问题描述
为什么下面的代码输出4?
Why does the following code output 4?
char** pointer = new char*[1];
std::cout << sizeof(pointer) << "
";
我有一个指针数组，但它的长度应该是 1，不是吗?
I have an array of pointers, but it should have length 1, shouldn't it?
推荐答案
pointer 是一个指针.它是指针的大小，在您的系统上为 4 个字节.
pointer is a pointer. It is the size of a pointer, which is 4 bytes on your system.
*pointer 也是一个指针.sizeof(*pointer) 也将是 4.
*pointer is also a pointer. sizeof(*pointer) will also be 4.
**pointer 是一个字符.sizeof(**pointer) 将是 1.注意 **pointer 是一个字符，因为它被定义为 char**.new`ed 数组的大小永远不会进入这个.
**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.
注意 sizeof 是一个编译器操作符.它在编译时呈现为常量.任何可以在运行时更改的内容(例如新数组的大小)都无法使用 sizeof 来确定.
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.
注 2:如果您已将其定义为:
Note 2: If you had defined that as:
char* array[1];
char** pointer = array;
现在 pointer 的值与以前基本相同，但现在您可以说:
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.
这篇关于指针数组和sizeof混淆的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！