跳过变量初始化格式错误还是会导致未定义的行为？

Question

考虑这段代码：

void foo()
{
    goto bar;
    int x = 0;
    bar: ;
}

GCC 和 Clang reject it，因为跳转到 bar: 绕过了变量初始化。 MSVC 根本没有抱怨（除了在 bar: 之后使用 x 会导致警告）。

我们可以用 switch:

做类似的事情

void foo()
{
    switch (0)
    {
        int x = 0;
        case 0: ;
    }
}

现在所有三个编译器 emit errors。

这些片段格式不正确吗？还是它们会导致 UB？

我曾经认为两者都是错误的，但我找不到标准的相关部分。 [stmt.goto] doesn't say anything about this, and neither does [stmt.select].

Answer 1

当初始化是非空的时，它是病式的。

[stmt.dcl]

3 It is possible to transfer into a block, but not in a way that bypasses declarations with initialization (including ones in conditions and init-statements). A program that jumps 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 vacuous initialization ([basic.life]). In such a case, the variables with vacuous initialization are constructed in the order of their declaration.

初始化程序使初始化非空。相比之下，这个

void foo()
{
    goto bar;
    int x; // no initializer
    bar: ;
}

将是合式的。尽管有关使用具有不确定值的 x 的通常注意事项将适用。

Answer 2

来自goto statement：

If transfer of control enters the scope of any automatic variables (e.g. by jumping forward over a declaration statement), the program is ill-formed (cannot be compiled), unless all variables whose scope is entered have

scalar types declared without initializers

class types with trivial default constructors and trivial destructors declared without initializers

cv-qualified versions of one of the above

arrays of one of the above

跳过变量初始化格式错误还是会导致未定义的行为？

Is jumping over a variable initialization ill-formed or does it cause undefined behaviour?

c++

goto

initializer

variable-declaration

language-lawyer