Question

程序要求用户输入enter函数中的分子和分母，然后需要简化然后显示它。我试过运行它，我的程序坏了。

有关如何执行此操作的任何提示？

我仍在努力学习如何做结构。

    #define _CRT_SECURE_NO_WARNINGS
    #include <stdio.h>

    struct Fraction
    {
        int numerator;
        int denominator;
    };

    void enter(struct Fraction *choice)
    {
        printf("Numerator: \n");
        scanf("%d", choice->numerator);

        printf("Denominator: \n");
        scanf("%d", choice->denominator);
    }

    void simplify(struct Fraction *reduce)
    {
        reduce->numerator = reduce->numerator / reduce->numerator;
        reduce->denominator = reduce->denominator / reduce->denominator;
    }

    void display(const struct Fraction *show)
    {
        printf("%d / %d", show->numerator, show->denominator);
    }

    int main(void)
    {
        struct Fraction f;

        printf("Fraction Simplifier\n");
        printf("===================\n");

        enter(&f);
        simplify(&f);
        display(&f);
    }

Answer 1

问题1

行

    scanf("%d", choice->numerator);
    scanf("%d", choice->denominator);

需要：

    scanf("%d", &choice->numerator);
    scanf("%d", &choice->denominator);
    //         ^^ Missing

问题2

以下几行：

    reduce->numerator = reduce->numerator / reduce->numerator;
    reduce->denominator = reduce->denominator / reduce->denominator;

相当于：

    reduce->numerator = 1.0;
    reduce->denominator = 1.0;

您需要使用代码来计算分子和分母的GCD，然后使用：

    double gcd = get_gcd(reduce->numerator, reduce->denominator);
    reduce->numerator = reduce->numerator/gcd;
    reduce->denominator = reduce->denominator/gcd;

结构新手，为什么我的程序没有运行？

1 个答案: