c++ - new int[100] 和 new int[100]() 的区别；

Question

作为标题

#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;
}

有什么区别？我试过两个输出都是零。

另外还有什么关于智能指针，如何确保它可以零初始化。

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

Answer 1

new int[100] 执行 default initialization ，所有元素都将被初始化为不确定的值。请注意，从中读取会导致 UB ，一切皆有可能。

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

new int[100]() 执行 value initialization ，因为所有元素的效果都是 zero-initialized到 0。

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

4) otherwise, the object is zero-initialized.

编辑

std::make_unique采用第二种方式进行初始化。

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]())

附言:std::make_unique_for_overwrite采用第一种方式。

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])