Ruminations

The expression used in a sizeof operator is an unevaluated expression in C and C++. This can make for some surprising situations if you are unaware of it. For instance:

#include <stdio.h>

int main( void ) {
  int a = 12;
  int b = sizeof( ++a );
  printf( "%d\n", a );
  return 0;
}

This code will print 12 instead of 13 because the expression ++a is unevaluated.

So what does it mean for an expression to be unevaluated? Basically, it means that the expression is used at compile time only as a way to determine a type, which is then used to evaluate the result of the sizeof operator. So in the above example, a’s type is determined, and the resulting type for ++ is determined, but no code is generated to execute the ++.

Here’s an abusive example that demonstrates this:

#include <stdio.h>

class c {
public:
  double operator++();
};

int main(int argc, char *argv[]){
  c c1;
  int i = sizeof( ++c1 );
  int j = sizeof( c1 );
  ::printf( "%d, %d\n", i, j );
  return 0;
}

If you run this code, you will see 8, 1 printed and will not get a link error (because of the failure to define operator++)!