指针和多维数组 [英] pointers and multi-dimensional arrays
问题描述
可能重复：
如何在C ++中使用数组？
是2d数组是双指针吗？
二维数组和指针
我知道这是一个非常基本的问题，但没有数量的谷歌清除这对我。这就是为什么在这里发布。
在c ++中考虑声明 int x [10];
I know that this is a very basic question but no amount of googling cleared this for me. That's why am posting it here.
In c++ consider the declaration int x[10];
这是一个1维数组其中x是基本指针，它包含数组的第一个元素的地址。所以 x 给我的地址和 * x 给出第一个元素。
This is a 1-dimensional array with x being the base pointer that is it contains the address of the first element of the array. So x gives me that address and *x gives the first element.
类似的声明
int x[10][20];
这里是什么样的变量 x 当我做
what kind of variable is x here. When i do
int **z = x;
编译器说它不能转换 int（*）[ code>到 int ** .And为什么 cout cout<< * x; 给出相同的值
如果我声明一个指针数组为
the compiler says it cannot convert int (*)[20] to int **.And why does cout<<x; and cout<<*x; give the same value??
And also if i declare an array of pointers as
int *p[10];
那么 x p （在他们的类型）？因为当一个声明 int x [10] 和 int * p > x 到 p 但是在二维数组的情况下不是这样？为什么？
有人可以为我清除这个或者提供一个很好的资源材料。
then is there a difference between x and p ( in their types) ?? because when one declares int x[10] and int *p then it is valid to assign x to p but it is not so in case of two dimensional arrays? why?
Could someone please clear this for me or else provide a good resource material on this.
推荐答案
数组和指针同样的事情。在C和C ++中，多维数组只是数组数组，没有涉及指针。
Arrays and pointers aren't the same thing. In C and C++, multidimensional arrays are just "arrays of arrays", no pointers involved.
int x[10][20];
是一个10个数组的数组，每个数组包含20个元素。如果在上下文中使用 x ，它将衰减为指向其第一个元素的指针，那么最终将获得一个指向这些20个元素数组之一的指针 - int（*）[20] 。请注意，这样的事情不是指向指针的指针，因此无法进行转换。
Is an array of 10 arrays of 20 elements each. If you use x in a context where it will decay into a pointer to its first element, then you end up with a pointer to one of those 20-element arrays - that's your int (*)[20]. Note that such a thing is not a pointer-to-a-pointer, so the conversion is impossible.
int *p[10];
是一个10指针的数组，所以是不同于x。
is an array of 10 pointers, so yes it's different from x.
特别是，你可能有麻烦，因为你似乎认为数组和指针是一样的 - 你的问题说：
In particular, you may be having trouble because you seem to think arrays and pointers are the same thing - your question says:
这是一个1维数组，x是基本指针，它包含数组的第一个元素的地址。所以x给我的地址和* x给出第一个元素。
This is a 1-dimensional array with x being the base pointer that is it contains the address of the first element of the array. So x gives me that address and *x gives the first element.
这不是真的。一维 x 是一个数组，它只是在一些上下文中数组衰减为指向其第一个元素的指针。
Which isn't true. The 1-dimensional x is an array, it's just that in some contexts an array decays into a pointer to its first element.
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