我想用C ++转换这段代码 [英] I want to convert this code in C++

问题描述

我想用c ++转换这段代码

i want to convert this code in c++

void makeMove(int *x, int *y)
{
    // Take the input move
    printf("Enter your move, (row, column) -> ");
    scanf("%d %d", x, y);
    return;
}

我的尝试：

What I have tried:

void makeMove(int *x, int *y)
{
 // Take the input move
 cout << "Enter your move, (row, column) -> ";
 cin >>x >>y;
 return;
}

推荐答案

我认为它没有优势，但是你去了：



这应该有效：

I see no advantage in it, but here you go:

This should work:

cin >> *x >> *y;//de ref the pointers

最好使用说出名字，并指出一个前缀p

It is better to work with "speaking names" and give pointers a prefix "p"

void makeMove(int *px, int *py)

在更复杂的场景中你会理解的更好。

In more complex scenarios you will understand that better.

使用 C ++ 你可以使用引用：

With C++ you might use references:

void makeMove(int & x, int & y)
{
 // Take the input move
 cout << "Enter your move, (row, column) -> ";
 cin >> x >> y;
}

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