c - 在 C 中从函数返回 scanf 值

Question

我正在尝试在我的计算器程序中编写一个子函数来验证操作数。我希望函数执行的操作是将 scanf 输入返回到 main() 函数。这是我用于子函数的代码:

void checkOperand(float f1, float f2)
{
    printf("Please enter two numbers to add, separated by a space: \n\n");
    while( (scanf("%f %f",&f1,&f2)) !=2 ) 
        {
            // you will enter this loop when there is wrong input from the user and there may be garbage characters inputted.
            // eat up each character until buffer is clear indicated by new line
            while(getchar() != '\n')
            {
                continue;
            }
            printf("\nError reading your input. Please try again: \n\n");
        }

}       

然后，在 main() 函数中，这里是我尝试调用它的方式:

int main(int argc, const char * argv[])
{
    int nChoice = 6; // initialize with an invalid choice
    float fNums1;
    float fNums2;
    float result;

    printf("Welcome to My Handy Calculator:\n\n\t1. Addition\n\t2. Subtraction\n\t3. Division\n\t4. Multiplication\n\t5. Exit\n\n");

    //scanf returns number of successful translations. If user inputs characters instead of numbers, it will not return 1 as you are
    //scanning one value. This is the way to trap the wrong user input.

     nChoice = checkMenuOption(nChoice);

    switch ((nChoice))
    {
        case 1:
            checkOperand(fNums1,fNums2);
            result = fNums1 + fNums2;
            printf("\n\nResult of adding %5.2f and %5.2f is %5.2f \n\n",fNums1,fNums2,result);
            break;
    }
}

但是 scanf 函数的值(f1 和 f2)不会返回到主函数，当执行计算时，它说:

"Result of adding 0 and 0 is 0."

Answer 1

您的参数是按值传递的。更改您的参数以通过指针传递:

void checkOperand(float *f1, float *f2)

和

while( (scanf("%f %f",f1,f2)) !=2 ) 

在你的调用代码中，

        checkOperand(&fNums1,&fNums2);