Which one will execute faster, if (flag==0) or if (0==flag)?

Question

Interview question: Which one will execute faster, if (flag==0) or if (0==flag)? Why?

Answer 1

Well there is a difference when flag is a user defined type

struct sInt
{
    sInt( int i ) : wrappedInt(i)
    {
        std::cout << "ctor called" << std::endl;
    }

    operator int()
    {
        std::cout << "operator int()" << std::endl;
        return wrappedInt;
    }

    bool operator==(int nComp)
    {
        std::cout << "bool operator==(int nComp)" << std::endl;
        return (nComp == wrappedInt);
    }

    int wrappedInt;
};

int 
_tmain(int argc, _TCHAR* argv[])
{
    sInt s(0);

    //in this case this will probably be faster
    if ( 0 == s )
    {
        std::cout << "equal" << std::endl;
    }

    if ( s == 0 )
    {
        std::cout << "equal" << std::endl;
    }
}

In the first case (0==s) the conversion operator is called and then the returned result is compared to 0. In the second case the == operator is called.

Answer 2

If at all there was a difference, what stops compiler to choose the faster once? So logically, there can't be any difference. Probably this is what the interviewer expects. It is actually a brilliant question.

Answer 3

Well, I am agreeing completely with all said in the comments to the OP, for the exercise sake:

If the compiler is not clever enough (indeed you should not use it) or the optimization is disabled, x == 0 could compile to a native assembly jump if zero instruction, while 0 == x could be a more generic (and costly) comparison of numeric values.

Still, I wouldn't like to work for a boss who thinks in these terms...

Answer 4

Same code for amd64 with GCC 4.1.2:

        .loc 1 4 0  # int f = argc;
        movl    -20(%rbp), %eax
        movl    %eax, -4(%rbp)
        .loc 1 6 0 # if( f == 0 ) {
        cmpl    $0, -4(%rbp)
        jne     .L2
        .loc 1 7 0 # return 0;
        movl    $0, -36(%rbp)
        jmp     .L4
        .loc 1 8 0 # }
 .L2:
        .loc 1 10 0 # if( 0 == f ) {
        cmpl    $0, -4(%rbp)
        jne     .L5
        .loc 1 11 0 # return 1;
        movl    $1, -36(%rbp)
        jmp     .L4
        .loc 1 12 0 # }
 .L5:
        .loc 1 14 0 # return 2;
        movl    $2, -36(%rbp)
 .L4:
        movl    -36(%rbp), %eax
        .loc 1 15 0 # }
        leave
        ret

Answer 5

As others have said, there is no difference.

0 has to be evaluated. flag has to be evaluated. This process takes the same time, no matter which side they're placed.

The right answer would be: they're both the same speed.

Even the expressions if(flag==0) and if(0==flag) have the same amount of characters! If one of them was written as if(flag== 0), then the compiler would have one extra space to parse, so you would have a legitimate reason at pointing out compile time.

But since there is no such thing, there is absolutely no reason why one should be faster than other. If there is a reason, then the compiler is doing some very, very strange things to generated code...

Answer 6

Surely no difference in terms of execution speeds. The condition needs to be evaluated in both cases in the same way.