How does a preprocessor macro work?

Question

#define B 100+B
main()
{
    int i= B;
}

I know it's wrong, but just out of curiosity, when I compile it I get this weird error:

"B was not declared in this scope".

Why is it so? If this error was because the compiler removes the macro after its substitution then how does the following code worked fine, when B must have been removed before it was made available to A ?

#define B 100
#define A 100+B
main()
{
    int i= B;
    int j =A;
}

Placing spaces on only one side of the `=` operator is so ugly… — , Aug 05 '12 at 18:14
@RadekSlupik But at least they're on different sides in the two lines, which makes it look kind of funny. — Daniel Fischer, Aug 05 '12 at 18:15
@Daniel Fischer: then perhaps you will like this one: http://blog.aerojockey.com/post/iocccsim — Karoly Horvath, Aug 05 '12 at 18:25

score 14 · Accepted Answer · answered Aug 05 '12 at 18:09

14

Here's the output of the preprocessor:

gcc -E x.c
# 1 "x.c"
# 1 "<built-in>"
# 1 "<command-line>"
# 1 "x.c"

main()
{
    int i= 100+B;
}

As you can see, it did the substituion. Now comes the compile step which fails because there's no B declared.

The other code is fine, here's the output:

main()
{
    int i= 100;
    int j =100+100;
}

answered Aug 05 '12 at 18:09

Karoly Horvath

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5

*Preprocessor*, not *precompiler* – Praetorian Aug 05 '12 at 18:10
@cirronimbo: if you have visual studio: http://stackoverflow.com/questions/277258/c-c-source-file-after-preprocessing – Karoly Horvath Aug 05 '12 at 20:03
@KarolyHorvath: yeah, I saw that, but unfortunately I work on code::blocks 10.5, not on visual studio, and I really have no clue how to generate this preprocessor output even after looking at the above code, in code::blocks. – cirronimbo Aug 06 '12 at 16:16
@cirronimbo: that's just an IDE, a front-end. use a command shell, and type the appropriate command for the compiler (gcc/msvc). – Karoly Horvath Aug 06 '12 at 16:42

score 9 · Answer 2 · answered Aug 05 '12 at 18:13

Macro expansion isn't done recursively, if the macro name appears in the replacement text, it isn't expanded again. So with

#define B 100 + B

a replacement of B yields the token sequence 100 + B and B is not expanded again (if it were, you'd have an infinite recursion). So the compiler sees a reference to the undeclared variable B after the preprocessor finished.

But in

#define B 100
#define A 100 + B

when the macro A is expanded, the macro name B appears in the replacement text. Then B is expanded and the compiler sees 100 + 100 which contains no references to undeclared variables.

This answer should get a tick. Just sayin'. – Cat Plus Plus Aug 05 '12 at 18:16 — Cat Plus Plus, Aug 05 '12 at 18:16

score 0 · Answer 3 · edited Aug 05 '12 at 18:23

0

macro replacement are simple text operation.you can debug this type of problem in a simple step by step compile.

use cc -E filename.c -O filename.i

for generate extended c code

vi filename.i for reading pure/extended c code

edited Aug 05 '12 at 18:23

Johan Lundberg

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answered Aug 05 '12 at 18:18

rajesh6115

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How does a preprocessor macro work?

3 Answers3