char a[] = ?string? 和有什么区别?和 char *p = ?string?;? [英] What is the difference between char a[] = ?string?; and char *p = ?string?;?

问题描述

正如标题所说，

char a[] = ?string?; and 
char *p = ?string?;  

这个问题是在面试中问我的.我什至不明白这句话.

This question was asked to me in interview. I even dont understand the statement.

char a[] = ?string?

这里是什么 ? 运算符?它是字符串的一部分还是具有某些特定含义?

Here what is ? operator? Is it a part of a string or it has some specific meaning?

推荐答案

第一个是数组，另一个是指针.

The first one is array the other is pointer.

数组声明char a[6]; 要求留出六个字符的空间，以名称a 表示.也就是说，有一个名为 a 的位置，可以坐六个字符.另一方面，指针声明 char *p; 请求一个存放指针的地方.指针的名称为 p，并且可以指向任何地方的任何字符(或连续的字符数组).

The array declaration char a[6]; requests that space for six characters be set aside, to be known by the name a. That is, there is a location named a at which six characters can sit. The pointer declaration char *p; on the other hand, requests a place which holds a pointer. The pointer is to be known by the name p, and can point to any char (or contiguous array of chars) anywhere.

声明

 char a[] = "string";
 char *p = "string"; 

将导致可以这样表示的数据结构:

would result in data structures which could be represented like this:

     +---+---+---+---+---+---+----+
  a: | s | t | r | i | n | g |  |
     +---+---+---+---+---+---+----+
     +-----+     +---+---+---+---+---+---+---+ 
  p: |  *======> | s | t | r | i | n | g | |    
     +-----+     +---+---+---+---+---+---+---+ 

认识到像 x[3] 这样的引用会根据 x 是数组还是指针生成不同的代码，这一点很重要.鉴于上面的声明，当编译器看到表达式 a[3] 时，它发出代码以从位置 a 开始，将三个元素移过它，然后获取字符那里.当它看到表达式 p[3] 时，它发出代码从位置 p 开始，在那里获取指针值，向指针添加三个元素大小，最后获取指向的字符.在上面的例子中，a[3] 和 p[3] 碰巧都是字符 l，但编译器得到的结果不同.

It is important to realize that a reference like x[3] generates different code depending on whether x is an array or a pointer. Given the declarations above, when the compiler sees the expression a[3], it emits code to start at the location a, move three elements past it, and fetch the character there. When it sees the expression p[3], it emits code to start at the location p, fetch the pointer value there, add three element sizes to the pointer, and finally fetch the character pointed to. In the example above, both a[3] and p[3] happen to be the character l, but the compiler gets there differently.

来源:comp.lang.c 常见问题列表·问题 6.2

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