C程序计算字和行的标准输入 [英] C Program that counts Words and Lines in Standard input

问题描述

我是新来的C语言编程，我目前正试图教我如何建立一个C程序，可以计算字数和线条输入流中，并打印两总计到标准输出。

I am new to C programming and I am currently trying to teach myself how to create a C program that can count words and lines in the input stream and print the two totals to the standard output.

什么我其实想做的是有计划计算行数和计数取决于一个词的定义，我觉得我过的单词数。

What I am actually trying to do is to have the program count the number of lines and count the number of words depending on the definition of a word in which I feel that I am off.

我想的话，排除空格，制表符，换行符，连字符或冒号。虽然有作为小数的程序输出结果（文字和线条）。

I want the words to exclude blanks, tabs, newlines, hyphens, or colons. While having the program output the results (words and lines) as decimals.

#include<stdio.h>

int main()
{
    int iochar;
    int words;
    int lines;

    printf("Enter something here:\n\n");

    while ((iochar = getchar ()) !=EOF)
    {
        if((iochar == ' ') || (iochar == '\t') || (iochar == '\n'))

        putchar(iochar);
    }

    return 0;
}

我是完全关闭这个程序？

Am I totally off on this program?

推荐答案

如果你的问题是如何解决的编译错误，这很简单。在最后增加一个右括号。

If your question is how to fix the compile error, that's simple. Add one more closing brace at the end.

不过你的程序仍然会做只有一个一次通过循环，只打印一个字符当且仅当用户键入一个空格，制表符或换行符。无论什么类型的用户，该程序将终止运行。我怀疑这是你想要的。

But your program will still do only one pass through the loop and will print only one character if and only if the user types a space, tab or newline. No matter what the user types, the program will then terminate. I doubt that's what you wanted.

我怀疑这是你想要的结果：

I suspect this is what you intended:

while ((iochar = getchar ()) !=EOF)
{
    if((iochar == ' ') || (iochar == '\t') || (iochar == '\n'))
    {
        putchar(iochar);
    }
}
return 0;

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