Why is token replacement done before macro expansion?

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Considering the following snippet:

#include <stdio.h>

#define MY_MACRO(\
   arg) \
   arg

#define MY_MACRO2(t1, t2) t1##t2

#define MY_ a

#define MACRO b

int main() {
    printf("%d\n", MY_MACRO2(MY_,MACRO)(45));
    return 0;
}

It turns out to compile and display 45, however, if MY_ and MACRO were expanded before substitution, this code should not compile. The reason why I notice this is when I read in the C standard the following:

6.10.3.1 (but also in C++ standard)

After the arguments for the invocation of a function-like macro have been identified,argument substitution takes place.A parameter in the replacement list, unless preceded by a # or ## preprocessing token or followed by a ## preprocessing token (see below), is replaced by the corresponding argument after all macros contained therein have been expanded. Before being substituted, each argument’s preprocessing tokens are completely macro replaced as if they formed the rest of the preprocessing file; no other preprocessing tokens are available

So if all macros contained in the arguments were expanded before replacement, why don't we end up with ab(45)?

c-preprocessor c c++
1 Answers

To let constructions like X(X()) work. Note that while X() is expanded the X macro is disabled to avoid infinite recursions. Expanding arguments before expanding the macro let's one use X in the arguments. A practical application of X(X()):

#define TEN(x) x x x x x x x x x x
#define HUNDRED(x) TEN(TEN(x))
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