类型不完整的函数参数和返回值 [英] Incomplete types as function parameters and return values
问题描述
以下代码使用 clang ++ 5.0.0 和 g ++ 7.2成功编译 (带有-std=c++17 -Wall -Wextra -Werror -pedantic-errors -O0
编译标志):
The following code compiles successfully both with clang++ 5.0.0 and g++ 7.2 (with the -std=c++17 -Wall -Wextra -Werror -pedantic-errors -O0
compilation flags):
struct Foo;
struct Bar
{
Foo get() const;
void set(Foo);
};
struct Foo
{
};
Foo Bar::get() const
{
return {};
}
void Bar::set(Foo)
{
}
int main()
{
Bar bar{};
(void)bar.get();
bar.set(Foo{});
}
使用不完整的类型作为函数参数和返回值是否有效? C ++在上面怎么说?
Is it valid to use incomplete types as function parameters and return values? What does the C++ say on it?
推荐答案
In a function definition, you cannot use incomplete types: [dcl.fct]/12:
在函数定义的上下文中,参数的类型或函数定义的返回类型不应是不完整的(可能是 cv 限定的)类类型,除非删除了该函数. /p>
The type of a parameter or the return type for a function definition shall not be an incomplete (possibly cv-qualified) class type in the context of the function definition unless the function is deleted.
但是函数 declaration 没有这样的限制.在定义Bar::get
和Bar::set
时,Foo
是完整类型,因此程序很好.
But a function declaration has no such restriction. By the time you define Bar::get
and Bar::set
, Foo
is a complete type, so the program is fine.
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