c - C语言中如何输入新行直到遇到空行

Question

这是我的代码，我正在尝试创建一个程序，使用函数对字符进行计数，然后在遇到空行时确定字符的平均值。该程序应该允许用户输入多行，直到遇到空行，但我似乎不能。

#include <stdio.h>
#include <Windows.h>

int main()
{
    char str[1000];
    int Digits, Char, SpecialChar, linecount = 0;
    int counter;
    int total;
    int average;

    Digits = Char = SpecialChar = 0;

    printf("Please type in your words here: ");
    gets(str);

    for (counter = 0; str[counter] != NULL; counter++)
    {

        if (str[counter] >= '0' && str[counter] <= '9')
            Digits++;
        else if ((str[counter] >= 'A' && str[counter] <= 'Z') || (str[counter] >= 'a' && str[counter] <= 'z'))
            Char++;
        else
            SpecialChar++;    
    }

    while (str[counter] != '\n')
    {
        if (str[counter] = '\n')
        {
            linecount ++;
        }
    }

    total = Digits + Char + SpecialChar;
    average = total / linecount;

    printf("\nDigits: %d \nCharacters: %d \nSpecial Characters: %d \nLine: %d", Digits, Char, SpecialChar, linecount);
    printf("\nTotal no. of characters = %d", total);
    printf("\nAverage no. of characters = %d", average);

    Sleep(5000000);
    return 0;
}

Answer 1

据我所知，函数gets在“\n”之后被中断。另外，使用 fgets 时，您必须注意字符串上的“\0”添加。这意味着

The programme is suppose to allow the user to enter multiple lines until an empty line is encountered but I can't seem to.

使用这种方式永远无法完成。因为 gets 不是推荐的函数，所以我以您可能正在搜索的方式稍微编辑了您的代码。

需要提及的是，在您阅读此代码之前我发现这可能是一个逻辑错误

for (counter = 0; str[counter] != NULL; counter++)

这看起来很奇怪，因为 fgets 总是会记录“\n”字符。所以，下一个条件

if (str[counter] = '\n')

永远不会是真的

我在您的代码中看到了一些其他错误，但不是主要错误。因此，我认为该建议足以任命他们

while (fgets(str, 1000, stdin) && str[0] != '\n'){ //I dont know if checking the first element of the string is redundancy, 
//because, the I think the fgets function will return NULL if you just press enter, as the first character

    for (counter = 0; str[counter] != '\n'; counter++{

        if (str[counter] >= '0' && str[counter] <= '9')
            Digits++;
        else if ((str[counter] >= 'A' && str[counter] <= 'Z') || (str[counter] >= 'a' && str[counter] <= 'z'))
            Char++;
        else
            SpecialChar++;    
    }

    linecount ++;  //new line is someting that you will always reach, so, 
                  //there is no reason for any condition

}