错误:请求成员在某个非结构或联合的地方[复制]

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Compile error: with request not something structure or union error 2 answers
编译错误:请求时不需要任何结构或错误2的答案。

I'm having trouble with my code. My program is a program to simplify fractions. So my problem is this:

我的代码有问题。我的程序是一个简化分数的程序。所以我的问题是:

I declare the structure Fraction. And then I declare structure Fraction f in my main function.

我声明结构分数。然后我在主函数中声明结构f。

But when I try to use any member of the structure fraction (i.e. f.num, f.den) it says it's not a member of a structure or union. I have like 10 errors all saying the same thing for my program.

但是当我尝试使用结构分数的任何成员时(即f。它说它不是一个结构或联盟的成员。我有10个错误，对我的程序说的都是一样的。

The error (verbatim): error: request for member "num" in something not a structure of union

错误(一字不差):错误:请求成员“num”在某种非联合结构中。

#include <stio.h>

struct Fraction{
    int num;
    int den;
    int lownum;//lownum = lowest numerator.
    int lowden;//lowden = lowest denominator
    int error;
};

void enter(struct Fraction *f);
void simplify(struct Fraction *f);
void display(const struct Fraction *f);


int main(void){
    struct Fraction f;

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

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

void enter(struct Fraction *f){
    printf("Please enter numerator : \n");
    scanf("%d", &f.num);

    printf("please enter denominator : \n");
    scanf("%d", &f.den);

    printf("%d %d", f.num, f.den);
}

void simplify(struct Fraction *f){
    int a;
    int b;
    int c;
    int negative; //is fraction positive?

    a = f.num;
    b = f.den;

    if (a/b < 0){
        negative = 1;
    }

    if(b == 0){
        f.error = 1;
    }

    if(a < 0){
        a = a * -1;
    }

    if(b < 0){
        b = b * -1;
    }

    //euclids method
    if(a < b){
        c = a;
        a = b;
        b = c;
    }
    while(b != 0){
        c = a % b;
        a = b;
        b = c;
    }

    f.lownum = f.num / a;
    f.lowden = f.den / a;

    if(negative = 1){
        f.lownum = f.lownum * -1;
    }
}

void display (const struct Fraction *f){
    if (f.error != 1){
        printf("%d / %d", f.lownum, f.lowden);
    }else{
        printf("error");    
    }
}

2 个解决方案

#1

在

void simplify(struct Fraction *f)

f is a pointer to struct Fraction. Therefore, instead of

f是struct分数的指针。因此,而不是

a = f.num;

you have to write

你必须写

a = (*f).num;

which can be shortened to the equivalent notation:

可以简化为等价符号:

a = f->num;

The same applies to all other references to struct Fraction *f in your functions.

这同样适用于函数中struct分数*f的所有其他引用。

#2

You use a pointer (struct Fraction *f). So you have to acces members with the -> operator:

使用一个指针(struct分数*f)。所以你必须加入->操作人员:

f->num

#1

在

void simplify(struct Fraction *f)