Nested Macros Evaluation??

**ganesh bala** · 02-12-2009

Code:

#include "stdio.h"
#define a(x,y) x##y
#define b(x) #x
#define c(x) b(x)
int main()
{
 printf("%s\t",c(a(34,56)));
 printf("%s\n",b(a(34,56)));
 return 0;
}
 
Output:
3456 a(34,56)

Doubt:
According to my understanding c(a(34,56)) should be expanded to c(3456) and then to b(3456) followed by "3456"
and b(a(34,56)) should be expanded to b(3456) followed by "3456". So the output should be:
3456 3456 but this is not so why?
If a statement contains a macro within macro then what would be the order of expansion in macros? Is it that first the inner one would be expanded followed by the outer one or vice versa?

**matsp** · 02-12-2009

No, because b() expands to the input as a string - not to the expansion of the input. The #x is the "magic" part here - the # means "immediately, translate x to a string", not "expand x, then translate to a string".

--
Mats

**ganesh bala** · 02-12-2009

Thanks matsp..

Still i m having doubt....

Code:

that means outerone will be exec'd first.. if so, 

first one should also leads to a(34,56)... how 3456 is printed ??..

c(a(34,56)) =>b(a(34,56)) =>a(34,56)

**matsp** · 02-12-2009

so in the c case, x is expanded before b() is expanded, which means that you get 3456 before you get to b().

--
Mats

**brewbuck** · 02-12-2009

Only one level of variable substitution is conducted at each expansion. This means that this:

Code:

#define TEST foo
#define A(X) #X

A(TEST)

Will expand to "TEST", not "foo". To expand to "foo" you must force another substitution:

Code:

#define TEST foo
#define A(X) AA(X)
#define AA(X) #X

A(TEST)

Now expands to "foo"

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