c - Leal instruction in for loop -

i'm reading book computer systems: programmer's perspective (2nd edition) , practice problem 3.23 little confused me:

a function fun_b has following overall structure:

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

the gcc c compiler generates following assembly code:

x @ %ebp+8 1 movl 8(%ebp), %ebx 2 movl $0, %eax 3 movl $0, %ecx .l13: 5 leal (%eax,%eax), %edx 6 movl %ebx, %eax 7 andl $1, %eax 8 orl  %edx, %eax 9 shrl %ebx   shift right 1 10 addl $1, %ecx 11 cmpl $32, %ecx 12 jne .l13 

reverse engineer operation of code , following:

a. use assembly-code version fill in missing parts of c code.

my solution.

int fun_b(unsigned x) {    int val = 0;    int i;    ( = 0 ;i < 32;i++) {       val  += val; //because leal (%eax,%eax), edx  --> %edx = %eax + %eax       val = val | x & 0x1;       x >>= 1;    }    return val; } 

book's solution.

int fun_b(unsigned x) {   int val = 0;   int i;   (i = 0; < 32; i++) {     val = (val << 1) | (x & 0x1);     x >>= 1;   }  return val; } 

please, explain me why leal function has non typical behavior in function. , dont understand how assembly code yielding statement val = (val << 1) | (x & 0x1)

in code:

val  += val; val = val | x & 0x1; 

here, val += val equivalent (val*2) equal val left shifted 1.

but think solution correct if assembly code like:

x @ %ebp+8 1 movl 8(%ebp), %ebx 2 movl $0, %eax 3 movl $0, %ecx .l13: 5 addl  %eax, %eax 6 movl %ebx, %edx 7 andl $1, %edx 8 orl  %edx, %eax 9 shrl %ebx   # shift right 1 10 addl $1, %ecx 11 cmpl $32, %ecx 12 jne .l13 

because if val + val separate statement, compiler places in eax register rather in edx (i'm not sure case always). so, code have given, possible solutions are:

val = (val << 1) | (x & 0x1);

or

val = (val + val) | (x & 0x1);

or

val = (val * 2) | (x & 0x1);