placement new的作用就是在operator new分配好的內(nèi)存上執(zhí)行對(duì)象的構(gòu)造,那么new operator使用了operator new來(lái)申請(qǐng)內(nèi)存之后是使用了placement new嗎?如果沒(méi)有話是怎么構(gòu)造的呢?和placement new有關(guān)系嗎?
new
is not an operator
!In c++, new
and operator
are both keywords. new int(x)
is a(n) (new-)expression. operator new is a function. new operator in your title is new-expression indeed. new-expression will invoke oeprator new function.
placement new的作用就是在operator new分配好的內(nèi)存上執(zhí)行對(duì)象的構(gòu)造,
Yes, that's true. To help you understand, here is a sample:
char* ptr = new char[sizeof(T)]; // ptr use a new-expression(newchar [sizeof(T)] to allocate memory, char is guaranteed to be sizeof 1
T* tptr = new(ptr) T; // Here is placement new
In a nutshell, placement new will use already allocated memory to construct the object. Where is the already allocated memory from? Either from new expression(free store) or allocated from activation record
like int buffer[10]
, both ok.
那么new operator使用了operator new來(lái)申請(qǐng)內(nèi)存之后是使用了placement new嗎?如果沒(méi)有話是怎么構(gòu)造的呢?和placement new有關(guān)系嗎?
Above is my answer to your questions
BTW, from the case int buffer[10]
, we can see pre-new-expression is not a must for placement new(however, note that placement new itself is a new-expression, which will invoke operator new function because all it does is just construct here). If your question is "will placement new always be after operator new/new-expression", it will be a good question.
One year ago, I was confused about how to combine operator new with the constructor then asked a question, FrankHB answered my question: https://tieba.baidu.com/p/508... Now, look back to this question, it is a X-Y question, what really confused me was how does new expression invoke constructor, so it is not related to placement new. Hope it will also inspire you.
所以我認(rèn)為或許自己和您一年前的疑問(wèn)相似,內(nèi)存申請(qǐng)和構(gòu)造函數(shù)這兩個(gè)過(guò)程是如何結(jié)合的呢?
the wordcombination
(結(jié)合) is not properly now(I also make such mistake as said above), let me re-organize my wording:
new expression does two things:
You and I(one year ago) are both confused about how does compiler initialize the object(via constructor or else) after allocating. Yes, I mentioned compiler, because C++ standard guarantee new will do the two things, but didn't how to, so, its compiler's work. So, it is not c++'s work, just compiler's. Now, we can see the assembly:
struct Foo
{
Foo(int i) {}
};
int main()
{
auto *a = new Foo(1);
}
-O0:
Foo::Foo(int):
push rbp
mov rbp, rsp
mov QWORD PTR [rbp-8], rdi
mov DWORD PTR [rbp-12], esi
pop rbp
ret
main:
push rbp
mov rbp, rsp
push rbx
sub rsp, 24
mov edi, 1
call operator new(unsigned long)
mov rbx, rax
mov esi, 1
mov rdi, rbx
call Foo::Foo(int)
mov QWORD PTR [rbp-24], rbx
mov eax, 0
add rsp, 24
pop rbx
pop rbp
ret
codes related the new expression is follows: a
mov edi, 1
call operator new(unsigned long)
mov rbx, rax
mov esi, 1
mov rdi, rbx
call Foo::Foo(int)
mov QWORD PTR [rbp-24], rbx
Now, it is clear enough, right? assemble calls two procedures, oeprator new and Foo::Foo(int)
So, your question is how the two combined?
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