C ++数组作为函数参数 [英] C++ arrays as function arguments

问题描述

• 我可以通过阵列功能，就像我将与原语，如int和布尔办？


• 我可以按值传递呢？


• 如何函数知道数组的大小是通过呢？

解决方案

  

我可以通过阵列功能一样
  我将与原语如int做
  和布尔？

可以，但只能使用指针（即：由参考）

。

  

我可以按值传递呢？

没有。您可以创建支持类，但普通阵列没有。

  

如何功能知道它是通过在阵列的大小？

它没有。这是一个理由去使用的东西，像矢量＆lt; T＆GT; 而不是 T *

澄清

一个函数可以接受引用或指向特定大小的数组：

 无效FUNC（CHAR（* P）[13]）
{
    为（中间体n = 0时; Nδ13 ++ n）的
        的printf（％C，（* p）的[N]）;
}诠释的main（）
{
    所以char a [13] =你好，世界;
    FUNC（＆放大器;一个）;    焦B〔5] =糟糕;
    //下一行不会编译
    // FUNC（和b）;    返回0;
}
 

我是pretty肯定这不是什么OP不过是寻找。

• Can I pass arrays to functions just as I would do with primitives such as int and bool?
• Can I pass them by value?
• How does the function know of the size of the array it is passed?

解决方案

Can I pass arrays to functions just as I would do with primitives such as int and bool?

Yes, but only using pointers (that is: by reference).

Can I pass them by value?

No. You can create classes that support that, but plain arrays don't.

How does the function know of the size of the array it is passed?

It doesn't. That's a reason to use things like vector<T> instead of T *.

Clarification

A function can take a reference or pointer to an array of a specific size:

void func(char (*p)[13])
{
    for (int n = 0; n < 13; ++n)
        printf("%c", (*p)[n]);
}

int main()
{
    char a[13] = "hello, world";
    func(&a);

    char b[5] = "oops";
    // next line won't compile
    // func(&b);

    return 0;
}

I'm pretty sure this is not what the OP was looking for, however.

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