pointers - 为什么指向int的指针是使用 “*int”而不是 “&int”初始化的？

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例如，当初始化一个指向int的指针时，我们使用:

var pointer *int

为什么语法不是:

var pointer &int

对我来说，第二个版本更有意义，因为它看起来像“变量“指针”是一个int的内存地址”

或者换一种说法，我发现混淆“*”既用于定义内存地址类型(如上)，又用于解引用，例如。
*指针= 123

这看起来是否令人困惑，还是我在这里错过了一些东西？

最佳答案

&是address operator，不是类型。

For an operand x of type T, the address operation &x generates a pointer of type *T to x.

但这确实是*也用作指针间接指向:

For an operand x of pointer type *T, the pointer indirection *x denotes the variable of type T pointed to by x

如Tour of Go所示，*引用类型和运算符。

这个thread from 2016提出了类似的问题。

The pointer syntax is just copied from C, of course.
See also Go's Declaration Syntax
If we had used '&', then a Go pointer would look like a C++ reference.

和:

*int defines a pointer type and then can be used with or without the * on the variable.

With the * you access the value pointed to.

Without, you access the memory address pointed to.

& retrieves a memory address from a variable.

An address is not a pointer and vice versa.
You define a pointer using a memory address.
var x int = 10    // declare an int variable `x` holding the value 10.
var y *int = &x  // Create a pointer `y` *using* the memory address of `x`.
*y will have the value 10.
y will be the memory address of x

从到Pointers in Go 中的“Dave Cheney”中查看更多内容:

Go pointers, like C pointers, are values that, uh, point to other values. This is a tremendously important concept and shouldn’t be considered dangerous or something to get hung up on.

A pointer is a value that points to the memory address of another variable.

但:

you cannot alter the address p points to unless you assign another address to it
No: var p *int; p++ possible.

(另请参见“Go at Google: Language Design in the Service of Software Engineering”)

Once a value is assigned to a pointer, with the exception of nil, Go guarantees that the thing being pointed to will continue to be valid for the lifetime of the pointer.

关于pointers - 为什么指向int的指针是使用 “*int”而不是 “&int”初始化的？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/60236794/

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