如何为字符串动态分配内存空间并从用户那里获取该字符串? [英] How to dynamically allocate memory space for a string and get that string from user?

问题描述

我想使用 C 程序读取用户的输入.我不想使用数组，

I want to read input from user using C program. I don't want to use array like,

char names[50];

因为如果用户给出长度为10的字符串，那么剩下的空格就浪费了.

because if the user gives string of length 10, then the remaining spaces are wasted.

如果我使用字符指针，

char *names;

然后我需要以这种方式分配内存，

then I need to allocate memory for that in such a way of,

names = (char *)malloc(20 * sizeof(char));

在这种情况下，也有可能会浪费内存.

In this case also, there is a possibility of memory wastage.

所以，我需要的是为一个与字符串长度完全相同的字符串动态分配内存.

So, what I need is to dynamically allocate memory for a string which is of exactly same as the length of the string.

让我们假设，

如果用户输入是"stackoverflow"，那么分配的内存应该是14(即字符串的长度=13和1个额外的空间用于'').

If the user input is "stackoverflow", then the memory allocated should be of 14 (i.e. Length of the string = 13 and 1 additional space for '').

我怎样才能做到这一点?

How could I achieve this?

推荐答案

一次读取一个字符(使用getc(stdin))并增长字符串(realloc) 随你.

Read one character at a time (using getc(stdin)) and grow the string (realloc) as you go.

这是我前段时间写的一个函数.请注意，它仅用于文本输入.

Here's a function I wrote some time ago. Note it's intended only for text input.

char *getln()
{
    char *line = NULL, *tmp = NULL;
    size_t size = 0, index = 0;
    int ch = EOF;

    while (ch) {
        ch = getc(stdin);

        /* Check if we need to stop. */
        if (ch == EOF || ch == '
')
            ch = 0;

        /* Check if we need to expand. */
        if (size <= index) {
            size += CHUNK;
            tmp = realloc(line, size);
            if (!tmp) {
                free(line);
                line = NULL;
                break;
            }
            line = tmp;
        }

        /* Actually store the thing. */
        line[index++] = ch;
    }

    return line;
}

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