问题描述

以下字符串是怎么连接的,求解释

//program 6.4 joining strings
#define STDC_WANT_LIB_EXT1 1
#include
#include
int main(void)
{
char preamble[] = "the joke is:nn";
char str[][40] = {
"My dog hasn't got any nose.n",
"How does your dog smell then?n",
"My dog smells horrible.n"
};

unsigned int strCount = sizeof(str)/sizeof(str[0]);
unsigned int length = 0;
for(unsigned int i = 0; i < strCount ; ++i)
    length += strnlen_s(str[i], sizeof(str[i]));
char joke[length + strnlen_s(preamble, sizeof(preamble)) + 1];

if(strncpy_s(joke, sizeof(joke), preamble, sizeof(preamble)))
{
    printf("error copying preamble to joke.n");
    return 1;
}

for(unsigned int i = 0;  i < strCount ;  ++i)
{
    if(strncat_s(joke, sizeof(joke), str[i], sizeof(str[i])))
    {
        printf("error copying string str[%u].", i);
        return 2;
    }

}
    printf("%s", joke);
        return 0;

}
这段代码到底是怎么组合字符的,搞了半天只看懂一点!
希望大神能给详细解释解释 (快到指针了,好激动)

解决方案

#define STDC_WANT_LIB_EXT1 1
#include <stdio.h>
#include <string.h>
int main(void)
{
    char preamble[] = "the joke is:nn";
    char str[][40] =
    {
        "My dog hasn't got any nose.n",
        "How does your dog smell then?n",
        "My dog smells horrible.n"
    };
    // 计算出str数组中的str个数，本程序中str值为3
    unsigned int strCount = sizeof(str)/sizeof(str[0]);
    unsigned int length = 0;
    // strnlen_s函数返回字串的长度
    for (unsigned int i = 0; i < strCount ; ++i)
        length += strnlen_s(str[i], sizeof(str[i]));
            // length为str数组中的字符串拼接后总长度
    char joke[length + strnlen_s(preamble, sizeof(preamble)) + 1];

    //  先strncpy_s函数拷贝，再strncat_s函数连接，作用区域都是joke
    if (strncpy_s(joke, sizeof(joke), preamble, sizeof(preamble)))
    {
        printf("error copying preamble to joke.n");
        return 1;
    }

    for (unsigned int i = 0;  i < strCount ;  ++i)
    {
        if (strncat_s(joke, sizeof(joke), str[i], sizeof(str[i])))
        {
            printf("error copying string str[%u].", i);
            return 2;
        }

    }
    printf("%s", joke);
    return 0;
}