c++:为什么我们可以在switch的情况下声明一个变量

Question

int i;
i = 2;
switch(i)
{
    case 1: 
        int k;
        break;
    case 2:
        k = 1;
        cout<<k<<endl;
        break;
}

我不知道上面的代码为什么有效。

在这里，我们永远无法进入案例 1，但为什么我们可以在案例 2 中使用 k？

Answer 1

其实有2个问题:

<强>1。为什么我可以在 case 标签之后声明一个变量？

这是因为在 C++ 中标签必须是这样的:

N3337 6.1/1

labeled-statement:

...

• attribute-specifier-seqopt case constant-expression : statement

...

在 C++ 中，声明语句 也被视为语句(与 C 相对):

N3337 6/1:

statement:

...

• declaration-statement

...

<强>2。为什么我可以跳过变量声明然后使用它？

因为:N3337 6.7/3

It is possible to transfer into a block, but not in a way that bypasses declarations with initialization. A program that jumps (The transfer from the condition of a switch statement to a case label is considered a jump in this respect.)

from a point where a variable with automatic storage duration is not in scope to a point where it is in scope is ill-formed unless the variable has scalar type, class type with a trivial default constructor and a trivial destructor, a cv-qualified version of one of these types, or an array of one of the preceding types and is declared without an initializer (8.5).

因为 k 是标量类型，并且在声明点没有初始化，所以可以跳过它的声明。这在语义上是等价的:

goto label;

int x;

label:
cout << x << endl;

但是，如果 x 在声明时被初始化，这将不起作用:

 goto label;

    int x = 58; //error, jumping over declaration with initialization

    label:
    cout << x << endl;