指针上的此操作如何工作？

Question

  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?

Answer 1

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*)

Answer 2

  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.

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

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.

Answer 7

Line A de-reference pointer q typed as int * , ie 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; .