Lets look at this simple functions:
// set_test.cpp
#include <stdint.h> #include <bitset> const int size = 128; typedef uint8_t byte_set[size]; bool any_in_byteset(uint8_t* data, size_t size, byte_set set) { for (auto i=0u; i < size; i++) if (set[data[i]]) return true; return false; } typedef std::bitset<size> bit_set; bool any_in_bitset(uint8_t* data, size_t size, bit_set set) { for (auto i=0u; i < size; i++) if (set[data[i]]) return true; return false; }
The file was compiled with g++ -std=c++11 -O3 set_test.cpp; Assembly code of the core of any_in_byteset:
28: 0f b6 10 movzbl (%eax),%edx 2b: 83 c0 01 add $0x1,%eax 2e: 80 3c 11 00 cmpb $0x0,(%ecx,%edx,1) 32: 75 0c jne 40 34: 39 d8 cmp %ebx,%eax 36: 75 f0 jne 28Statement if (set[data[i]]) return true are lines 28, 2e and 32, i.e.: load from memory, compare and jump. Instructions 2b, 34 and 36 handles the for loop.
Now, look at assembly code of any_in_bitset:
5f: 0f b6 13 movzbl (%ebx),%edx 62: b8 01 00 00 00 mov $0x1,%eax 67: 89 d1 mov %edx,%ecx 69: 83 e1 1f and $0x1f,%ecx 6c: c1 ea 05 shr $0x5,%edx 6f: d3 e0 shl %cl,%eax 71: 85 44 94 18 test %eax,0x18(%esp,%edx,4) 75: 75 39 jne b0
All these instructions implements the if statement! Again we have load from memory (5f), but checking which bit is set require much more work. Input (edx) is split to lower part --- i.e. bit number (67, 6c) and higher part --- i.e. word index (6c). Last step is to check if a bit is set in a word --- GCC used variable shift left (6f), but x86 has BT instruction, so in a perfect code we would have 2 instructions less.
However, as we see simple access in the bitset is much more complicated than simple memory fetch from byteset. For short sets memory fetches are well cached and smaller number of instruction improves performance. For really large set cache misses would kill performance, then bitset is much better choice.
Brak komentarzy:
Prześlij komentarz