什么是空白？

std::is_void的说明指出:

提供等于 true 的成员常量值，如果 T 是 < strong > type void、 const void、 volatile 挥发性空洞

那么什么是 const void或者 volatile void呢？

这个答案 说明 const void返回类型将是无效的(不过是在 VC + + 2015上编译的)

const void foo() { }

如果按照标准，const void是无效的(VC 是错误的)-那么什么是 const void？

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Types can be the result of templates; a template might state const T, and be instantiated with T as void.

The linked answer is misled, or rather, limited in view in that it regards the special case of a non-template type, and even then const void might be meaningless, but it is valid code.

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最佳答案

const void is a type which you can form a pointer to. It's similar to a normal void pointer, but conversions work differently. For example, a const int* cannot be implicitly converted to a void*, but it can be implicitly converted to a const void*. Likewise, if you have a const void* you cannot static_cast it to an int*, but you can static_cast it to a const int*.

const int i = 10;
void* vp = &i;                           // error
const void* cvp = &i;                    // ok
auto ip = static_cast<int*>(cvp);        // error
auto cip = static_cast<const int*>(cvp); // ok
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As void, const void is a void type. However, if const void is a return type, the const is meaningless (albeit legal!), because [expr]/6:

If a prvalue initially has the type “cv T”, where T is a cv-unqualified non-class, non-array type, the type of the expression is adjusted to T prior to any further analysis.

However, it is a valid type itself and occurs in e.g. C-standard library functions, where it's used to ensure const-correctness of argument pointers: int const* cannot be converted to void*, but void const*.


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