does make_unique value initializes char array

Question

For example -

#include <memory>

int main(){
    const auto bufSize = 1024;
    auto buffer = std::make_unique<char[]>(bufSize);
}

Is the buffer here already filled with '\0' characters or will I have to manually fill it to avoid garbage values.

And what would be the possible way to do this, will std::memset(&buffer.get(), 0, bufSize) suffice?

Solution 1

All of the make_* functions use value-initialization for the type if you don't provide constructor parameters. Since the array-form of make_unique doesn't take any parameters, it will zero-out the elements.

Solution 2

Yes, all the elements will be value initialized by std::make_unique.

The function is equivalent to:

unique_ptr<T>(new typename std::remove_extent<T>::type[size]())

and

value initialization

This is the initialization performed when a variable is constructed with an empty initializer.

Syntax

new T (); (2)

and

The effects of value initialization are:

3) if T is an array type, each element of the array is value-initialized;
4) otherwise, the object is zero-initialized.

then for each element of type char, they'll be value-initialized (zero-initialized) to '\0'.

Solution 3

According to cppreference, yes:

2) Constructs an array of unknown bound T. This overload only participates in overload resolution if T is an array of unknown bound. The function is equivalent to:

unique_ptr<T>(new typename std::remove_extent<T>::type[size]())
                                       value initialization ^

Value initialization indicated by me.