C语言 初始化“指向整数数组的指针”
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Initializing "a pointer to an array of integers"
提问by EnterKEY
int (*a)[5];
How can we Initialize a pointer to an array of 5 integers shown above.
我们如何初始化一个指向上面显示的 5 个整数数组的指针。
Is the below expression correct ?
下面的表达正确吗？
int (*a)[3]={11,2,3,5,6};
回答by Grijesh Chauhan
Suppose you have an array of int of length 5e.g.
假设你有一个长度为 int 的数组，5例如
int x[5];
Then you can do a = &x;
然后你可以做 a = &x;
int x[5] = {1};
int (*a)[5] = &x;
To access elements of array you: (*a)[i](== (*(&x))[i]== (*&x)[i]== x[i]) parenthesis needed because precedence of []operator is higher then *. (one common mistake can be doing *a[i]to access elements of array).
要访问数组元素，您需要：(*a)[i](== (*(&x))[i]== (*&x)[i]== x[i]) 括号，因为[]运算符的优先级高于*。（一个常见的错误可能是*a[i]访问数组元素）。
Understand what you asked in question is an compilation time error:
了解您所问的是编译时错误：
int (*a)[3] = {11, 2, 3, 5, 6};
It is not correct and a type mismatch too, because {11,2,3,5,6}can be assigned to int a[5];and you are assigning to int (*a)[3].
它不正确并且类型也不匹配，因为{11,2,3,5,6}可以分配给int a[5];并且您正在分配给int (*a)[3].
Additionally,
此外，
You can do something like for one dimensional:
你可以为一维做类似的事情：
int *why = (int p[2]) {1,2};
Similarly, for two dimensional try this(thanks @caf):
同样，对于二维试试这个（感谢@caf）：
int (*a)[5] = (int p[][5]){ { 1, 2, 3, 4, 5 } , { 6, 7, 8, 9, 10 } };
回答by Lundin
{11,2,3,5,6}is an initializer list, it is not an array, so you can't point at it. An array pointer needs to point at an array, that has a valid memory location. If the array is a named variable or just a chunk of allocated memory doesn't matter.
{11,2,3,5,6}是一个初始化列表，它不是一个数组，所以你不能指向它。数组指针需要指向具有有效内存位置的数组。如果数组是一个命名变量或者只是一块分配的内存并不重要。
It all boils down to the type of array you need. There are various ways to declare arrays in C, depending on purpose:
这一切都归结为您需要的数组类型。有多种方法可以在 C 中声明数组，具体取决于目的：
// plain array, fixed size, can be allocated in any scope
int array[5] = {11,2,3,5,6};
int (*a)[5] = &array;
// compound literal, fixed size, can be allocated in any scope
int (*b)[5] = &(int[5]){11,2,3,5,6};
// dynamically allocated array, variable size possible
int (*c)[n] = malloc( sizeof(int[n]) );
// variable-length array, variable size
int n = 5;
int vla[n];
memcpy( vla, something, sizeof(int[n]) ); // always initialized in run-time
int (*d)[n] = &vla;
回答by Lundin
int a1[5] = {1, 2, 3, 4, 5};
int (*a)[5] = &a1;
回答by Stefano Falasca
int vals[] = {1, 2};
int (*arr)[sizeof(vals)/sizeof(vals[0])] = &vals;
and then you access the content of the array as in:
然后您访问数组的内容，如下所示：
(*arr)[0] = ...
