Question

I have an assembly code which is 100 times these two instructions :

    movl    %eax, -16(%rbp)
    movl    -12(%rbp), %eax

that’s corresponding to this c code loop:

 int i;
    int a=5, b;
    for (i=0 ; i < sptr->numberOfIterations ; i += 100){
        b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;
        b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;
        b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;
        b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;
        b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;
        b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;
        b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;
        b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;
        b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;
        b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a;b=a; // 100 assignments 
    }

why does the operation of b=a; go to two instructions ? and how comes that I calculate the number of cycles it takes (each b=a;) is one cycle?

I compiled it with g++

Answer 1

You should give the compiler the chance to optimize the code. With the proper optimization flags -O3 -march=native my compiler (gcc) is able to reduce all of that to the following line:

movl    $5, %eax

no loop, repetition of the code, nothing.

So your compiler may or may not do optimizations and a lot of stuff that you don’t see. And processors are different. Mine here is able to put the constant 5 into an immediate and doesn’t create the variable a at all.