CODEWITHSUNDEEP
cmacrosc-preprocessor
Qoobee
Qoobee

Reputation: 312

Strange behavior of Macro-expansion

Here's the code:

#include <stdio.h>
#include <stdio.h>
#define VAL1(a,b)    a*b
#define VAL2(a,b)    a/b
#define VAL3(a,b)    ++a%b
int main()
{
    int a = 1;
    int b = 2;
    int c = 3;
    int d = 3;
    int e = 5;

    int result = VAL2(a,d)/VAL1(e,b)+VAL3(c,d);  // result = 1
    //int result = a/d/e*b+++c%d;                // result = 0

    printf("%d\n", result);

    return 0;
}

Why aren't the results of two statements the same?

Upvotes: 2

Views: 144

Answers (3)

David Schwartz
David Schwartz

Reputation: 182769

In one case you have + ++ and in the other case you have ++ +. + ++ and ++ + are different streams of tokens. Macro pasting doesn't change tokenization because it's tokens that are pasted.

If you punch your program into a C pre-processor, you'll get this out for that line:

int result = a/d/e*b+ ++c%d;

Notice that the preprocessor had to insert a space because one is mandatory between a + token and a ++ token.

Upvotes: 6

tomahh
tomahh

Reputation: 13661

I tried to compile it with GCC 4.7, and the two expressions gave the same results.

If you want to see how macro will expand, you can use cpp, which is the C Preprossessor, and will give you the output after preprocessor expression are executed, here the output is

int main()
{
    int a = 1;
    int b = 2;
    int c = 3;
    int d = 3;
    int e = 5;

    int result = a/d/e*b+++c%d;  
    int result2 = a/d/e*b+++c%d;  

    printf("%d %d\n", result, result2);

    return 0;
}

Upvotes: 0

bash.d
bash.d

Reputation: 13207

If you really think it is necessary to use such macros, you'll need to use parenthesis () to avoid side-effects:

#define VAL1(a,b)    ((a)*(b))
#define VAL2(a,b)    ((a)/(b))
#define VAL3(a,b)    ((++a)%(b))

Upvotes: -1

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