CODEWITHSUNDEEP
c++pointers
qazwsx
qazwsx

Reputation: 26868

How does this operation on pointers work?

  int x = 4;
  int* q = &x;                 // Is it always equivalent to int *q = &x;  ?
  cout << "q = " << q << endl; // output: q = 0xbfdded70
  int i = *q;                  // A
  int j = *(int*)q;            // B, when is this necessary?
  cout << "i = " << i << endl; // output: i = 4
  cout << "j = " << j << endl; // output: j = 4

My question is what does lines A and B do, and why the outputs are both 4?

Upvotes: 1

Views: 136

Answers (7)

marcinj
marcinj

Reputation: 49976

It is a basic usage of pointers, in A you dereference pointer (access the variable to which a pointer points)":

int i = *q; // A

while B is doing exactly the same but it additionally casts pointer to the same type. You could write it like that:

int j = *q; // B

there is no need for (int*)

Upvotes: 3

Summer_More_More_Tea
Summer_More_More_Tea

Reputation: 13356

Line A de-reference pointer q typed as int *, i.e. a pointer points to an int value.

Line B cast q as (int *) before de-reference, so line B is the same as int j = *q;.

Upvotes: 0

Drew Dormann
Drew Dormann

Reputation: 63735

  int x = 4;

x is 4

  int* q = &x;

q is the memory location of x (which holds 4)

  cout << "q = " << q << endl; // output: q = 0xbfdded70

There's your memory location.

  int i = *q; // A

i is the value at memory location q

  int j = *(int*)q; // B

j is the value at memory location q. q is being cast to an int pointer, but that's what it already is.

Upvotes: 2

sepp2k
sepp2k

Reputation: 370112

Line A takes the value that q points to and assigns it to i. Line b casts q to the type int* (which is q's type already, so that cast is entirely redundant/pointless), then takes the value that q points to and assigns it to j.

Both give you 4 because that's the value that q points to.

Upvotes: 0

SlavaNov
SlavaNov

Reputation: 2485

A: Dereference - takes a pointer to a value (variable or object) and returns the value

B: Cast to int* and than dereference

The result is the same because the pointer is already to int. That's it.

Upvotes: 1

Ivaylo Strandjev
Ivaylo Strandjev

Reputation: 70931

Lines A and B are equivelent as q is already an int* and therefor (int*)q equals q. int i = *q; yelds that i becomes the value of the integer pointed to by q. If you want to make i to be equal to the adress itself remove the asterisk.

Upvotes: 1

Alok Save
Alok Save

Reputation: 206508

int i = *q; // A

Dereferences a pointer to get the pointed value

int j = *(int*)q; // B

type casts the pointer to an int * and then dereferences it.

Both are same because the pointer is already pointing to an int. So typecasting to int * in second case is not needed at all.

Further derefenecing yields the pointed integer variable value in both cases.

Upvotes: 1

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