删除的构造函数必须是私有的吗？ [英] Must a deleted constructor be private?

问题描述

class A
{
public:
    A() = default;
    A(const A&) = delete;
};

class A
{
public:
    A() = default;

private:
    A(const A&) = delete;
};

在任何情况下，这两个定义是否总是相同的？

Are these two definitions always identical to each other in any cases?

推荐答案

它们是不同的只是生产诊断。如果您将其设置为 private ，则会报告额外的和多余的访问冲突：

They are different only wrt the produced diagnostics. If you make it private, an additional and superfluous access violation is reported:

class A
{
public:
    A() = default;
private:
    A(const A&) = delete;
};

int main()
{
    A a;
    A a2=a;
}

会导致GCC 4.8输出以下附加 ：

results in the following additional output from GCC 4.8:

main.cpp: In function 'int main()':
main.cpp:6:5: error: 'A::A(const A&)' is private
     A(const A&) = delete;
     ^
main.cpp:12:10: error: within this context
     A a2=a;
          ^

因此我建议始终使删除的方法 public 。

hence my recommendation to always make deleted methods public.

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