分配字符指针数组 [英] Allocating a char pointer array

问题描述

我有一些文件，我正在读取的内容到一个字符数组，我想把从每个文件读取的内容存储在一个字符数组中，这样我可以访问单独的字符数组内容）按索引，我不知道如何做的语法。

I've got a number of files that I am reading the contents into a char array, I would like to store the contents I read from each file in an array of characters so that I could access the separate character arrays (file contents) by index, I am not sure of the syntax on how to do this.

我的理解是，我需要将我模式数组中的每个内存地址设置为每个char数组一个内存地址;到存储在patternData中的字符数组，但我不知道如何设置这个。

My understanding is that I need to set each of the memory addresses in my patterns array to one memory address per char array; to the char array stored in patternData but I do not know how to set this. What is the syntax that I am missing that I need to get this to happen?

我尝试过的

我会认为如果我存储的类型 char * 那么我需要一个类型 char ** 存储单独的字符数组数组。

I would have thought that If I was storing a type of char* then I would need a type char** to store the separate arrays of char arrays.

我将使用以下符号访问 char * 将模式索引的内存地址设置为

I would access a char* by using the following notation to set the memory address of the pattern index to

&patterns[INDEX] = &pData;

但这不起作用。有很多字符指针数组问题，但我不确定正确的方法做一个简单的赋值 pData 到模式的索引。

However this does not work. There is a plethora of "char pointer array" questions but I'm not sure of the correct way to do this simple assignment of pData to an index of patterns.

char *tData;
int tLength;
char *pData;
int pLength;

char **patterns;

void ReadFromFile(FILE *f, char **data, int *length) //This is what is passed into function


    int main(int argc, char **argv)
    {
        for (int i = 1; i <= 5; i++)
        {
            FILE *f;
            char fileName[1000];
            sprintf(fileName, "file%d.txt", i);
            f = fopen(fileName, "r");
            if (f == NULL)
                return 0;

            ReadFromFile(f, &pData, &pLength); //works to here, contents of file assigned to pData
            fclose(f);

            &patterns[i - 1] = &pData;

        }

        return 0;

    }

推荐答案

不正确：

&patterns[i - 1] = &pData;

您尝试分配给接受地址运算符的结果，这是不可能的因为它不是一个左值。

You are trying to assign to the result of the "take an address" operator, which is not possible because it's not an lvalue.

作业应该如下：

patterns[i - 1] = pData;

但您需要在您之前分配模式做这个。您可以使用

but you need to allocate patterns before you do this. You can use

patterns = malloc(sizeof(char*)*5);

或简单地声明 patterns 五个：

char *patterns[5];

这假设您的 ReadFromFile 函数 char * ，并将其分配给 pData 指向的地址。注意，你需要 free 所有通过 malloc / calloc / realloc 。

This assumes that your ReadFromFile function allocates a char*, and assigns it to the address pointed to by pData. Note that you need to free all pointers that were obtained through malloc/calloc/realloc.

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