#include <iostream> using namespace std;int main() {int a;char *name = NULL;a = strlen(name);return 0; }

以上程序編譯沒問題，運(yùn)行將會(huì)報(bào)錯(cuò)。

原因就是name為NULL，strlen參數(shù)不能為NULL，為探究原因，我查看了glibc對(duì)strlen函數(shù)的實(shí)現(xiàn)代碼如下:

#include <string.h> #include <stdlib.h>#undef strlen#ifndef STRLEN #define STRLEN strlen #endifsize_t STRLEN(const char *str) {const char *char_ptr;const unsigned long int *longword_ptr;unsigned long int longword, himagic, lomagic;//Handle the first few characters by reading one character at a time.//Do this until CHAR_PTR is aligned on a longword boundary.for (char_ptr = str; ((unsigned long int) char_ptr & (sizeof(longword) - 1)) != 0; ++char_ptr)if (*char_ptr == '\0') return char_ptr - str;longword_ptr = (unsigned long int *) char_ptr;/*Bits 31, 24, 16, and 8 of this number are zero. Call these bits the"holes." Note that there is a hole just to the left of each byte, with an extra at the end:bits: 01111110 11111110 11111110 11111111bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDDThe 1-bits make sure that carries propagate to the next 0-bit.The 0-bits provide holes for carries to fall into.*/himagic = 0x80808080L;lomagic = 0x01010101L;if (sizeof(longword) > 4) {/* 64-bit version of the magic. *//* Do the shift in two steps to avoid a warning if long has 32 bits. */himagic = ((himagic << 16) << 16) | himagic;lomagic = ((lomagic << 16) << 16) | lomagic;}if (sizeof(longword) > 8) abort();/* Instead of the traditional loop which tests each characters,we will test a longword at a time. The tricky part is testingif *any of the four* bytes in the longword in question are zero. */for (;;) {longword = *longword_ptr++; //strlen函數(shù)為什么不能傳入空指針原因!if (((longword - lomagic) & ~longword & himagic) != 0) {/*Which of the bytes was the zero? If none of them were, it wasa misfire; continue the search. */const char *cp = (const char *)(longword_ptr - 1);if (cp[0] == 0)return cp - str;if (cp[1] == 0)return cp - str + 1;if (cp[2] == 0)return cp - str + 2;if (cp[3] == 0) return cp - str + 3;if (sizeof(longword) > 4) {if (cp[4] == 0) return cp - str + 4;if (cp[5] == 0)return cp - str + 5;if (cp[6] == 0)return cp - str + 6;if (cp[7] == 0)return cp - str + 7;}}} }

從第47行可見，如果傳入NULL，則

longword = *longword_ptr++;

longword_ptr為NULL，從而*longword_ptr是不可訪問的。

注:*longword_ptr++等效于*(longword_ptr++), 即先求*longword_ptr,再對(duì)longword_ptr+1。

?

轉(zhuǎn)自:https://blog.csdn.net/a3192048/article/details/80933476

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