如何在类的成员函数中调用复制构造函数？

Question

这是我得到的:

void set::operator =(const set& source)
{
    if (&source == this)
        return;

    clear();

    set(source);
}

这是我得到的错误:

vset.cxx:33:错误:声明'source'会影响参数

我该如何正确地做到这一点？

Answer 1

您正在寻找复制交换习语:

set& set::operator=(set const& source)
{
    /* You actually don't need this. But if creating a copy is expensive then feel free */
    if (&source == this)
        return *this;

    /*
     * This line is invoking the copy constructor.
     * You are copying 'source' into a temporary object not the current one.
     * But the use of the swap() immediately after the copy makes it logically
     * equivalent.
     */
    set tmp(source);
    this->swap(tmp);

    return *this;
}

void swap(set& dst) throw ()
{
    // swap member of this with members of dst
}