如何在没有Goto语句的情况下返回程序的开头

Question

#include <stdio.h>
#include <math.h>

long factcalc(int num1);

int main(void) 
{
    int num1;
    long factorial;
    int d;
    int out;

    printf("Please enter a number that is greater than 0");
    scanf_s("%d", &num1);

    if (num1 < 0) {
        printf("Error, number has to be greater than 0");
    } else if (num1 == 0) {
        printf("\nThe answer is 1");
    } else {
        factorial = factcalc(num1);
        printf("\nThe factorial of your number is\t %ld", factorial);
    }

    return 0;
}

long factcalc(int num1) 
{
    int factorial = 1;
    int c;

    for (c = 1; c <= num1; c++)
    factorial = factorial * c;

    return factorial;
}

我想知道,我怎么做到这样,程序一直要求用户输入,直到用户输入'-1'？因此,即使在计算了一个数字的阶乘之后,它仍然要求更多的数字,直到用户输入-1,同样适用于显示错误消息等的情况.提前致谢.

Answer 1

通过引入无限循环可以很容易地实现.

#include <stdio.h>
#include <math.h>

#ifndef _MSC_VER
#define scanf_s scanf
#endif

long factcalc(int num1);

int main(void)
{
    int num1;
    long factorial;
    int d;
    int out;

    for (;;) {
        printf("Please enter a number that is greater than 0");
        scanf_s("%d", &num1);
        if (num1 == -1) {

            break;
        }

        else if (num1 < 0) {

            printf("Error, number has to be greater than 0");
        }

        else if (num1 == 0) {

            printf("\nThe answer is 1");
        }

        else {

            factorial = factcalc(num1);
            printf("\nThe factorial of your number is\t %ld", factorial);
        }
    }

    return 0;
}

long factcalc(int num1) {

    int factorial = 1;
    int c;

    for (c = 1; c <= num1; c++)
        factorial = factorial * c;

    return factorial;
}