strcpy和printf一个多维char数组C.
说我有一个arrays
char messages[10][2][50]; strcpy的正确语法是什么，以便将数据放入其中一个字符串（大多数为50的内部字符数组），然后通过％s将其提供给printf的相应约定？
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。
你可以填写下面的空白吗？
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arrays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...
并且输出就是这样
消息1第1部分，m2 p2
m2，p2
m3，p3
等等
我刚写了一个快速的程序来展示你所问过的事情……在声明中加载它们，strncpy到其中一个，然后打印出来。
希望能帮助到你
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Tommi Kyntola，我在评论中谈到了
#include #include // 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; }
你可以省略最大的订阅（在你的例子中是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
我会用：
假设您要复制到char* new_buff
memcpy(new_buff, messages, 10*2*50);
你可以做一个3嵌套循环并使用strncpy 。
不要使用strcpy …它是不安全的
正如已经指出的，最好使用strncpy，或者如下面的示例中所示，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]);
这将是
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]);
试着搞定这个概念。