C pointer explanation

[g.eckert]

Am I correct in description of the following code

int array[5]={0};
int *ptr=array;

...

scanf("%d", &*(ptr + i) );/* Store input at the address of the value at (ptr + i)*/

Does the comment correcty describe what I am doing??

[ZuK]

scanf("%d", &*(ptr + i) );/* Store input at the address of the value at (ptr + i)*/

Taking the address of a dereferenced pointer doesn't make much sense
I'd use

scanf("%d", ptr + i );/* Store input at the address of the value at (ptr + i)*/

Kurt

[g.eckert]

I swear I used

scanf("%d", (ptr + i) );

and the compiler complained about something. It is working now though

[Paul McKenzie]

I swear I used

scanf("%d", (ptr + i) );

and the compiler complained about something. It is working now though

If it's 'C' you're compiling and not C++, the compiler should complain that scanf() has not been defined, but in 'C", it will allow the compile to be successful (since missing function declarations are OK in 'C"). You are missing the include for <stdio.h> in your sample.

That's why when posting code, you better post everything, including the headers that you included. If it really is 'C' and not C++ that you're compiling, a missing header can make the difference between a program that compiles OK and works when run and a program that compiles OK and fails to run correctly.

Regards,

Paul McKenzie

[g.eckert]

Yes it is ' C ' that I am using. I could not find the "Plain C" board.

I was looking for clarification on what I was doing with the pointers and derefrencing. Making sure that I was describing the statement correctly. Thanks guys.

[Lindley]

Taking the address of a dereferenced pointer doesn't make much sense

Not for a pointer specifically, no. However, if you were dealing with a more general iterator in a C++ templated method, the &* trick would ensure you had an actual address to work with.