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… — user1203803
– user1203803, Commented Aug 5, 2012 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
– Daniel Fischer, Commented Aug 5, 2012 at 18:15
@Daniel Fischer: then perhaps you will like this one: blog.aerojockey.com/post/iocccsim — Karoly Horvath
– Karoly Horvath, Commented Aug 5, 2012 at 18:25

Karoly Horvath · Accepted Answer · 2012-08-05 18:09:32Z

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 5, 2012 at 18:09

Karoly Horvath

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4 Comments

Praetorian Over a year ago

Preprocessor, not precompiler

Karoly Horvath Over a year ago

@cirronimbo: if you have visual studio: stackoverflow.com/questions/277258/…

cirronimbo Over a year ago

@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.

Karoly Horvath Over a year ago

@cirronimbo: that's just an IDE, a front-end. use a command shell, and type the appropriate command for the compiler (gcc/msvc).

Daniel Fischer · Accepted Answer · 2012-08-05 18:13:41Z

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.

Johan Lundberg · Accepted Answer · 2012-08-05 18:23:01Z

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

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