explain the pointers to pointers issue

Question

I'm new into C and I'm trying to get a grip on pointers. I came across this expression

char *foo = *(char **)bar;

what exactly is going on? I understand that * is pointer ** is pointer to pointer, but I don't really get what is *(char **)bar

edit: worth to mention that bar is declared as const void *bar

Ed Swangren · Accepted Answer · 2014-01-10 02:23:01Z

8

*(char**)bar;

Cast bar to a pointer to pointer to char and dereference, which results in a pointer to char (i.e., a char*).

Remember that pointers are merely indirection. When you dereference a pointer you "drop a star", i.e.:

char ***p = ...;
p    -> char***
*p   -> char**
**p  -> char*
***p -> char

edited Jan 10, 2014 at 2:23

answered Jan 10, 2014 at 0:47

Ed Swangren

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Eric Lippert · Accepted Answer · 2014-01-10 17:49:41Z

3

Beginners and experts alike can benefit from getting the basics really solid.

A "variable of type t" is a storage location which can be used to store or fetch a value of type t.
A "pointer to t" is a value.
Applying the dereferencing operator * to a "pointer to t" produces a "variable of type t".

So what have we got?

void *bar = whatever;
char *foo = *(char **)bar;

bar is a variable of type "pointer to void".
foo is a variable of type "pointer to char".
casting bar to char** reads the variable bar and fetches a value of type "pointer to void". That value is then converted into a value of type "pointer to pointer to char".
Dereferencing that value produces a variable of type "pointer to char".
The value of that variable is then fetched, and the value is assigned to variable foo.

Make sense?

edited Jan 10, 2014 at 17:49

answered Jan 10, 2014 at 1:00

Eric Lippert

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1 Comment

webduvet Over a year ago

Yes thanks, bar in this case was of type void* hence the cast to char**

Ned · Accepted Answer · 2014-01-10 00:52:44Z

1

I don't know what type bar is without the cast, but here goes:

bar is being cast into a pointer to a pointer to a char.

The * before the (char **) is "dereferencing" the value to its right. In this case, it is returning the char * pointer held in the location pointer to by bar.

The result of the "dereferencing" is a char * -- pointer to a character -- which is assigned to the foo variable of the same type.

answered Jan 10, 2014 at 0:52

Ned

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