Question

好的，所以我在这里遇到了一个小问题。不要介意他们为我提供的代码中的注释。

我的问题在于功能。我希望它进行测试，以确保再次[0] = y或n。如果它不循环，直到我输入正确的数字。

现在它做了什么：它无休止地循环，无论我投入什么。

我确实错过了一些东西，我确信它确实如此。

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

//functions called
float wages_loop();

//start main 
int main(void)
{
    char status[10], another[10];
    char buffer[80];

    float wages, other_income, interest, dividends, test;
    int dependents;
    int single_acc = 0, mj_acc = 0, ms_acc = 0, sh_acc = 0;

    printf("Would you like to start: ");
    gets_s(another);

    while (another[0] = 'y')
    {
        //printf("What is your Status: ");
        //gets_s(status);


        wages = wages_loop();



        //printf("\n How much in Other Income. ");
        //gets_s(buffer);
        //other_income = atof(buffer);

        //printf("\n How much in interest. ");
        //gets_s(buffer);
        //interest = atof(buffer);

        //printf("\n How much in Dividends. ");
        //gets_s(buffer);
        //dividends = atof(buffer);

        //printf("\n How many Dependents. ");
        //gets_s(buffer);
        //dependents = atoi(buffer);

        printf("\n\n\t\t Your wage is: %.2f \n", wages);
        system("pause");
    } //end loop

    printf("\n\n\t\t\t Number of Singles filleing: %i \n", single_acc);

    return 0;

}//end main


float wages_loop()
{
    char again[10];
    char buffer[80];
    float wages, total_wages = 0;

    printf("\n How much in Wages. ");
    gets_s(buffer);
    wages = atof(buffer);

    total_wages = wages + total_wages;

    printf("\n Do you have any more wages. (y or n)");
    gets_s(again);

    if (again[0] != 'y' || 'n')
    {
        while (again[0] != 'y' || 'n')
        {
            printf("\n\n INCORRCT ANSWER. \n\n");
            printf("\n Do you have any more wages. (y or n)");
            gets_s(again);
        }
    }

    while (again[0] = 'y')
    {
        printf("\n Enter Wages: ");
        gets_s(buffer);
        wages = atof(buffer);

        total_wages = wages + total_wages;

        printf("\n Do you have any more wages. ");
        gets_s(again);
    }

    return total_wages;
}

Answer 1

while (another[0] = 'y')

这是作业，而不是平等。将=更改为==

编辑：你在这里再次做到了：

while (again[0] = 'y')

此外：

(again[0] != 'y' || 'n')

应该是

(again[0] != 'y' && again[0] != 'n')

因为'n'本身将始终返回true

（感谢Jonathan Henson）

EDIT2：正如Sam在评论中指出的那样，你没有在循环中设置another[0]，所以即使你将运算符更改为{{{}也是如此1}}，添加一个语句再次获取用户输入。

EDIT3：当log0指出时，您可以通过调高编译器的警告级别来避免此问题。

无尽的循环：为什么这个无穷无尽

1 个答案: