C++ Allocate dynamic array inside a function -

so need allocate array of int inside function. array declared before calling function (i need use array outside function) , size determined inside function. possible ? have been trying lot of thing nothing worked far.

thanks guys ! here code :

void fillarray(int *array) {   int size = ...//calculate size here   allocate(array, size);   //.... }  void alloc(int * &p, int size) {   p = new int[size]; }  int main() {   //do stuff here   int *array = null;   fillarray(array);   // stuff filled array  } 

if have understood correctly did not declare array before calling function. seems declared pointer int instead of array. otherwise if indeed declared array may not change size , allocate memory in function.

there @ least 3 approaches task. first 1 looks like

int *f() {     size_t n = 10;      int *p = new int[n];      return p; } 

and functionn called like

int *p = f(); 

the other approach declare parameter of function having type of pointer pointer int. example

voif f( int **p ) {     size_t n = 10;      *p = new int[n]; } 

and function can called like

int *p = null;  f( &p ); 

the third approach use reference pointer function parameter. example

voif f( int * &p ) {     size_t n = 10;      p = new int[n]; } 

and function called like

int *p = null;  f( p ); 

the better way use standard class std::vector<int> instead of pointer. example

void f( std::vector<int> &v ) {    size_t n = 10;     v.resize( n ); } 

and function can called like

std::vector<int> v;  f( v ); 

also use smart pointer std::unique_ptr