的getchar（）！= EOF [英] getchar() != EOF

问题描述

我是从C语言程序设计书上运行下面的程序：

I am running the following program from the C Programming Language book:

#include <stdio.h>
main()
{
  int c;
  while((c=getchar()) != EOF)
    putchar(); 
}

当我运行这个程序，我得到一个无法解释的行为。如果从按以下顺序在命令行我输入的字符： {'H'，'E'，'L'，'L'，'O'，'\\ n'，'^ D'} 然后我得到以下响应打印到屏幕上：你好，在 \\ n 输入，程序退出一旦 ^ D 中输入。

When I run this program, I get an unexplained behavior. If I input characters from the command line in the following sequence: {'h', 'e', 'l', 'l', 'o', '\n', '^D'} then I get the following response printed to screen: hello, after \n is input, and the program quits once ^D in entered.

然而，当我更改顺序如下： {'H'，'E'，'L'，'L'，'O'，'^ D'} 然后我得到打印到屏幕以下响应：你好，但程序不会退出。难道不应该退出，一旦我进入 ^ D ？我必须输入 ^ D 第二次为程序退出。或者我已经进入 ^ D 后以下 \\ n 程序只退出。我不明白为什么当我进入 ^ D 程序不会退出不管。有什么想法？

However, when I change the sequence as follows: {'h', 'e', 'l', 'l', 'o', '^D'} then I get the following response printed to screen: hello, but the program does not quit. Shouldn't it quit once I enter ^D? I have to enter ^D a second time for the program to quit. OR the program only quits after I have entered ^D following \n. I don't understand why the program doesn't quit no matter when I enter ^D. Any thoughts?

我在UNIX系统上运行。

I am running on a UNIX system.

推荐答案

当你输入^ D输入缓冲区清空，您键入到现在为止被发送到你的程序的一切（（'传递结束），而不实际发送^ D字符）。它类似于输入换行符，然而，在此情况下，换行字符本身被发送了。程序考虑其输入时，它读取零字符关闭。出现这种情况，当你键入新行后跟^ D或两个连续的^ D。

When you type ^D ('end-of-transmission') the input buffer is flushed and everything you typed until now is sent to your program (without actually sending ^D character). It is similar to typing newline character, however, in this case the newline character itself is sent too. A program considers its input as closed when it reads zero characters. This happens when you type newline followed by ^D or two consecutive ^D.

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