Question

C++中传递结构体的问题

程序如下：
#include <iostream>

struct box
{
char maker[40];
float height;
float width;
float length;
float volume;
};

void pass_by_value(box a, box b);

int main()
{
using namespace std;

box b1;

cout << "Please input the maker of the box: ";
cin >> b1.maker;
cout << "Please input the height of the box: ";
cin >> b1.height;
cout << "Please input the width of the box: ";
cin >> b1.width;
cout << "Please input the length of the box: ";
cin >> b1.length;
cout << "Please input the volume of the box: ";
cin >> b1.volume;

box b2 = {'o', 0, 0, 0, 0};

pass_by_value(b1, b2);

cout << b2.maker << endl;
cout << b2.height << endl;
cout << b2.width << endl;
cout << b2.length << endl;
cout << b2.volume << endl;

return 0;
}

void pass_by_value(box a, box b)
{
strcpy(b.maker, a.maker);
b.height = a.height;
b.width = a.width;
b.volume = a.volume;
}
但是最后输出的结果仍然是原来b2的值，而不是b1的值，请问这是为什么？

Answer 1

函数传进去是只是形参，不会改变原结构的
你可以用指针，修改后的代码如下：
#include <iostream>

struct box
{
char maker[40];
float height;
float width;
float length;
float volume;
};

void pass_by_value(box *a, box *b);　　// 参数改为结构指针

int main()
{
using namespace std;

box b1;

cout << "Please input the maker of the box: ";
cin >> b1.maker;
cout << "Please input the height of the box: ";
cin >> b1.height;
cout << "Please input the width of the box: ";
cin >> b1.width;
cout << "Please input the length of the box: ";
cin >> b1.length;
cout << "Please input the volume of the box: ";
cin >> b1.volume;

box b2 = {'o', 0, 0, 0, 0};

pass_by_value(&b1, &b2);　　　　　　// 调用函数时要用&取地址作为参数

cout << b2.maker << endl;
cout << b2.height << endl;
cout << b2.width << endl;
cout << b2.length << endl;
cout << b2.volume << endl;

return 0;
}

void pass_by_value(box *a, box *b)
{
strcpy(b->maker, a->maker);
b->height = a->height;
b->width = a->width;
b->volume = a->volume;
}

Answer 2

在pass_by_value里面的a和b是main函数里的一份拷贝，无论怎么变都不会影响main函数中原来的结构体的，这就是传值（pass by value）的含义。

追问

那应该怎么修改呢

追答

传递指针，修改指针指向的结构体，就能达到修改原结构体的效果了。zxypsxf 的回答就是这个意思。