switch case: error: case label does not reduce to an integer constant

Question

int value;

const int signalmin = some_function();

switch(value)
{
   case signalmin:
   break;
}

I read the value of some_function and use that int value to do a switch case on. The C99 compiler gives back:

error: case label does not reduce to an integer constant

But I cannot use a #define because the int value is being read before the switch executes.

Answer 1

switch labels must be constant expressions, they have to be evaluated at compile time. If you want to branch on run-time values, you must use an if.

A const-qualified variable is not a constant expression, it is merely a value you are not allowed to modify.

The form of integer constant expressions is detailed in 6.6 (6) [C99 and the n1570 draft of the C2011 standard]:

6 An integer constant expression 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.

The restriction that only sizeof expressions whose result is an integer constant are allowed rules out sizeof expressions whose operand is a variable length array.

Answer 2

Let me chip in with an example. The following was tested on gcc version 4.6.3 with the flags -std=c99 -pedantic set:

#define SOME_HARDCODED_CONSTANT 0 //good
int foo(int i, int b){
    const int c=0; //bad
    int a=0; //bad

    switch(i){
        case c:     //compile error
        case a:     //compile error.
        case (b+a): //compile error
        case SOME_HARDCODED_CONSTANT: //all good
        case 5: //all good
    }
}

As others have noted, case arguments cannot be evaluated at runtime. Use an if-else block to do that.

Answer 3

In C. all case labels must be compile time constants. In C, the const qualifier does not create a compile-time constant, it merely designates that a run-time variable is read-only.

A switch is not the appropriate control structure for what you're trying to do.

Answer 4

I am using the code below, and it is working fine.

I am getting an error using case "+": i.e. with double-quotes, so try writing case '+': with single quotes.

#include <stdio.h>

int main() {
    char x;
    int a = 20, b = 10;
    scanf("%c", &x);
    switch (x) {
        case '+':
            printf("%d", a + b);
            break;
        case '-':
            printf("%d", a - b);
            break;
        case '*':
            printf("%d", a * b);
            break;
        case '/':
            printf("%d", a / b);
            break;
        default:
            printf("sorry");
    }
    return 0;
}

Answer 5

In C, variables aren't to be used in the switch case labels instead constant expressions are only allowed there.

Answer 6

On OSX, clang seems to take constants as case labels without complaints.

#include <stdio.h>

#define SOME_HARDCODED_CONSTANT 0 //good for sure
int foo(int i, int b){ 
    const int c=1; //no problem!!!

    switch(i){
        case SOME_HARDCODED_CONSTANT: //all good
            printf("case SOME_HARDCODED_CONSTANT\n"); break;
        case c:     //no compile error for clang
            printf("case c\n"); break;
        case 5: //all good
            printf("case 5\n"); break;
    }   
    return i+b;
}

int main() {
    printf("test foo(1,3): %d\n", foo(1,3));
}

Output:

$> cc test.c -o test; ./test 
case c
test foo(1,3): 4