2010-02-01 90 views

回答

9

方法A:

return (the_bitset >> start_bit).to_ulong(); 

方法B(比方法A在我的機器上快了100倍):

unsigned long mask = 1; 
unsigned long result = 0; 
for (size_t i = start_bit; i < end_bit; ++ i) { 
    if (the_bitset.test(i)) 
     result |= mask; 
    mask <<= 1; 
} 
return result; 
+0

'operator []'應該更快,因爲沒有像'test()'那樣的邊界檢查。 – 2017-05-15 21:21:18

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