Order of evaluation in C [closed]

Question

The code below gives an answer (further below), that I do not understand.

#include <stdio.h>

int fA (int x) {
   int w = x;

   printf("%d", x);

   if (x > 4) 
      w += fA(x - 2);

   if (x > 2)
      w += fA(x - 4);

   printf("%d", x);
   return w;
}

int fB (int x) {

   if (x < 1) 
      return 1;

   int w = x;

   if (x > 2) 
      w = w * fB(x - 1);

   if (x > 1) 
      w= w + fA(x - 1);

   return w;
}

int main (void) {
   printf("\n %d %d \n", fA(6), fB(3));
   return 0;
}

it prints

112264004226
12 11

The question is why? In my opinion it should starts with 6. Thanks!

Answer 1

There is no guarantee that parameters to a function will be evaluated in any particular order.

So when you call printf with fA(6) and fB(3) as parameters, the compiler is free to call either one before the other.

In this particular case, fB(3) was evaluated first. But if you use a different compiler, it might evaluate fA(6) first.

Answer 2

There's no defined order. This depends on compiler, for example on:

# gcc -v
Configured with: --prefix=/Applications/Xcode.app/Contents/Developer/usr --with-gxx-include-    dir=/usr/include/c++/4.2.1
Apple LLVM version 7.0.2 (clang-700.1.81)
Target: x86_64-apple-darwin14.5.0
Thread model: posix

this produces

640042261122
12 11

Moreover - the same compiler may decide order to be different for optimization