我的C程序错误地计算了字符 [英] My C program counts characters incorrectly

问题描述

大家好，所以我写了一些代码，该代码应该计算用户在控制台中键入的字符，使用getchar（）和while循环，直到键入EOF字符为止，但这会增加 count 应该的变量。例如，我输入3个字符，然后输入EOF字符（在本例中为'z'），最后输出我输入了6个字符，如果我输入4个字符+'z'，则表示8，如果5则表示10

Hy everyone, so I wrote some code that should count characters that are typed in console by user, using getchar() and a while loop until EOF character is typed, but it adds more to the count variable that it should. For example, I enter 3 characters, and then EOF character(in this case 'z') and at the end It outputs that I entered 6 characters, if I enter 4 chars + 'z' it says 8, if 5 it says 10. It displays x2 number of charaters it should.

#include <stdio.h>
#define END 'z'

int main()
{
    printf("Hello:\n");
    int count = 0;
    int c;

    while ((c = getchar()) != END)
    {
        count++;
    }

    printf("You entered %d charaters.", count);
}

为什么会这样？ ：/

Why is that so? :/

推荐答案

每次使用getchar（）输入字符，然后按 enter，再输入一个char这是换行符。

Every time you enter a character with getchar() and after that press "enter", you enter one more char which is a newline character.

while ((c = getchar()) != EOF)
{
    if (c=='\n')
        continue;
    count++;
}

这将解决您的问题。

我已经对您和我的代码进行了一些测试，只是看是否是问题所在。输出在这里：

I have done some tests with your and my code, just to see if that was the problem. The output is here:

使用您的代码输出：

Hello:
a
s
d
df
You entered 9 charaters.

Hello:
asdf

You entered 5 charaters.

输出带有我的代码：

Hello:
a
s
d
f
You entered 4 charaters

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