Question

说我有一个数组

char messages[10][2][50];

strcpy的正确语法是什么，为了将数据放入其中一个字符串（大多数为50的内部字符数组），然后是相应的约定，通过％s将它提供给printf？

就此而言，我是否以正确的顺序声明了数组下标？它是10批，成对（2）字符串。每个字符串宽50个字符。

01{{50 chars},{50 chars}}
02{{50 chars},{50 chars}}
...
09{{50 chars},{50 chars}}
10{{50 chars},{50 chars}}

各种互联网资源似乎与哪些下标省略冲突，无论我尝试什么似乎都会产生意想不到的结果。

e.g。你能填补下面的空白吗

strcpy(message???, "Message 1 Part 1");
strcpy(message???, "m1 p2");
strcpy(message???, "m2 p1");
strcpy(message???, "m2 p2");
strcpy(message???, "m3 p1");
strcpy(message???, "m3 p1");
//So on...

int i;
for(i=0;i<10;i++)
    printf("%s, %s\n", message???, message???);

这样阵列具有并保持的结构：

01{{"Message 1 Part 1\0"},{"m1 p2\0"}}
02{{"m2 p1\0"},{"m2 p2\0"}}
01{{"m3 p1\0"},{"m3 p2\0"}}
//So on...

并且输出

消息1第1部分，m2 p2

m2，p2

m3，p3

等等

Answer 1

我刚刚编写了一个快速程序来展示你所询问的事情......在声明中加载它们，strncpy到其中一个，然后将它们打印出来。

希望有所帮助

编辑：我有点讨厌魔术数字，所以我几乎完全删除了它们编辑：我添加了替代品Tommi Kyntola和我在评论中谈到的

#include <stdio.h>
#include <string.h>

// safe string copy macro, terminates string at end if necessary
// note: could probably just set the last char to \0 in all cases
// safely if intending to just cut off the end of the string like this

#define sstrcpy(buf, src, size) strncpy(buf, src, size); if(strlen(src) >= size) buf[size-1] = '\0';

#define MSGLIMIT 10
#define MSGLENGTH 30
#define MSGFIELDS 2
#define MSGNAME 0
#define MSGTEXT 1


int main(void) {
    char messages[MSGLIMIT][MSGFIELDS][MSGLENGTH] = { {"bla", "raa"},
                                                      {"foo", "bar"}
                                                    };
    int i;

    char *name1 = "name16789012345678901234567890";
    char *text1 = "text16789012345678901234567890";

    char *name2 = "name26789012345678901234567890";
    char *text2 = "text26789012345678901234567890";

    char *name3 = "name36789012345678901234567890";
    char *text3 = "text36789012345678901234567890";


    // doesn't set last char to \0 because str overruns buffer
    // undocumented result of running this, but likely to just get the name2 string
    // as that'll be the very next thing in memory on most systems

    strncpy(messages[2][MSGNAME], name1, MSGLENGTH); // 2 because it's the next empty one
    strncpy(messages[2][MSGTEXT], text1, MSGLENGTH);

    // alternative suggested by Tommi Kyntola
    // printf family are more complicated and so cost more cpu time than strncpy
    // but it's quick and easy anywhere you have string.h and fine most of the time

    snprintf(messages[3][MSGNAME], MSGLENGTH, "%s", name2);
    snprintf(messages[3][MSGTEXT], MSGLENGTH, "%s", text2);

    // uses the define macro at the top of the page to set the last char to \0 if
    // otherwise not set by strncpy, adds a little weight but still the better option
    // if performance of this section of code is important

    sstrcpy(messages[4][MSGNAME], name3, MSGLENGTH);
    sstrcpy(messages[4][MSGTEXT], text3, MSGLENGTH);


    for(i = 0; i < 5; i++) // 5 because that's how many I've populated
            printf("%s : %s\n", messages[i][MSGNAME], messages[i][MSGTEXT]);

    return 0;
}

Answer 2

你可以省略最大的订阅（在你的例子中它是10），因为它可以由编译器根据剩余的订阅来计算。要传递50个元素的图层，请使用指针：（* messages）[10] [2] - 将指针放在50个元素的图层上

Answer 3

我会用：

假设您要复制到char* new_buff

memcpy(new_buff, messages, 10*2*50);

你可以做3个嵌套循环并使用strncpy。

不要使用strcpy ...它是不安全的

Answer 4

正如已经指出的，最好使用strncpy，或者如下例所示，使用断言，以防止可能的缓冲区溢出。 strcpy与strncpy的性能略有提升。

    #define FIRST_OF_PAIR 0
#define SECOND_OF_PAIR 1

    int message_num = 7;

    char messages[10][2][50];

    char *string = "hello";

    assert(strlen(string) < 50);
    assert(message_num > 0 &&  message_num < 10);

    strcpy(messages[message_num][SECOND_OF_PAIR], "Hello");

    printf("%s", messages[message_num][SECOND_OF_PAIR]);

Answer 5

将是

strcpy(message[0][0], "Message 1 Part 1");
strcpy(message[0][1], "m1 p2");
strcpy(message[2][0], "m2 p1");
strcpy(message[2][1], "m2 p2");
strcpy(message[3][0], "m3 p1");
strcpy(message[3][1], "m3 p2");


for(i=0;i<10;i++)
    printf("%s, %s\n", message[i][0], message[i][1]);

尝试获得这个概念。

strcpy和printf一个多维char数组C.

5 个答案: