C preprocessor - token pasting - confusing result. Why is that?

Question

I thought this program would print the value -12--2=-10. When I run it, it prints 0.

I cannot figure out why? Any hints?

#include <stdio.h> 
#define        ALPHA(x,y)        x##2-y 

int main(void) {
    int i = -1;
    int i2 = -2;
    printf("%d", ALPHA(i, i2));
    return 0;
}

Hint : preprocessing happens before compilation, and variable names are just tokens at this point. The value they get much later at runtime are irrelevant. — Quentin, Jul 22 '15 at 12:12

score 7 · Accepted Answer · answered Jul 22 '15 at 12:14

The preprocessing phase is done before any compilation and is done on text. It has no notion of variables or types (that is the compile phase), let alone actual values (runtime).

So, what you are doing is:

1) ALPHA(i, i2)
2) i##2-i2
3) i2-i2

So you end up with printf("%d", i2-i2) which prints zero.

score 6 · Answer 2 · edited Jul 22 '15 at 12:19

6

The preprocessor will output that as:

#include <stdio.h> 

int main(void) {
    int i = -1;
    int i2 = -2;
    printf("%d", i2-i2);
    return 0;
}

So it will print zero

edited Jul 22 '15 at 12:19

Quentin

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answered Jul 22 '15 at 12:15

Rowland Shaw

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score 6 · Answer 3 · answered Jul 22 '15 at 12:52

Compilation process is a bit different in C as compared to other programming languages. In C, 3 phases are involved in getting an .exe file from .src file.

xyz.c -> PREPROCESSOR -> tmp.c(temporary) -> COMPILER -> xyz.obj-> LINKER -> xyz.exe

Basically, a preprocessor reads your code line by line and if it is a preprocessing statement, then only it performs the preprocessing directive and outputs your code to the Compiler in pure textual form.

In case of your code,preprocessor will send this code to the compiler for compilation :

 //includes stdio.h from include folder

    int main(void)
    {
        int i = -1;
        int i2 = -2;
        printf("%d", i2 - i2);
        return 0;
    }

So,when the compiler will compile this code,it will give the result for print as 0 only. This is the reason you are getting 0 printed, when you are running the code. Hope this will help you.

score 5 · Answer 4 · answered Jul 22 '15 at 12:13

5

ALPHA(i, i2) becomes i2-i2

Per a comment above, pre-processing is textual replacement BEFORE compilation happens.

answered Jul 22 '15 at 12:13

Michael Kohne

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score 2 · Answer 5 · answered Jul 22 '15 at 12:31

You mixed up the i and 1 characters. Try a different text editor font.

i ## 2 produces i2 which happens to be valid in your program, with a value of -2.

1 ## 2 is needed to get the expected 12.

That doesn't account for the negative sign you seem to expect, but I still like this theory.

C preprocessor - token pasting - confusing result. Why is that?

5 Answers5