(s) here
void setup()
int x_key = A0; //This var only exists inside the setup function
servo1.attach(2);
pinMode(A0, INPUT);
pinMode(2, OUTPUT);
void loop()
if (analogRead(A0) < 512)
x_position++;
servo1.write(x_position);
edited May 13 at 8:52
Sim Son
41118
answered May 12 at 19:31
Duncan CDuncan C
2,3492720
2
Surely servo1 should be global as well?
– copper.hat
May 13 at 1:02
Yup, that too. I was in a bit of a hurry when I made that post.
– Duncan C
May 15 at 0:14
add a comment |
Yes, that is how C and C++ (and most other C-like languages) work. Variables have "scope". Any variable define inside a pair of curly braces (between a and a ) is only visible inside those braces.
If you want to reference a variable in both setup() and loop(), you have to make it a global variable, defined at the top of your code.
#include "Servo.h"
Servo servo1;
int x_position = 90; //Define your global var(s) here
void setup()
int x_key = A0; //This var only exists inside the setup function
servo1.attach(2);
pinMode(A0, INPUT);
pinMode(2, OUTPUT);
void loop()
if (analogRead(A0) < 512)
x_position++;
servo1.write(x_position);
edited May 13 at 8:52
Sim Son
41118
answered May 12 at 19:31
Duncan CDuncan C
2,3492720
Yes, that is how C and C++ (and most other C-like languages) work. Variables have "scope". Any variable define inside a pair of curly braces (between a and a ) is only visible inside those braces.
If you want to reference a variable in both setup() and loop(), you have to make it a global variable, defined at the top of your code.
#include "Servo.h"
Servo servo1;
int x_position = 90; //Define your global var(s) here
void setup()
int x_key = A0; //This var only exists inside the setup function
servo1.attach(2);
pinMode(A0, INPUT);
pinMode(2, OUTPUT);
void loop()
if (analogRead(A0) < 512)
x_position++;
servo1.write(x_position);
edited May 13 at 8:52
Sim Son
41118
answered May 12 at 19:31
Duncan CDuncan C
2,3492720
edited May 13 at 8:52
Sim Son
41118
edited May 13 at 8:52
Sim Son
41118
edited May 13 at 8:52
Sim Son
41118
41118
answered May 12 at 19:31
Duncan CDuncan C
2,3492720
answered May 12 at 19:31
Duncan CDuncan C
2,3492720
answered May 12 at 19:31
Duncan CDuncan C
2,3492720
2,3492720
2
Surely servo1 should be global as well?
– copper.hat
May 13 at 1:02
Yup, that too. I was in a bit of a hurry when I made that post.
– Duncan C
May 15 at 0:14
add a comment |
2
Surely servo1 should be global as well?
– copper.hat
May 13 at 1:02
Yup, that too. I was in a bit of a hurry when I made that post.
– Duncan C
May 15 at 0:14
2
2
Surely servo1 should be global as well?
– copper.hat
May 13 at 1:02
Surely servo1 should be global as well?
– copper.hat
May 13 at 1:02
Yup, that too. I was in a bit of a hurry when I made that post.
– Duncan C
May 15 at 0:14
Yup, that too. I was in a bit of a hurry when I made that post.
– Duncan C
May 15 at 0:14
add a comment |
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You need to select your code and tap the
(code formatting) button. I did it for you.– Duncan C
May 12 at 19:29
3
Of course it's undeclared. The variable is in setup. It's not in loop.
– Majenko♦
May 12 at 19:30
1
Because the variable is only valid, where it is declared, to use the official term: it's scope. When the program moves out of the scope of this variable, it will get thrown away and be no longer available. You can declare it in global scope, meaning outside of any function
– chrisl
May 12 at 19:32
This is more a question about C/C++, not about Arduino
– chrisl
May 12 at 19:33
@chrisl true, but the OP doesn't know enough to know where to ask the question. We all have to start somewhere. I just wish people would do some self-study before running to the internet looking for somebody else to answer their questions for them.
– Duncan C
May 12 at 19:35