指针数组和大小混淆 [英] Pointer array and sizeof confusion

问题描述

为什么以下代码会输出 4 ？

  char ** pointer = new char * [1]; 
 std :: cout<< sizeof（pointer）<< \\\
; 
  

我有一个指针数组，但它应该有长度1，指针是一个指针。

解决方案

它是一个指针的大小，在您的系统上是4个字节。

*指针也是一个指针。 sizeof（* pointer）也将为4。

**指针是一个char。 sizeof（** pointer）将为1.请注意**指针是一个字符，因为它被定义为 char ** 。

请注意， sizeof 是一个编译器运算符。它在编译时被渲染为常量。在运行时可以更改的任何内容（如新数组的大小）不能使用 sizeof 来确定。

注意2：如果你定义为：

  char * array [1] 
 char ** pointer = array; 
  

现在指针之前，但现在你可以说：

  int arraySize = sizeof //数组的总空间大小
 int arrayLen = sizeof（array）/ sizeof（array [0]）; // number of element == 1这里。 
  

Why does the following code output 4?

char** pointer = new char*[1];
std::cout << sizeof(pointer) << "\n";

I have an array of pointers, but it should have length 1, shouldn't it?

解决方案

pointer is a pointer. It is the size of a pointer, which is 4 bytes on your system.

*pointer is also a pointer. sizeof(*pointer) will also be 4.

**pointer is a char. sizeof(**pointer) will be 1. Note that **pointer is a char because it is defined as char**. The size of the array new`ed nevers enters into this.

Note that sizeof is a compiler operator. It is rendered to a constant at compile time. Anything that could be changed at runtime (like the size of a new'ed array) cannnot be determined using sizeof.

Note 2: If you had defined that as:

char* array[1];
char** pointer = array;

Now pointer has essencially the same value as before, but now you can say:

 int  arraySize = sizeof(array); // size of total space of array 
 int  arrayLen = sizeof(array)/sizeof(array[0]); // number of element == 1 here.

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