初始化＆QUOT;一个指针整数＆QUOT的阵列; [英] Initializing "a pointer to an array of integers"

问题描述

  INT（*一）[5];
 

我们如何初始化一个指针，上述5个整数数组。

是低于前pression是否正确？

  INT（*一）[3] = {11,2,3,5,6};
 

解决方案

假设你有长度的一个int数组 5 例如

  INT X [5];
 

然后，你可以做 A =安培; X;

  INT X [5] = {1};
 INT（*一）[5] =＆放大器; X;
 

要访问数组的元素你：（*一）[我] （== （*（＆安培; X））[I] == （*安培; X）[我] == X [I] ）括号必要的，因为 [] 运算符的precedence高则 * 。 （一个常见的​​错误可以做 * A [I] 来访问数组的元素）。

明白你在问的问题是一个编译时错误：

  INT（*一）[3] = {11，2，3，5，6};
 

这是不正确的类型不匹配也是如此，因为 {11,2,3,5,6} 可分配给 int类型的[5]; 和你分配给 INT（*一）[3] 。

此外，

您可以做一些像一维：

 为int *为什么=（INT [2]）{1,2};
 

同样，对于二维试试这个（感谢@ CAF ）：

  INT（*一）[5] =（INT [] [5]）{{1，2，3，4，5}，{6，7，8，9 ，10}};
 

 int (*a)[5];

How can we Initialize a pointer to an array of 5 integers shown above.

Is the below expression correct ?

int (*a)[3]={11,2,3,5,6}; 

解决方案

Suppose you have an array of int of length 5 e.g.

int x[5];

Then you can do 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).

Understand what you asked in question is an compilation time error:

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].

Additionally,

You can do something like for one dimensional:

int *why = (int[2]) {1,2};

Similarly, for two dimensional try this(thanks @caf):

int (*a)[5] = (int [][5]){ { 1, 2, 3, 4, 5 } , { 6, 7, 8, 9, 10 } };

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