Hello, I've been looking for the GCC built-in to swap two octets within a 16-bit word. I looked at the list of GCC built-ins, and found bswap32 and bswap64, but no bswap16. http://gcc.gnu.org/onlinedocs/gcc/Other-Builtins.html int32_t __builtin_bswap32 (int32_t x) Returns x with the order of the bytes reversed; for example, 0xaabbccdd becomes 0xddccbbaa. Byte here always means exactly 8 bits. int64_t __builtin_bswap64 (int64_t x) Similar to __builtin_bswap32, except the argument and return types are 64-bit. On x86, GCC is smart enough to optimize the operation to a single instruction. $ cat swap16.c #include <stdint.h> uint16_t swap16(uint16_t n) { return (n >> 8) | (n << 8); } $ gcc -O3 -fomit-frame-pointer -S swap16.c _swap16: movzwl 4(%esp), %eax rolw $8, %ax ret Is my swap16 function recognized as a bswap16 operation internally, therefore hand-written assembly code is provided? On my platform (SH-4) the operation is not optimized: $ sh-superh-elf-gcc -O3 -S swap16.c _swap16: extu.w r4,r4 mov r4,r0 shlr8 r4 shll8 r0 or r4,r0 rts extu.w r0,r0 Even though the operation could be done in a single swap.b instruction (and possibly an extu.w for ABI compliance) So, if I understand the situation correctly, the bswap16 built-in may not be needed because GCC can recognize the C pattern for bswap16, but no optimized assembly code has been provided for my platform. Is that correct? -- Regards.
