在 C 中获取用户输入的最简单方法是什么? [英] What is the simplest way of getting user input in C?

问题描述

在 C 中似乎有很多方法可以获取用户输入.

There seem to be a LOT of ways you can get user input in C.

需要很少代码的最简单方法是什么?

What is the easiest way that requires little code?

基本上我需要显示这个:

Basically I need to display this:

Enter a file name: apple.text

基本上我需要向用户询问文件名.所以我需要一些东西来获取用户将输入的那个词.

Basically I need to ask the user for a file name. So I need something that just gets that one word that the user will be inputting.

推荐答案

最简单的正确"方法可能就是这个，摘自 Bjarne Stroustrup 的论文 将标准 C++ 作为一种新语言学习.

The simplest "correct" way is probably this one, taken from Bjarne Stroustrup's paper Learning Standard C++ As A New Language.

(注意:我更改了 Bjarne 的代码以检查 isspace() 而不是行尾.另外，由于@matejkramny 的评论，使用 while(1) 而不是 while(true) ......只要我们足够异端来编辑 Stroustrup 的代码，我也会用 C89 注释代替 C++ 风格.:-P)

(Note: I changed Bjarne's code to check for isspace() instead of just end of line. Also, due to @matejkramny's comment, to use while(1) instead of while(true)...and so long as we're being heretical enough to edit Stroustrup's code, I've subbed in C89 comments instead of C++ style too. :-P)

#include <stdio.h>
#include <ctype.h>
#include <stdlib.h>

void quit() /* write error message and quit */
{
    fprintf(stderr, "memory exhausted
");
    exit(1);
}

int main()
{
    int max = 20;
    char* name = (char*) malloc(max); /* allocate buffer */
    if (name == 0) quit();

    printf("Enter a file name: ");

    while (1) { /* skip leading whitespace */
        int c = getchar();
        if (c == EOF) break; /* end of file */
        if (!isspace(c)) {
             ungetc(c, stdin);
             break;
        }
    }

    int i = 0;
    while (1) {
        int c = getchar();
        if (isspace(c) || c == EOF) { /* at end, add terminating zero */
            name[i] = 0;
            break;
        }
        name[i] = c;
        if (i == max - 1) { /* buffer full */
            max += max;
            name = (char*) realloc(name, max); /* get a new and larger buffer */
            if (name == 0) quit();
        }
        i++;
    }

    printf("The filename is %s
", name);
    free(filename); /* release memory */
    return 0;
}

涵盖:

• 跳过空格直到到达字符输入处
• 动态扩展字符串缓冲区以适应任意大小的字符串
• 无法分配内存的处理情况

是否有更简单但破碎的解决方案，甚至可能运行得更快一点?绝对！！

Are there simpler but broken solutions, which might even run a bit faster? Absolutely!!

如果你使用 scanf 进入一个没有读取大小限制的缓冲区，那么你的输入超过了缓冲区的大小，它会产生安全漏洞和/或崩溃.

If you use scanf into a buffer with no limit on the read size, then your input exceeds the size of the buffer, it will create a security hole and/or crash.

将读取的大小限制为例如文件名的 100 个唯一字符似乎比崩溃更好.但情况可能更糟；例如，如果用户的意思是 (...)/dir/foo/bar.txt 但你最终误解了他们的输入并覆盖了一个名为 bar.t 的文件他们关心.

Limiting the size of the reading to, say, only 100 unique characters of a filename might seem better than crashing. But it can be worse; for instance if the user meant (...)/dir/foo/bar.txt but you end up misinterpreting their input and overwriting a file called bar.t which perhaps they cared about.

最好在处理这些问题时尽早养成良好的习惯.我的观点是，如果您的需求证明某些接近金属和类 C"的东西是合理的，那么考虑跳到 C++ 是非常值得的.它旨在精确地管理这些问题 - 使用强大且可扩展但仍然表现良好的技术.

It's best to get into good habits early in dealing with these issues. My opinion is that if your requirements justify something close-to-the-metal and "C-like", it's well worth it to consider the jump to C++. It was designed to manage precisely these concerns--with techniques that are robust and extensible, yet still perform well.

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