what is the difference between new int() and new int[]

Question

int main()
{
  int *y = new int(5);
  cout << "y = " << y << " &y = " << &y << " *y = " << *y << endl;
  int *p = new int[5];
  cout << "p = " << p << " &p = " << &p << " *p =" << *p << endl;
}

one used [] and another one used (), what's the different and how it works? Can someone help to explain to me? Thanks!

Thanks!! That help me a lot. The int *y = new int(5) mean that the value 5 is given to *y. But int[5] mean that make a 5 int numbers spaces. — Jay, Sep 17 '17 at 01:56

score 1 · Answer 1 · edited Sep 17 '17 at 07:32

1

You output so many things in your std::cout but most of them don't matter.

try

std::cout << sizeof(int(5)) << " " << sizeof(int[5]) << "\n";

which will output

4 20

since int(5) means a single integer initialised with 5. So sizeof yields the size of an int.

On the other hand, int[5] is what you expect: an array of 5 integers, so sizeof yields 5*sizeof(int).

Two more remarks:

creating good old arrays with new definitely is something a beginner should refrain from. If you need dynamically sized indexable storage, use std::vector. There is no speed panalty.
refrain from using namespace std; simply type the std::, that makes the code more readable and clearly separates your stuff from library stuff.

edited Sep 17 '17 at 07:32

Remy Lebeau

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answered Sep 17 '17 at 02:03

DrSvanHay

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Help me to deeper understand! – Jay Sep 17 '17 at 02:48
"*there is no speed penalty*" - not true. Speed is *slightly* slower with a `std::vector`. `new int[5]` allocates memory but does not fill it. `vector(5)` (or `vector.resize(5)`) allocates memory and zero-initializes it. Also, `operator[]` on a raw pointer is just a simple dereference of the pointer plus an offset, whereas `operator[]` on a `std::vector` is a function call that has to manipulate the call stack. Speed differences are negligible, but they do exist. – Remy Lebeau Sep 17 '17 at 07:39
@RemyLebeau The compiler will inline the call to the operator. – DrSvanHay Sep 17 '17 at 15:49

score 1 · Answer 2 · answered Sep 17 '17 at 07:26

int *y = new int(5)

This allocates a single int and sets its value to 5, then assigns y to point at the allocated int.

int *p = new int[5]

This allocates an array of 5 ints, whose values are initially undefined, then assigns p to point at the first int in the allocated array.

what is the difference between new int() and new int[]

2 Answers2