今天到I 公司去面試，面試方式比較特殊，沒(méi)有筆試，就是2 個(gè)面試官，一人一句輪番發(fā)問(wèn)，涉及面很廣，涉及到操作系統(tǒng)(MMU 、page out 、process/thread 、semaphore 、interrupt), OOP( 多態(tài)、design pattern) 、數(shù)據(jù)結(jié)構(gòu)( 排序、二叉查找樹(shù)) 、計(jì)算機(jī)網(wǎng)絡(luò)(OSI 5 層) 、C 語(yǔ)言(big/small endian) 、英語(yǔ)口語(yǔ)等等，問(wèn)了大約一個(gè)小時(shí)左右。

所有問(wèn)題都是口頭表述，只在紙上寫了一個(gè)memcpy 程序，用C 語(yǔ)言實(shí)現(xiàn)，腦子一發(fā)蒙，既然寫成了strcpy ，真該死。

回家了查詢了一下memcpy 定義，如下：

Void *memcpy(void *dest, const void *src, unsigned int count);

查詢msdn, 發(fā)現(xiàn)Remark 如下：

memcpy copies count bytes from src to dest ; If the source and destination overlap, the behavior of memcpy is undefined. Use memmove to handle overlapping regions.

以上描述針對(duì)dest 和 src 所指的內(nèi)存地址有重疊的情況，內(nèi)存地址重疊情況，memcpy 函數(shù)處理步驟未定，而memmove 對(duì)重疊情況給予處理；

在winXP+visual c++2005 測(cè)試 memcpy 函數(shù)，程序如下：

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#include "stdafx.h"

#include <string.h>

int _tmain (int argc , _TCHAR * argv [])

{

?????? char s [16] = "aabbcc" ;

?????? char d [16] = {0};

??????

?????? memcpy (s +2, s , 4);

?????? printf ("%s" , s );

?????? return 0;

}

結(jié)果輸出 “aaaabb”, 由此可見(jiàn)windows 平臺(tái)的c 運(yùn)行時(shí)MSVCRT 的memcpy 函數(shù)對(duì)重疊部分做了處理，同memmove 的實(shí)現(xiàn)。//notes: 如果重疊部分不做處理，應(yīng)該輸出”aaaaaa”

下面我們用c 語(yǔ)言來(lái)實(shí)現(xiàn)memcpy 函數(shù), 首先我們寫出不對(duì)內(nèi)存重疊的處理函數(shù)，如下：

void *memcpy_no_handle_overlap (void *dest , void *src , unsigned int count )

{

?????? if ((NULL ==dest ) || (NULL ==src ))

????????????? return NULL ;

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?????? char *d = (char *)dest ;

?????? char *s = (char *)src ;

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?????? //Do normal (Upwards) Copy

?????? while (count -- > 0)

????????????? *d ++ = *s ++;

?????? return dest ;

}

測(cè)試程序如下：

int _tmain(int argc, _TCHAR* argv[])

{

?????? char s[16] = "aabbcc";

?????? char d[16] = {0};

??????

?????? memcpy_no_handle_overlap(s+2, s, 4);

?

?????? printf("%s", s);

?????? return 0;

}

輸出結(jié)果”aaaaaa”

下面討論處理memory overlapping 情況，如下圖：

判斷overlapping 條件如下：

If ( (dest <= src) ||??????????????? // green region 1

?? (dest >=src+count) )?????????? // green region 2

{

?????? // no memory overlapping

}

Else? // red region 3

{

?????? // there is overlapping

}

Overlapping 的處理：

我們可以看到memcpy_no_handle_overlap 函數(shù)，是從低地址依次賦值到高地址；在處理overlapping 時(shí)，如果我們采用同樣的方法( 低地址到高地址) ，高地址的值將會(huì)被覆蓋，所以我們應(yīng)該從高地址依次到低地址賦值，如下圖：

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? 函數(shù)代碼如下:

void *memcpy_handle_overlap(void *dest, void *src, unsigned int count)

{

?????? if ((NULL==dest) || (NULL==src))

????????????? return NULL;

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?????? char *d = (char *)dest;

?????? char *s = (char *)src;

?

?????? //Check for overlapping buffers:

?????? if ( (d<=s) || (d>=s+count) )

?????? {?????

????????????? //Do normal (Upwards) Copy

????????????? while (count-- > 0)

???????????????????? *d++ = *s++;

?????? }

?????? else

?????? {

????????????? //Do Downwards Copy to avoid propagation

????????????? while (count > 0)

????????????? {

???????????????????? *(d+count-1) = *(s+count-1);

???????????????????? --count;

????????????? }

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?????? }

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?????? return dest;

}

測(cè)試代碼：

int _tmain(int argc, _TCHAR* argv[])

{

?????? char s[16] = "aabbcc";

?????? char d[16] = {0};

??????

?????? memcpy_handle_overlap(s+2, s, 4);

?

?????? printf("%s", s);

?????? return 0;

}

輸出結(jié)果為: “aaaabb “

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最后測(cè)試代碼如下：

int _tmain(int argc, _TCHAR* argv[])

{

?????? char s[16] = "aabbcc";

?????? memcpy_no_handle_overlap(s+2, s, 4);

?????? printf("memcpy(ignore memory overlapping): %s/n", s);

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?????? strcpy(s, "aabbcc");

?????? memcpy_handle_overlap(s+2, s, 4);

?????? printf("memcpy(handle memory overlapping): %s/n", s);

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???? strcpy(s, "aabbcc");

?????? memcpy(s+2, s, 4);

?????? printf("memcpy( MSVCRT ): %s/n", s);

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?????? strcpy(s, "aabbcc");

?????? memmove(s+2, s, 4);

?????? printf("memmove( MSVCRT): %s/n", s);

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?????? return 0;

}

輸出結(jié)果為：

memcpy(ignore memory overlapping): aaaaaa

memcpy(handle memory overlapping): aaaabb

memcpy( MSVCRT ): aaaabb

memmove( MSVCRT): aaaabb

總結(jié)

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