C：char指针数组没有发挥预期的动态 [英] C: array of char pointers not working as expected dynamically

问题描述

我从我的code下面的代码片段，我想使用动态分配的的char * 数组来保存字符串从标准输入到来。

I have the below snippets from my code where I am trying to use a dynamically allocated char * array to hold strings coming from stdin.

char **reference
reference = calloc(CHUNK, sizeof(char *));

我使用的是临时的静态数组首先存储从标准输入，然后根据一定的条件下将其复制到字符数组的字符串* 。我在运行时内存分配到个人的char * 。

I am using a temporary static array to first store the string from stdin, then based on a certain condition copy it to the array of char * . I am allocating memory to individual char * in runtime.

                        reference[no_of_ref]=malloc(strlen(temp_in) + 1);
                        reference[no_of_ref++]=temp_in;
//                   printf(" in temp : %s , Value : %s   , Address of charp : %p\n",temp_in,reference[no_of_ref-1],reference[no_of_ref-1]);
                        memset(&temp_in,'\0',sizeof(temp_in));
                        pre_pip = -1;
                }

        /*If allocated buffer is at brim, extend it by CHUNK bytes*/
                if(no_of_ref == CHUNK - 2)
                        realloc(reference,no_of_ref + CHUNK);

所以 no_of_ref 持有终于收到字符串总数。例如20，但是，当我打印整个参考阵列看到每个字符串，我得到的排在最后，得到印刷20次。相同的字符串

so no_of_ref holds the total number of strings finally received. e.g 20. But when I print the whole reference array to see each string, I get the same string that came last , getting printed 20 times.

推荐答案

下面的code的引入问题：

Here of your code introduces the problem:

reference[no_of_ref]=malloc(strlen(temp_in) + 1);
reference[no_of_ref++]=temp_in;

这是因为在C指针分配影响指针和指针只，不会做它的内容什么。您应该使用之类的东西的memcpy 或的strcpy ：

That's because assignment of pointers in C affects pointers and only pointers, which won't do anything with its contents. You shall use things like memcpy or strcpy:

reference[no_of_ref]=malloc(strlen(temp_in) + 1);
strcpy(reference[no_of_ref], temp_in);
no_of_ref+=1;

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