我们可以使用sizeof运算符找到数组的大小, 如下所示。
//Finds size of arr[] and stores in 'size'
int size = sizeof(arr)/sizeof(arr[0]);
我们可以不使用sizeof运算符来做同样的事情吗?
方法1(写我们自己的sizeof)
给定一个数组(你不知道数组中元素的类型), 找到数组中元素的总数而不使用sizeof运算符?
一种解决方案是编写我们自己的sizeof运算符(请参见这个有关详细信息)
//C++ program to find size of an array by writing our
//sizeof
#include <bits/stdc++.h>
using namespace std;
//User defined sizeof macro
# define my_sizeof(type) ((char *)(&type+1)-(char*)(&type))
int main()
{
int arr[] = {1, 2, 3, 4, 5, 6};
int size = my_sizeof(arr)/my_sizeof(arr[0]);
cout <<"Number of elements in arr[] is "
<<size;
return 0;
}
输出:
Number of elements in arr[] is 6
方法2(使用指针hack)
与上述解决方案相比, 以下解决方案非常简短。数组A中的元素数可以使用表达式找到
int size = *(&arr + 1) - arr;
//C++ program to find size of an array by using a
//pointer hack.
#include <bits/stdc++.h>
using namespace std;
int main()
{
int arr[] = {1, 2, 3, 4, 5, 6};
int size = *(&arr + 1) - arr;
cout <<"Number of elements in arr[] is "
<<size;
return 0;
}
输出:
Number of elements in arr[] is 6
这是如何运作的?
指针算法在这里发挥了作用。我们不需要将每个位置明确转换为字符指针。
&arr ==> Pointer to an array of 6 elements.
[See this for difference between &arr
and arr]
(&arr + 1) ==> Address of 6 integers ahead as
pointer type is pointer to array
of 6 integers.
*(&arr + 1) ==> Same address as (&arr + 1), but
type of pointer is "int *".
*(&arr + 1) - arr ==> Since *(&arr + 1) points
to the address 6 integers
ahead of arr, the difference
between two is 6.
如果发现任何不正确的地方, 或者想分享有关上述主题的更多信息, 请发表评论。
来源:
https://www.srcmini02.com/68583.html