strncpy和strlen的可能的实现

#include <stdio.h>
#include <stdlib.h>
//为避免与标准库中的函数发生混淆,我将它们命名为stringNCopy和stringLength
char *stringNCopy(char *dest, const char *src, size_t count)
{
    for (size_t i = 0; i < count; ++i)
    {
        dest[i] = src[i];
        if (!src[i])
        {
            break;
        }
    }
    return dest;
}
size_t stringLength(const char *str)
{
    size_t count = 0;
    while (*str++)
    {
        ++count;
    }
    return count;
}
int main()
{
    char *dest = malloc(sizeof(char) * 100);
    const char *src = "hello world\n世界你好";
    stringNCopy(dest, src, stringLength(src) + 1);
    printf("%s\n", dest);
    free(dest);
    char arr[50];
    char *ret = stringNCopy(arr, "programming 程序设计", 10000);
    printf("%s\n%s\n", arr, ret);
    ret = stringNCopy(arr, "Programmer", 1);
    printf("%s\n%s\n", arr, ret);
    char test[2];
    const char *str = "测试\0Test";
    ret = stringNCopy(test, str, 100); //非法的操作,可通过编译,但是运行之后可能会报错(vc++下编译并运行之后会报错,Windows下gcc编译后运行无报错)
    printf("%s\n%s\n", test, ret);
    printf("%u\n%u\n%u\n", stringLength(""), stringLength("\0\0\0"), stringLength("字符串 String"));
    return 0;
}

posted on 2018-02-13 18:16  布伊什  阅读(397)  评论(0编辑  收藏  举报