用C数组的元素数量++ | 用C数组的元素数量

本文介绍了用C数组的元素数量++的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

比方说，我有一个数组改编。当以下不给数组元素的数目：的sizeof（ARR）/ sizeof的（ARR [0]）

我只能1箱子件事：数组包含不同派生类型的数组类型的元素

我的权利，并在那里（我几乎可以肯定存在的必须的定）等这样的情况？

对不起，琐碎的问题，我是一个Java开发我比较新的C ++。

谢谢！

解决方案

One thing I've often seen new programmers doing this:

void f(Sample *arr)
{
   int count = sizeof(arr)/sizeof(arr[0]); //what would be count? 10?
}

Sample arr[10];
f(arr);

So new programmers think the value of count will be 10. But that's wrong.

Even this is wrong:

void g(Sample arr[]) //even more deceptive form!
{
   int count = sizeof(arr)/sizeof(arr[0]); //count would not be 10
}

It's all because once you pass an array to any of these functions, it becomes pointer type, and so sizeof(arr) would give the size of pointer, not array!

EDIT:

The following is an elegant way you can pass an array to a function, without letting it to decay into pointer type:

template<size_t N>
void h(Sample (&arr)[N])
{
    size_t count = N; //N is 10, so would be count!
    //you can even do this now:
    //size_t count = sizeof(arr)/sizeof(arr[0]);  it'll return 10!
}
Sample arr[10];
h(arr); //pass : same as before!

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