CODEWITHSUNDEEP
cconcatenationc-preprocessor
Ely
Ely

Reputation: 11174

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

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;
}

Upvotes: 3

Views: 226

Answers (5)

Prerna Jain
Prerna Jain

Reputation: 1250

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.

Upvotes: 6

Potatoswatter
Potatoswatter

Reputation: 137930

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.

Upvotes: 2

Rowland Shaw
Rowland Shaw

Reputation: 38128

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

Upvotes: 6

mtijanic
mtijanic

Reputation: 2902

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.

Upvotes: 7

Michael Kohne
Michael Kohne

Reputation: 12044

ALPHA(i, i2) becomes i2-i2

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

Upvotes: 5

Related Questions