基礎函數的實作（strcpy，strcat, strcmp, atoi, itoa）【轉】

  strcpy:

char *strcpy(char *strDestination, const char *strSource);

{

assert(strDestination && strSource);

char *cp=strDestination;

while(*cp++ = *strSource++);

return strDestination;

}

VERSION 2:

char* strcpy(char * dst, const char * src)

{

char * cp = dst;

while( *cp++ = *src++ )

;                           

return( dst );

}

VERSION 3:

char *srcpy(char *dest,const char *source)

{

assert((dest!=NULL)&&(source!=NULL));

char *address = dest;

while(*source!='\0')

{

*dest++=*source++;

}

*dest = '\0';

return address;

}

strcat:

VERSION 1:

char * strcat(char * dest, const char * src)

{

char *tmp = dest;

while (*dest)

dest++;

while ((*dest++ = *src++) != '\0')

;

return tmp;

}

strcmp:

VERSION 1:

int strcmp ( const char* src, const char* dst )

{

int ret = 0 ;

while( !(ret = *(unsigned char *)src - *(unsigned char *)dst) && *dst)

++src, ++dst;

if ( ret < 0 )

ret = -1 ;

else if ( ret > 0 )

ret = 1 ;

return( ret );

}

VERSION 2:

int strcmp(const char *dest, const char *source)  

{  

assert((NULL != dest) && (NULL != source));  

while (*dest && *source && (*dest == *source))  

{  

dest ++;  

source ++;  

}  

return *dest - *source;  

}

VERSION 3:

int strcmp(char *source, char *dest)

{

assert(source != NULL && dest != NULL);

while(*source++==*dest++)

{

if(*source=='\0'&&*dest=='\0')

return 0;       

}

return -1;

}

itoa:

#include "stdafx.h"

#include <iostream>

using namespace std;

void itoa(int num,char str[] )

{

int sign = num,i = 0,j = 0;

char temp[11];

if(sign<0)//判斷是否是一個負數

{

num = -num;

};

do

{

temp[i] = num%10+'0';       

num/=10;

i++;

}while(num>0);

if(sign<0)

{

temp[i++] = '-';//對于負數,要加以負号

}

temp[i] = '\0';

i--;

while(i>=0)//反向操作

{

str[j] = temp[i];

j++;

i--;

}

str[j] = '\0';

}

atoi:

int atoi(char s[])

{

int i = 0,sum = 0,sign;    //輸入的數前面可能還有空格或制表符應加判斷

while(' '==s[i]||'\t'==s[i])

{

i++;

}

sign = ('-'==s[i])?-1:1;

if('-'==s[i]||'+'==s[i])

{

i++;

}

while(s[i]!='\0')

{

sum = s[i]-'0'+sum*10;

i++;

}   

return sign*sum;

}