C++ vector at(#) 替换清理内存？

Question

The pains of compiled libraries is never knowing exactly how the memory is managed. I'm lead to believe that vector's elements are placed on the heap unless explicity told not to.

Being placed on the heap, it obviously needs to be deleted when it is no longer used, which seems to happen when the vector object is deleted.

The question is when the at(#) or operator[] is called does it delete the memory being replaced?

For example:

std::vector<string> secretlyAnArray(5);
secretlyAnArray.at(0) = std::string("Does this memory leak?");
secretlyAnArray.at(0) = std::string("When I overwrite the object?")

Happy to learn better methods to replace data at a specific index of a vector, or just pointing at the documentation that explains it.

Edit: After the helpful comments of Anis and Daniel; Is it correct in saying that:

// Given: 
secretlyAnArray.at(0) = std::string("Does this memory leak?");

// This will call the equavant code of for an L value:
string::operator=(const std::string& other)

//Given
secretlyAnArray.at(0) = std::move(std::string("Does this memory leak?"));

//This will call the equavalent code for an R value:
string::operator=(const std::string&& other)

Answer 1

std::vector<std::string> secretlyAnArray(5);
secretlyAnArray.at(0) = std::string("Does this memory leak?");
secretlyAnArray.at(0) = std::string("When I overwrite the object?")

This code assigns an object of type std::string to another object of type std::string . The std::vector itself isn't at all involved in this assignment. Here, it just provides a reference to the destination (assigned-to) object.

What does assignment do for a given type depends on its definition. There is no generic answer. For instance, with std::string , when you copy/move-assign, the original content of the destination object (its string of characters) needs to be "destroyed". std::string does it for you, so, you don't need to care about it. It means that if the string is "long" (with regard to short string optimization ; SSO), the memory will be correctly deallocated, if needed.