strcpy 一题

题目要求:

Write a function about string copy, the strcpy prototype is "char *strcpy (char *strDest, const char *strSrc); "

Here strDest is destination string, strSrc is source string.

1) Write the function strcpy, don't call C/C++ string library.

2) Here strcpy can copy strSrc to strDest, but why we use char * as the return value of strcpy?

 

问题解决:

1)代码实现

#include <stdio.h>

#include <assert.h>



char *sstrcpy(char *strdst, const char *strsrc)

{

    assert(strdst != NULL && strsrc != NULL);



    char *ret_str = strdst;

    while ((*strdst++ = *strsrc++) != '\0');

    return ret_str;

}



int main(int argc, char **argv)

{

    char dst[20] = {0};

    char *src = "abcde";



    printf("res = %s\n", sstrcpy(dst, src));



    return 0;

}

 

2)返回字符串指针,是为了实现链式表达。Unix设计哲学思想。

比如

printf("res = %s\n", sstrcpy(dst, src));

 

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