Using variable for switch case statement

Question

# include <stdio.h>

int main(void)
{
    int var=1, x=1, y=2;
    switch(var)
    {
        case 'x':
            x++;
            break;
        case 'y':
            y++;
            break;
    }
    printf("%d %d",x,y);
    return 0;
}

here I am not getting the required output Can anyone explain it why ?

My Expected output is : 2,2

`var` is not `x` or `y`, it's `1`. What would your expected output be? — Joachim Isaksson, Jul 10 '13 at 21:15
@Chaos Maybe because someone didn't read their K&R C book? (not my downvote, btw). — Jens, Jul 10 '13 at 21:26
@Jens, the question is well formed, shows effort and it's obvious OP has misunderstood the concept, I personally don't think it deserves downvotes — Chaos, Jul 10 '13 at 21:30
@Chaos Well-formed with quite a bit of community contribution. Check the revision history, specifically the original post. I would have downvoted that too. — Substantial, Jul 10 '13 at 21:32
@AmanSingh: you should ask a new question instead of completely rewriting the old one, because now the comments and the answers make no sense at all. — DCoder, Jul 28 '13 at 11:25
I am not allowed to ask new question because of a number of downvotes in a silly question that's why i have edited the previous one — Aman Singh, Jul 28 '13 at 11:41

score 10 · Accepted Answer · answered Jul 10 '13 at 21:20

In a switch statement (in C), you can't use variables in case. You must use constant.

And also, case 'x': do not refer to the variable x but to a constant 'x' who is a char. You are not testing what you seem to want... In this case, you are testing case 121:, where 121 is the ascii code for the letter 'x'.

You can solve your problem with something like :

# include <stdio.h>

#define INIT_X 1
#define INIT_Y 2
// ^^^^^^^^^^^^^

int main(void)
{
    int var=1, x=INIT_X, y=INIT_Y;
    //         ^^^^^^^^^^^^^^^^^^
    switch(var)
    {
        case INIT_X:
        //   ^^^^^^
            x++;
            break;
        case INIT_Y:
        //   ^^^^^^
            y++;
            break;
    }
    printf("%d %d",x,y);
    return 0;
}

Got ya, Thank u Pierre for ur help – Aman Singh Jul 10 '13 at 21:26 — Aman Singh, Jul 10 '13 at 21:26

Jonathan Leffler · Answer 2 · 2013-07-10T21:28:22.400

8

You misunderstand the switch statement.

The switch statement compares the expression (often a simple variable) in the switch (expression) with a series of distinct compile-time constant values in the various case labels, and executes the code after that label. If the value does not match any of the explicit case labels, then the default label is used if it is present, or the whole switch is skipped if there is no default label.

In your code, you have var set to 1. Neither case 'x': nor case 'y': matches 1 (they'd be equivalent to case 120: and case 121: in most codesets based on ASCII), and there is no default, so the switch is skipped, and the output is 1 2 (not, as you appear to have expected, 2 2).

What is a compile-time constant?

The values in the case labels must be determinable by the compiler as the code is compiled, and must be constant expressions. That means that the expressions in the case labels cannot refer to variables or functions, but they can use basic computations on fixed (integral) values.

Given:

#include <math.h>
const int x = 3;               // C++ compilers treat this differently
enum { BIG_TIME = 60 };
#define HOURS(x) (3600 * (x))

case sin(x):     // Invalid - function of a variable
case x:          // Invalid - variable
case sin(0.0):   // Invalid - function
case 'x':        // Valid - character constant
case 'xy':       // Valid but not portable
case BIG_TIME:   // Valid - enumeration value names are constant
case HOURS(2):   // Valid - expands to (3600 * (2)) which is all constant

edited Jul 10 '13 at 21:28

answered Jul 10 '13 at 21:17

Jonathan Leffler

730,956
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Can you please explain about the compile-time constant ? – Aman Singh Jul 10 '13 at 21:22
and if i use just x and y instead of 'x' and 'y' ? – Aman Singh Jul 10 '13 at 21:24
2

It will not be a C valid expression even if some compiler will accept that. – Pierre Fourgeaud Jul 10 '13 at 21:25
Is your `case 'xy':` valid ? It seems that you did a mistake here. At least you will end with a warning like : `warning: multi-character character constant`... But your are right, it is a constant ;) – Pierre Fourgeaud Jul 10 '13 at 21:30
It is valid, but the interpretation of multi-character constants is implementation-defined. – Jonathan Leffler Jul 10 '13 at 21:31
@PierreFourgeaud: Life got in the way of editing the previous comment... It (a multi-character constant) is valid, but the interpretation of multi-character constants is implementation-defined. GCC chooses to generate the warning almost always. See also [Does anyone know of a C compiler that fails to compile this?](http://stackoverflow.com/questions/328215/does-anyone-know-of-a-c-compiler-that-fails-to-compile-this/). – Jonathan Leffler Jul 10 '13 at 21:37

score 2 · Answer 3 · answered Jul 10 '13 at 21:16

2

You cannot use variables in a case as the values must be compile-time constants. Also your code is incorrect as 'x' and 'y' are constants (literals), and hence do not refer to the variables x and y.

answered Jul 10 '13 at 21:16

Daniel Hedberg

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score 2 · Answer 4 · answered Jul 10 '13 at 21:21

2

using 'x' in quotes is actually using the constant ASCII code value for the character x, which is actually number value 120 (according to the ASCII chart). It does not use the variable you declared.

answered Jul 10 '13 at 21:21

Lochemage

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11

Using variable for switch case statement

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