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));

 

转载于:https://www.cnblogs.com/aqing1987/p/4613706.html