关于c++的new运算的问题

Question

偶然在网上看到这么一段代码，想问下其中的new (pv)Solution()是什么用法
class Solution {
public:
int a;
char c;
static int sb;
};
Solution x;
void main()
{
Solution s;
void *pv = operator new(sizeof(Solution));
Solution *ps = new (pv)Solution(); //这是什么用法？
Solution *ps2 = new Solution();
}

Answer 1

谢谢各位回答，在c++primer找到答案了

new (place_address) type
new (place_address) type (initializers)
new (place_address) type [size]
new (place_address) type [size] { braced initializer list }
where place_address must be a pointer and the initializers provide (a possibly empty)
comma-separated list of initializers to use to construct the newly allocated object.
When called with an address and no other arguments, placement new uses
operator new(size_t, void*) to “allocate” its memory. This is the version of
operator new that we are not allowed to redefine (§ 19.1.1, p. 822). This function
does not allocate any memory; it simply returns its pointer argument. The overall new
expression then finishes its work by initializing an object at the given address. In
effect, placement new allows us to construct an object at a specific, preallocated
memory address.
效果就是在pv指向的内存上调用构造函数