Sign extension with bitwise shift operation

Question

following this Q&A I tried to exam the answer so I wrote:

#include 

int main ()
{

        int t;int i;
        for (i=120;i<140;i++){         


        

Answer 1

The >> operator, right shift, is arithmetic right shift in most compilers, meaning divide-by-2.

So, if, e.g. int i ==-4 (0xfffffffc), then i>>1 == -2 (0xfffffffe).

Having said this, I would recommend you to check the assembly of your code.
e.g. x86 has 2 separate instructions - shr & sar, denoting logical shift & arithmetic shift respectively.
Generally, compilers use shr (logical shift) for unsigned variables & sar (arithmetic shift) for signed variables.

---

Below is the C code & corresponding assembly, generated with gcc -S:

a.c:

int x=10;
unsigned int y=10;

int main(){
    unsigned int z=(x>>1)+(y>>1);
    return 0;
}

a.s:

    .file   "a.c"
.globl x
    .data
    .align 4
    .type   x, @object
    .size   x, 4
x:
    .long   10
.globl y
    .align 4
    .type   y, @object
    .size   y, 4
y:
    .long   10
    .text
.globl main
    .type   main, @function
main:
    pushl   %ebp
    movl    %esp, %ebp
    subl    $16, %esp
    movl    x, %eax
    sarl    %eax ; <~~~~~~~~~~~~~~~~ Arithmetic shift, for signed int
    movl    y, %edx
    shrl    %edx ; <~~~~~~~~~~~~~~~~ Logical shift, for unsigned int
    addl    %edx, %eax
    movl    %eax, -4(%ebp)
    movl    $0, %eax
    leave
    ret
    .size   main, .-main
    .ident  "GCC: (Ubuntu/Linaro 4.5.2-8ubuntu4) 4.5.2"
    .section    .note.GNU-stack,"",@progbits