如何连接一个字符数组情况下，[0]和[1]在C一个字符指针？ [英] How to concatenate a char array case [0] and [1] to a char pointer in C?

问题描述

我想连接两个字符7和9，形成这个字符串79。

首先，我初始化的变量。 （限制：我只用CHAR类型，我不能再拍CHARAC一样的变量。）

 的char * CHARAC =（字符*）malloc的（的sizeof（char）的* 200）;
字符*合并=（字符*）malloc的（的sizeof（char）的* 200）;
 

然后我给的值CHARAC [0]和CHARAC [1]。

  CHARAC [0] =7;
CHARAC [1] ='9';
的printf（CHARAC [0]％C \\ N，CHARAC [0]）;
的printf（CHARAC [1]：％C \\ N，CHARAC [1]）;
 

我想连接CHARAC [0]和CHARAC [1]

 的strcpy（组合，（字符*）CHARAC [0]）;
strcat的（组合，（字符*）CHARAC [1]）;
的printf（组合：％S \\ n，合并）;
免费（组合）;
 

不过，code以上不起作用。我得到了错误信息：警告：转换为不同大小的整指针

阅读所有的意见和建议后，我得到的输出结果，我想。

下面是最后的code：

 的char * CHARAC =（字符*）malloc的（的sizeof（char）的* 200）;
        字符*合并=（字符*）malloc的（的sizeof（char）的* 200）;        CHARAC [0] =7;
        CHARAC [1] ='9';
        CHARAC [2] ='\\ 0';
        的printf（CHARAC [0]％C \\ N，CHARAC [0]）;
        的printf（CHARAC [1]：％C \\ N，CHARAC [1]）;
        的printf（CHARAC [2]：％C \\ N，CHARAC [2]）;        的strcpy（相结合，与放大器; CHARAC [0]）;
        的strcat（相结合，与放大器; CHARAC [1]）;
        的strcpy（相结合，与放大器; CHARAC [0]）;
        的strcat（相结合，与放大器; CHARAC [2]）;
        的printf（组合：％S \\ n，合并）;
        免费（组合）;
 

解决方案

他们已经串联原样。这是因为它们是对彼此相邻的存储器。您已声明他们在同一阵列中，并在相邻的索引。

  + --------------- +
| 7| 9|
+ --------------- +
   [0] [1]
 

这是你的存储阵列。七（ CHARAC [0] ）是旁边九（ CHARAC [1] ）。
正如评论所说，只需空终止阵列来完成的连接。

  CHARAC [2] ='\\ 0';
 

该阵列现在的printf（）友好的，因为它是空终止。

I want to concatenate two characters '7' and '9' to form this string "79".

First, I initialized the variables. (Restriction: I have to use char types only and I must not make another charac alike variable.)

char *charac = (char *) malloc(sizeof(char) * 200);
char *combined = (char *) malloc(sizeof(char) * 200);

Then I gave the values for charac[0] and charac[1].

charac[0] = '7';
charac[1] = '9';
printf("charac[0] : %c\n", charac[0] );
printf("charac[1] : %c\n", charac[1] );

I want to concatenate charac[0] and charac[1]

strcpy(combined, (char*)charac[0]);
strcat(combined, (char*)charac[1]);
printf("combined : %s\n", combined);
free(combined);

But the code above doesn't work. I got the error message: "warning: cast to pointer from integer of different size".

After reading all of your comments and suggestions, I obtain the output result I want.

Here is the final code:

char *charac = (char *) malloc(sizeof(char) * 200);
        char *combined = (char *) malloc(sizeof(char) * 200);

        charac[0] = '7';
        charac[1] = '9';
        charac[2] = '\0';
        printf("charac[0] : %c\n", charac[0] );
        printf("charac[1] : %c\n", charac[1] );
        printf("charac[2] : %c\n", charac[2] );

        strcpy(combined, & charac[0]);
        strcat(combined, & charac[1]);
        strcpy(combined, & charac[0]);
        strcat(combined, & charac[2]);
        printf("combined : %s\n", combined);
        free(combined);

解决方案

They are already concatenated as is. This is because they are right next to each other in memory. You have declared them in the same array, and at adjacent indexes.

+---------------+
|  '7'  |  '9'  |
+---------------+
   [0]     [1]

This is your array in memory. The seven (charac[0]) is right next to the nine (charac[1]). As suggested in the comments, simply null terminate the array to complete the concatenation.

charac[2] = '\0';

The array is now printf() friendly as it is null terminated.

这篇关于如何连接一个字符数组情况下，[0]和[1]在C一个字符指针？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！