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I was given a variation of this C code in an interview recently, and asked what would be returned by the function. 

#define fun(a) a*a

int main() {    
  return(fun(4+5));   
}

I've run it with a printf("%d"...) in place of the return and it prints 29. No other types than "%d" showed a result. Can someone explain what is going on here?

DYZ
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user2805260
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1 Answers1

11

a is expanded, not evaluated as a. So you get:

4+5*4+5

and with operator priorities; you get 4 + 20 + 5 => 29

better way (with extra outside parenthesis for protecting against outside operators too thanks to Thomas comment):

#define fun(a) ((a)*(a))

but all parenthesis of the world still don't protect against i++, function calls (with side effects, preferably)... so argument reusing in macros is always a problem (there's no syntax to declare & assign a temp variable either). Prefer inlined functions in that case.

Jean-François Fabre
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