如何修复此 do-while 循环,以便控制台程序在自动关闭之前提示用户?

How do I fix this do-while loop so console program prompts the user before automatically closing?

当我 运行 编写此代码时,一切 运行 都很顺利,直到最后一部分。在询问 "Would you like to repeat?" 问题后,控制台不会提示用户回答,而是结束编程。

如何编辑 do-while 循环的代码以便提示用户回答而不是自动关闭程序?我觉得这是格式说明符的问题,我对此很陌生并且一直遇到问题。谢谢!

#include <stdio.h>

int main(void)
{
    double num1, num2;
    char operation, repeat = "y";
    printf("This is a calculator.");

    do {
        printf("\nWould you like to multiply(*), divide(/), add(+) or subtract(-) the two numbers you will soon input? \n");
        scanf("%c", &operation);
        printf("Please enter the first number you would like to deal with. \n");
        scanf("%lf", &num1);
        printf("And the second?\n");
        scanf("%lf", &num2);

        switch (operation)
        {
        case '*':
            printf("The product of %1.2lf and %1.2lf is %1.2lf.\n",
                   num1, num2, num1 * num2);
            break;
        case '/':
            printf("The quotient of %1.2lf and %1.2lf is %1.2lf.\n",
                   num1, num2, num1 / num2);
            break;
        case '+':
            printf("The sum of %1.2lf and %1.2lf is %1.2lf.\n",
                   num1, num2, num1 + num2);
            break;
        case '-':
            printf("The difference of %1.2lf and %1.2lf is %1.2lf.\n",
                   num1, num2, num1 - num2);
            break;
        }
        printf("Would you like to repeat?(y/n)\n");
        scanf("%c", &repeat);
    } while (repeat == "y" || repeat == "Y");
}

stdin中有上次输入操作留下的换行符。你的

scanf("%c",&repeat);

读取该换行符,因为转换说明符 %c 不会跳过空白字符。使用

scanf(" %c", &repeat);

跳过前导空格。


在 C 和 C++ 中,单个字符用单引号引起来。

char ch;
ch == "A";

会将 ch 的值与字符串文字 "A".

的地址进行比较

所以...

while(repeat=="y"||repeat=="Y");

~>

while(repeat == 'y' || repeat == 'Y');

char operation, repeat="y";

~>

char operation, repeat = 'y';

你的编译器应该已经警告过你了。如果不是,您应该提高编译器的警告级别。


您可能还想检查未定义的被零除。


最后一件事:printf() 不关心 %lf 中的长度说明符 l,因为默认参数传播,它与 %f 相同。调用带有可变数量参数的函数中的 float 参数在传递给函数之前总是转换为 double。所以 printf().

只有 %f

PS: 正如 Cacahuete Frito 在评论中所说:

You should check the return value of scanf()

是的,你应该。永远不要相信用户。