Question

this is the actual code that I’m trying to convert to gcc style asm inline assembly code.

#include<iostream>
using namespace std;

int reverse(int num);

int main(){
    int num;
    cout << "enter number: ";
    cin >> num;
    cout << endl;
    cout << reverse(num);
    return 0;
}

int reverse(int num){
    if(num == 0 || num == 1){
        return num;
    }
    __asm
    {
        xor eax, eax
        xor ecx, ecx     
        mov ebx, num    
        clc             ; clear carry
not_found:
        inc ecx
        shl ebx, 1
        jnc not_found


        rcr eax, 1
        mov edx, ecx
again:
        shl ebx, 1
        rcr eax, 1
        inc ecx
        cmp ecx, 32
        jne again

        dec edx     
again2:     
        shr eax, 1
        dec edx
        cmp edx, 0
        jne again2
    }

}

As i was unable to compile the above code with gcc, I tried converting it to some thing that could be compiled successfully by a gcc compiler, but so far I could not manage to produce any meaningful result.

Answer 1

Based on the code in the comment of the OP, here is a modified example that shifts by one bit using inline assembly:

#include<iostream>
using namespace std;
int reverse(int num);
int main()
{
    int num;
    cout << "enter number: ";
    cin >> num;
    cout << endl;
    cout << reverse(num) << endl;
    return 0;
}

int reverse(int num)
{
    if (num == 0 || num == 1)
    {
        return num;
    }
    int temp = 0;
    asm( "shll $1, %0 \n\t" : "=r"(temp) : "0"(num));
    return temp;
}

Note that gcc puts the operands into registers if you ask for it using the r constraint so really no reason to mov them yourself (at least in this small sample).
Also, I have used 0 in the input constraint to signal that it should be the same register as the output, because that’s how shl works.

Still not sure about the num == 1 condition, but presumably that’s because we don’t know the actual function of the complete function.