c - intro to x86 assembly -

i'm looking on example on assembly in csapp (computer systems - programmer's perspective 2nd) , want know if understanding of assembly code correct.

practice problem 3.23

int fun_b(unsigned x) {    int val = 0;    int i;    ( ____;_____;_____) {    }    return val; } 

the gcc c compiler generates following assembly code:

x @ %ebp+8 // i've gotten far  1 movl 8(%ebp), %ebx  // ebx: x 2 movl $0, %eax  // eax: val, set 0 since eax return  // value stored , val being returned @ end 3 movl $0, %ecx  // ecx: i, set 0 4 .l13:  // loop 5 leal (%eax,%eax), %edx  // edx = val+val  6 movl %ebx, %eax  // val = x (?) 7 andl $1, %eax // x = x & 1 8 orl %edx, %eax  // x = (val+val) | (x & 1) 9 shrl %ebx shift right 1  // x = x >> 1 10 addl $1, %ecx  // i++ 11 cmpl $32, %ecx  // if < 32 jump loop 12 jne .l13  

there similar post on same problem solution, i'm looking more of walk-through , explanation of assembly code line line.

you seem have meaning of instructions figured out. comment on lines 7-8 wrong however, because assign eax val not x:

7 andl $1, %eax // val = val & 1 = x & 1 8 orl %edx, %eax  // val = (val+val) | (x & 1) 

putting c template be:

for(i = 0; < 32; i++, x >>= 1) {     val = (val + val) | (x & 1); } 

note (val + val) left shift, function doing shifting out bits x on right , shifting them in val right. such, it's mirroring bits.

ps: if body of for must empty can of course merge third expression.