Declaring array of int

Question

Is there any difference between these two declarations?

int x[10]; 

vs.

int* x = new int[10]; 

I suppose the former declaration (like the latter one) is a pointer declaration and both variables could be treated the same. Does it mean they are intrinsically the same?

Answer 1

#include<iostream>      int y[10];   void doSomething() {     int x[10];     int *z  = new int[10];     //Do something interesting      delete []z; }  int main() {     doSomething();  } 

‏‏‏‏‏‏‏

int x[10];  

- Creates an array of size 10 integers on stack.
- You do not have to explicitly delete this memory because it goes away as stack unwinds.
- Its scope is limited to the function doSomething()

int y[10]; 

- Creates an array of size 10 integers on BSS/Data segment.
- You do not have to explicitly delete this memory.
- Since it is declared global it is accessible globally.

int *z = new int[10]; 

- Allocates a dynamic array of size 10 integers on heap and returns the address of this memory to z.
- You have to explicitly delete this dynamic memory after using it. using:

delete[] z; 

Answer 2

The only thing similar between

int x[10]; 

and

int* x = new int[10]; 

is that either can be used in some contexts where a int* is expected:

int* b = x;   // Either form of x will work  void foo(int* p) {}  foo(x);      // Either form will work 

However, they cannot be used in all contexts where a int* is expected. Specifically,

delete [] x;  // UB for the first case, necessary for the second case. 

Some of the core differences have been explained in the other answers. The other core differences are:

Difference 1

sizeof(x) == sizeof(int)*10   // First case  sizeof(x) == sizeof(int*)     // Second case. 

Difference 2

Type of &x is int (*)[10] in the first case

Type of &x is int** in the second case

Difference 3

Given function

void foo(int (&arr)[10]) { } 

you can call it using the first x not the second x.

foo(x);     // OK for first case, not OK for second case.