创建在C简单的计算器 [英] Creating a simple calculator in C

问题描述

我试图写一个简单的C计算器脚本，只使用基本的+， - ，/，*。我有以下，但我不知道为什么它不能正常打印。

 ＃包括LT＆;＆stdio.h中GT;
＃包括LT＆;＆stdlib.h中GT;INT主要（无效）
{    //介绍瓦尔
    双数字1，数字，结果;
    字符符号; //操作*， - ，+ /    //允许用户交互
    的printf（请输入您的公式\\ n）;
    scanf函数（％F％C％F，＆安培;数字1，和放大器;符号，＆安培;数字2）;    开关（符号）{
        案例'+'：
            结果=数字1 +数字2;
            打破;
        默认：
            的printf（别的事我不知道）;
            打破;
    }    的getchar（）;
    返回0;
}
 

为什么结果不被打印？我在这里做得不对，

 结果=数字1 +数字2;
 

解决方案

  

为什么不被打印结果呢？

您正确地算出答案，的但不随地打印的

您需要有这样的：

 的printf（答：％F +％F =％F \\ N，数字1，数字，结果）;
 

如果没有打印语句，没有被打印出来。的

---

修改回应评论：

你做的printf 在您计算的结果？
就个人而言，我会放的printf就在之前的getchar（）;

有关更多的调试，只是你以后scanf函数，我会写：

 的printf（输入所接收：数字1是％F \\ n编号2为％F \\ nsymbol是％C \\ N，数字1，数字，符号）;
 

如果没有显示您所键入的输入，那么什么是错的与您如何收集输入。

I am trying to write a simple C calculator script, using only the basic +, -, /, *. I have the following, but I'm not sure why it's not printing correctly.

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

int main (void)
{

    //introduce vars
    double number1, number2, result;
    char symbol; //the operator *, -, +, /

    //allow user interaction
    printf("Enter your formula \n");
    scanf("%f %c %f", &number1, &symbol, &number2);

    switch (symbol) {
        case '+':
            result = number1 + number2;
            break;
        default:
            printf("something else happened i am not aware of");
            break;
    }

    getchar();
    return 0;
}

Why is the result not being printed? Am I doing something wrong here,

result = number1 + number2;

解决方案

"Why is the result not being printed?"

You calculate the answer properly, but do not print it anywhere.

You need to have something like:

printf("Answer: %f + %f = %f\n", number1, number2, result);

Without a print statement, nothing gets printed.

---

EDIT Responding to comment:

Did you do the printf after you calculate the result? Personally, I would put the printf just before the getchar();

For more debugging, just after your scanf, I would write:

printf("Input as received: number1 is %f\n number2 is %f\nsymbol is %c\n", number1, number2, symbol);

If that does not show the input that you typed, then something is wrong with how you gather input.

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