Why did GCC generate mov %eax,%eax and what does it mean?

Question

GCC 4.4.3 generated the following x86_64 assembly. The part that confuses me is the mov %eax,%eax. Move the register to itself? Why?

   23b6c:       31 c9                   xor    %ecx,%ecx        ; the 0 value for shift    23b6e:       80 7f 60 00             cmpb   $0x0,0x60(%rdi)  ; is it shifted?    23b72:       74 03                   je     23b77    23b74:       8b 4f 64                mov    0x64(%rdi),%ecx  ; is shifted so load shift value to ecx    23b77:       48 8b 57 38             mov    0x38(%rdi),%rdx  ; map base    23b7b:       48 03 57 58             add    0x58(%rdi),%rdx  ; plus offset to value    23b7f:       8b 02                   mov    (%rdx),%eax      ; load map_used value to eax    23b81:       89 c0                   mov    %eax,%eax        ; then what the heck is this? promotion from uint32 to 64-bit size_t?    23b83:       48 d3 e0                shl    %cl,%rax         ; shift rax/eax by cl/ecx    23b86:       c3                      retq    

The C++ code for this function is:

    uint32_t shift = used_is_shifted ? shift_ : 0;     le_uint32_t le_map_used = *used_p();     size_t map_used = le_map_used;     return map_used << shift; 

An le_uint32_t is a class which wraps byte-swap operations on big-endian machines. On x86 it does nothing. The used_p() function computes a pointer from the map base + offset and returns a pointer of the correct type.

Answer 1

In x86-64, 32-bit instructions implicitly zero-extend: bits 32-63 are cleared (to avoid false dependencies). So sometimes that's why you'll see odd-looking instructions. (Is mov %esi, %esi a no-op or not on x86-64?)

However, in this case the previous mov-load is also 32-bit so the high half of %rax is already cleared. The mov %eax, %eax appears to be redundant, apparently just a GCC missed optimization.