Question

#include <iostream>

using namespace std;

float distance(tocki *A, tocki *B);

int main()
{
    struct tocki{
        int x, y;
    };

    tocki A, B, C;

    cout << "x = ";
    cin >> A.x;
    cout << "y = ";
    cin >> A.y;

    cout << "x = ";
    cin >> B.x;
    cout << "y = ";
    cin >> B.y;

    cout << "x = ";
    cin >> C.x;
    cout << "y = ";
    cin >> C.y;

    cout << distance(&A, &B);

    return 0;
}

//distance between (x1,y1) i (x2,y2) e  d = sqrt((x2-x1)^2 - (y2-y1)^2);

float distance(tocki *A, tocki *B){
    return sqrt(pow(A.y - A.x, 2) - pow(B.y - B.x, 2));
}

我得到的错误是：

'tocki' was not declared in this scope
'A' was not declared in this scope
'tocki' was not declared in this scope
'B' was not declared in this scope

在这一行：

float distance(tocki *A, tocki *B);

那么，我到底做错了什么？我想传递一个struct to function并在main（）程序中获取该函数的结果。

Answer 1

将toki结构置于主函数之外

Answer 2

struct tocki应该在声明函数distance之前声明，因此编译器知道在检查参数类型时结构是否存在。此外，您应该使用A-> y ecc。，因为你已经传递了一个指向结构的指针。

在C ++中将struct传递给函数

2 个答案: