difference between new int[100] and new int[100]();

Question

As title

#include <iostream>
int main() {
    auto* a = new float[1000000];
    auto* b = new float[10]();
    for(auto i=0; i<1000000; i++){
        std::cout << "a" << a[i] << std::endl;
    }
    for(auto i=0; i<10; i++){
        std::cout << "b" << b[i] << std::endl;
    }
    return 0;
}

what's the difference? I had tried both output is zero.

In addition what's about smart pointer, how to make sure it can zero initialized.

std::unique_ptr<int[]> p = std::make_unique<int[]>(100);

Answer 1

new int[100] performs default initialization , all the elements would be initialized to indeterminate values. Note that reading from them leads to UB , anything is possible.

• otherwise, nothing is done: the objects with automatic storage duration (and their subobjects) are initialized to indeterminate values.

new int[100]() performsvalue initialization , as the effect all the elements would be zero-initialized to 0 .

3) if T is an array type, each element of the array isvalue-initialized ;

4) otherwise, the object is zero-initialized .

EDIT

std::make_unique takes the 2nd way for initializing.

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]())

PS:std::make_unique_for_overwrite takes the 1st way.

5) Same as (2), except that the array is default-initialized. 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])