In your code:
val += val;
val = val | x & 0x1;
Here, val += val
which is equivalent to (val*2)
which is effectively equal to val
left shifted by 1
.
But I think your solution is correct only if the assembly code was something like:
x at %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 by 1
10 addl $1, %ecx
11 cmpl $32, %ecx
12 jne .L13
Because if val + val
was a separate statement, compiler usually places it in eax register rather than in edx (i'm not sure this is the case always). So, for the code you have given, the possible solutions are:
val = (val << 1) | (x & 0x1);
or
val = (val + val) | (x & 0x1);
or
val = (val * 2) | (x & 0x1);