沒辦法, 現在的大公司面試就面這個, 你不得不研究下底層的實現.
要點:
1) 字長邊界對齊以便加快速度. 對齊時也要考慮機器的位數哦.
#include <stdio.h>
#include <string.h>
int my_strlen(const char * str)
{
const char * pstr = str;
// 先將地址邊界對齊 & 操作更高效, 避免了%取餘運算
//for (; (unsigned long)pstr%(sizeof(long)) != 0; ++pstr)
for (; (unsigned long)pstr & (sizeof(long)-1) != 0; ++pstr)
{
if (pstr[0] == '\0')
return pstr - str;
}
unsigned long himagic = 0x80808080;
unsigned long lomagic = 0x01010101;
// 如果是64位
if (sizeof(long) > 4)
{
himagic = ((himagic << 16) << 16) | himagic;
lomagic = ((lomagic << 16) << 16) | lomagic;
}
unsigned long longword = 0;
for (;; pstr+=sizeof(long))
{
longword = *(unsigned long*)pstr;
// (longword - lomagic) & ~longword 可測試出, 相減之後longword變量的那些bit發生了改變
// ... & himagic 可測試出, 變化的那些bit位中是否包含各個字節的最高位
if (((longword - lomagic) & ~longword & himagic) != 0)
{
if (pstr[0] == '\0')
return pstr - str;
if (pstr[1] == '\0')
return pstr - str + 1;
if (pstr[2] == '\0')
return pstr - str + 2;
if (pstr[3] == '\0')
return pstr - str + 3;
if (sizeof(long) > 4)
{
if (pstr[4] == '\0')
return pstr - str + 4;
if (pstr[5] == '\0')
return pstr - str + 5;
if (pstr[6] == '\0')
return pstr - str + 6;
if (pstr[7] == '\0')
return pstr - str + 7;
}
}
}
// never come to here...
return 0;
}
int main()
{
char * str = "helloworldaaaaa";
int ret = 0;
ret = my_strlen(str+1);
printf("len : %d\n", ret);
ret = strlen(str+1);
printf("len : %d\n", ret);
return 0;
}
輸出如下:
:!g++ -Wall -g strlen.cpp -o strlen && ./strlen
len : 14
len : 14