填充动态分配的字符串数组? [英] populating dynamically allocated array of strings?

问题描述

我有以下代码来填充字符串数组，但是每次更改值时，整个数组都会更改(而不是单个字符串)

I have the following code to populate an array of strings, but every time i change the value, the entire array changes (instead of a single string the array)

推荐答案

这是因为对于数组的所有位置，您都是指向char的同一指针.

That's because you are the same pointer to char for all positions of the array.

当您这样做:

words[i] = txt;

您正在分配一个指针.因此，每个word[i]都是相同的字符串(txt). 如果您真的想将单词读入缓冲区(如txt)，然后将其放入字符串数组，则需要将缓冲区字符串的内容复制到数组中的字符串，如下所示:

You are assigning a pointer. So every single word[i] is the same string (txt). If you really want to read the word into a buffer (like txt) and then put it into the array of strings, you need to copy the contents of the buffer string to the string in the array, like so:

strncpy(words[i], txt, MAX_WORD_LENGTH);

您的代码还有另一个问题，那就是字符串数组的分配. 应该是:

There's also another problem with your code, which is the allocation of the string array. It should be:

words = (char**)malloc(wordcount * sizeof(char*));

这是因为字符串数组是指向char指针(char**)的指针，并且数组的每个元素都是字符串(char*).现在您已经分配了一个char指针数组，但是还没有为每个字符串分配内存，这是我们接下来要做的:

That is because a string array is a pointer to a char pointer (char**), and each element of the array is a string (char*). Now you have allocated a array of char pointers, but you have not allocated the memory for each string, which is what we do next:

for (i = 0; i < wordcount; i++) {
    words[i] = (char*)malloc(MAX_WORD_LENGTH * sizeof(char));
}

如果您不想使用缓冲区而直接读入字符串数组，则代码将是这样的:

If you want to not use a buffer and read directly into the string array, your code would be something like this:

words = (char**)malloc(wordcount * sizeof(char*));
input = fopen(filename, "r");
while(!feof(input)) {
    words[i] = (char*)malloc(MAX_WORD_LENGTH * sizeof(char));
    fscanf(input, "%s", words[i]);
}

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