初始化块和指定范围的块之间有区别吗

发布于 2024-12-10 03:37:56 字数 367 浏览 0 评论 0原文

如果您编写一个类并使用类似的内容，

...
int x;
{ x = 2; }
...

这将是一个初始化块，对吧？

那么如何使用花括号来指定范围，以便它们像任何其他代码一样执行？基本上，它们不是构造函数的一部分。

例如，

如果您要在 switch 语句中使用块，它会像初始化块一样执行吗？

    switch(...)
    {
    case :
    { // this right here how does the compiler know the difference?
    ...
    break;
    }
    }

原文

If you writing a class and you use something like this

...
int x;
{ x = 2; }
...

This would be an initialisation block right?

So how would you use curly braces just to designate scope, so that they are executed like any other code? Basically so they are not part of the constructor.

if you were to use a block inside a switch statement, would this be executed like an initialisation block?

    switch(...)
    {
    case :
    { // this right here how does the compiler know the difference?
    ...
    break;
    }
    }

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只有影子陪我不离不弃 2024-12-17 03:37:57

初始化块仅适用于类范围内且前面没有函数原型的块（即不是函数体的块）。

在函数体内，您创建的任何块都会引入作用域，但不会被解释为初始化。

例子：

public class NameOfClass {
    {
        // This is an initalization block
    }

    AccessModifier ReturnType nameOfFunction(ParamType nameOfParam) {
        // This is a function body and is not an initialization

        {
          // This introduces scope, is not an initialization
        }
    }
}

Initialization blocks only apply to blocks that are in the scope of a class and are not preceded by a function prototype (i.e. blocks that are not the body of a function).

Within the body of a function, any blocks that you create will introduce scope but will not be interpreted as an initialization.

Example:

public class NameOfClass {
    {
        // This is an initalization block
    }

    AccessModifier ReturnType nameOfFunction(ParamType nameOfParam) {
        // This is a function body and is not an initialization

        {
          // This introduces scope, is not an initialization
        }
    }
}

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