为什么在使用typedef时将非const指针视为const? [英] Why is the non-const pointer being treated as a const when using typedef?

问题描述

typedef char* c;
const c ptr1 = "pointer";
++ptr1; /// error
const char* ptr2 = "pointer";
++ptr2; /// runs fine

现在ptr1应该是const char*类型，因此是非常量指针，那么为什么将其视为常量指针?

Now ptr1 should be of type const char* and thus a non-const pointer, then why is it being treated as a constant pointer ?

推荐答案

它们不相同.

第一个表示一个指向const-char的指针，第二个表示一个指向const-char的指针.

The first designates a const-pointer-to-char, the second is a pointer-to-const-char.

尝试从右到左阅读:

const char *p;  // p is a pointer to char which is const
char const *p;  // p is a pointer to const char (same as previous)
char * const p; // p is a const pointer to char
char const * const p; // p is a const pointer to const char

通过使用typedef typedef char* c，您将指向char的指针"的含义打包到单个别名c中:

By using the typedef typedef char* c you pack the meaning "pointer to char" into a single alias c:

const c p; // p is a const [c] --> p is a const [pointer to char]

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其他说明:

Typedef并不像宏那样就地扩展.

Typedefs are not in-place-expanded like macros are, i.e.

const c p;

真正成为

const [char*] p;

不是变成

const char* p; // Nope.

不要在您的脑海中像宏一样扩展它，使用typedef可以将char和*绑定在一起并形成一个原子.

Don't expand it like macros in your mind, with the typedefs, you've bound char and * together and formed an atom.

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