C++ print value of a pointer

Question

I have an array of double pointers, but every time I try do print one of the values the address gets printed. How do I print the actual value?

cout << arr[i] ? cout << &arr[i] ? they both print the address

Does anyone know?

Xever

Answer 1

If it's really an array of (initialized) double pointers, i.e.:

double *arr[] = ...
// Initialize individual values

all you need is:

cout << *arr[i];

Disc

Answer 2

cout << *(arr[i]) will print the value.

Oshaughnessy

Answer 3

cout << *(arr[i]);

Bright

@Potatoswatter. The compiler might not, and technically neither do I, because I can remember the operator precedence if I have to. But they certainly improve things. – Intransitive

@Potatoswatter: parens aren't needed for 5 + 112 % 65 ^ 1 / 5.5 && bar || foo^2 << 5 either, but (((5 + (112 % 65)) ^ (1 / 5.5)) && bar) || (foo^(2 << 5)) is much more clear – Bright

+1 for parenthesis, I always parenthesise that kind of expression. – Johnnajohnnie

Answer 4

If "arr" is declared as

double* arr[..];

Then you would use:

cout << *(arr[i])

Argil