Answers for "create int pointer"

C
4

pointers in cpp

#include <iostream>
using std::cout;

int main() {
  /* 
  Some things to keep in mind:
  	-you shouldn't circumvent the type system if you are creating raw ptrs
  	and don't need to "type pun" or cast (don't use void ptrs)
    -ptr types only reference memory (which are integers), not actual data, thus
    they should not be treated as data types
    char* is just 1 byte of mem, int* is just 4 bytes of mem, etc
    - '*' means that you are creating a pointer which "points" to the mem address
    of a variable
    - '&', in this case, means "get the mem address of this variable"
  */
  
  void* ptr; // a pointer that doesn't reference a certain size of memory
  int* int_ptr; // a pointer that points to data with
  				// only 4 bytes of memory (on stack)
  
  int a = 5; // allocates 4 bytes of mem and stores "5" there (as a primitive)
  ptr = &a; // can only access the memory address of 'a' (not the data there)
  
  int b = 45; 
  int_ptr = &b; // can access both memory address and data of 'b'
  
  cout << ptr << "\n"; // prints mem address of 'a'
  /*cout << *ptr << "\n"; <- this will error out; a void ptr cannot be 
  							 derefrenced */
  cout << *(int*)ptr << "\n"; // type punning to get around void ptr (extra work)
  
  cout << int_ptr << "\n"; // mem address of b
  cout << *int_ptr << "\n"; // data stored at b
  
  /* -- OUTPUTS -- */
  /*
  	some memory address (arbitrary) which contains 05 00 00 00 as its data
  	5
    some memory address (arbitrary) which contains 2D 00 00 00 as its data
    45
  */
  
  return 0; // you only need this if "main" isnt the linker entry point
  			// you also don't care
  
  // ur also probably wondering why I didn't using namespace std... cherno
}
Posted by: Guest on July-04-2020

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