i++ less efficient than ++i, how to show this?

Question

I am trying to show by example that the prefix increment is more efficient than the postfix increment.

In theory this makes sense: i++ needs to be able to return the

Answer 1

In the general case, the post increment will result in a copy where a pre-increment will not. Of course this will be optimized away in a large number of cases and in the cases where it isn't the copy operation will be negligible (ie., for built in types).

Here's a small example that show the potential inefficiency of post-increment.

#include 

class foo 
{

public:
    int x;

    foo() : x(0) { 
        printf( "construct foo()\n"); 
    };

    foo( foo const& other) { 
        printf( "copy foo()\n"); 
        x = other.x; 
    };

    foo& operator=( foo const& rhs) { 
        printf( "assign foo()\n"); 
        x = rhs.x;
        return *this; 
    };

    foo& operator++() { 
        printf( "preincrement foo\n"); 
        ++x; 
        return *this; 
    };

    foo operator++( int) { 
        printf( "postincrement foo\n"); 
        foo temp( *this);
        ++x;
        return temp; 
    };

};


int main()
{
    foo bar;

    printf( "\n" "preinc example: \n");
    ++bar;

    printf( "\n" "postinc example: \n");
    bar++;
}

The results from an optimized build (which actually removes a second copy operation in the post-increment case due to RVO):

construct foo()

preinc example: 
preincrement foo

postinc example: 
postincrement foo
copy foo()

In general, if you don't need the semantics of the post-increment, why take the chance that an unnecessary copy will occur?

Of course, it's good to keep in mind that a custom operator++() - either the pre or post variant - is free to return whatever it wants (or even do whatever it wants), and I'd imagine that there are quite a few that don't follow the usual rules. Occasionally I've come across implementations that return "void", which makes the usual semantic difference go away.