与指针声明数组大小 [英] Sizes of arrays declared with pointers

问题描述

char c[] = "Hello";
char *p = "Hello";

printf("%i", sizeof(c)); \\Prints 6
printf("%i", sizeof(p)); \\Prints 4

我的问题是：

为什么这些打印不同的结果？不 C [] 还声明指向数组的第一个字符（因此应该有大小4，因为它是一个指针）的指针？

Why do these print different results? Doesn't c[] also declare a pointer that points to the first character of the array (and therefore should have size 4, since it's a pointer)?

推荐答案

这听起来像你指针和数组之间的混淆。指针和数组（在这种情况下的char * 和的char [] ）是的不一样的东西。

It sounds like you're confused between pointers and arrays. Pointers and arrays (in this case char * and char []) are not the same thing.

• 阵列烧焦一个[​​SIZE] 表示，在位置值 A 是长度的数组尺寸


• 指针的char * A; 表示，在位置值 A 是一个指向一个字符。这可以用指针运算结合起来，表现得像一个数组（例如， A [10] 是过去无论 A 点）

• An array char a[SIZE] says that the value at the location of a is an array of length SIZE
• A pointer char *a; says that the value at the location of a is a pointer to a char. This can be combined with pointer arithmetic to behave like an array (eg, a[10] is 10 entries past wherever a points)

在内存中，它看起来像这样（从常见问题）：

In memory, it looks like this (example taken from the FAQ):

 char a[] = "hello";  // array

   +---+---+---+---+---+---+
a: | h | e | l | l | o |\0 |
   +---+---+---+---+---+---+

 char *p = "world"; // pointer

   +-----+     +---+---+---+---+---+---+
p: |  *======> | w | o | r | l | d |\0 |
   +-----+     +---+---+---+---+---+---+

这很容易混淆了指针和数组之间的区别，因为在很多情况下，一个数组引用衰变的指针到它的第一个元素。这意味着，在许多情况下，（当传递给函数调用如）阵列成为指针。如果您想了解更多，请的C常见问题本节详细介绍了。

一个重大现实不同的是，编译器知道数组有多长。使用上述的例子：

One major practical difference is that the compiler knows how long an array is. Using the examples above:

char a[] = "hello";  
char *p =  "world";  

sizeof(a); // 6 - one byte for each character in the string,
           // one for the '\0' terminator
sizeof(p); // whatever the size of the pointer is
           // probably 4 or 8 on most machines (depending on whether it's a 
           // 32 or 64 bit machine)

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