I have got a doubt with switch-case statements. I read that 'const variable' can be used in switch cases.
However following program is giving me an error case label does not reduce to an integer constant
#include <stdio.h>
int main()
{
int const var = 3;
int num=3;
switch (num)
{
case var:
printf("constant var");
break;
}
}
Where am I making the mistake ?
In C. all case labels must be compile time constants and const variables are read-only, but no constants. So, you can do this
#define var 3
instead of
int const var = 3;
according to the C standard(N1570 Committee Draft):
6.8.4.2 The switch statement
The expression of each case label shall be an integer constant expression and no two of the case constant expressions in the same switch statement shall have the same value after conversion. There may be at most one default label in a switch statement. (Any enclosed switch statement may have a default label or case constant expressions with values that duplicate case constant expressions in the enclosing switch statement.)
6.6 Constant expressions define integer constant expression
An integer constant expression117) shall have integer type and shall only have operands that are integer constants, enumeration constants, character constants, sizeof expressions whose results are integer constants, _Alignof expressions, and floating constants that are the immediate operands of casts. Cast operators in an integer constant expression shall only convert arithmetic types to integer types, except as part of an operand to the sizeof or _Alignof operator.
A variable is contain the constant value, while making a constant variable it doesn't change the objective of a variable. It always be a variable, and switch case takes only constant values not variables.
#include <stdio.h>
int main()
{
int const var = 3;
int num=3;
switch(num)
{
case var: // use a constant value(Not variable) here
printf("constant var");
break;
}
}
Your switch case must have condition like this, instead of var
#include <stdio.h>
int main()
{
int const var = 3;
int num=3;
switch(num)
{
case 3:
printf("constant var");
break;
}
}
Instead use #define var 3 like:
#include <stdio.h>
#define var 3
int main()
{
//const int var = 3;
int num=3;
switch(num)
{
case var:
printf("constant var");
break;
}
return 0;
}
In your code, switch case is looking for var constant instead of case 3
When you are trying to switch a case, the switching generally depends on the value which is assigned to the variable (in your case the variable is num whose value is 3). The value of this variable can even change. There is no rule that a variable whose value is getting switched (num) should be a constant integer. Switch cases are not meant for hard coded values. Depending upon the value of the variable that particular switch case will be executed and the value can even change (in you code it does'nt, because you have hard coded it as 3) Please try this code
#include <stdio.h>
int main()
{
//int const var = 3;
int num=3;
switch(num)
{
case 3: //If the value of num is "3", then execute this case.
printf("constant var");
break;
case 4: //If the value of num is "4", then execute this case.
printf("Test case");
break;
default: //If the value of num is anything other than "3" and "4", then execute this case
printf("Default case");
break;
}
return 0; //This should be included as you are returning an integer
}
Inside the switch case, you are supposed to mention the value (of num) depending on which the case will be executed. Also, please note that when you have declared the return type of your main() as integer, it is better you include a return 0 in the end. After you run the code with num=3, please try the same code with num=4 and num=5.
