C - how do I read * and -> together?

Question

If I have :

char *name; //this is in a struct

*row->name //row is able to get in the struct

How do I read *row->name and what is it returning?

I'll link the code I am reading: http://pastebin.com/MvLXkDCz

It's returning the first element `name` points to. It's like `row->name[0]` which should be used instead to avoid "*confusion*". — Iharob Al Asimi, May 18 '16 at 16:46
Learn about [*operator precedence*](http://en.cppreference.com/w/c/language/operator_precedence). — Some programmer dude, May 18 '16 at 16:47
@iharob if name is not initialized then what would get passed to sizeof(*row->name)? — don, May 18 '16 at 16:52
@don sizeof is calculated at compile time using only type information. — Matt, May 18 '16 at 16:53
@don As others said, [`sizeof` does not "*evaluate*" it's operand](http://stackoverflow.com/questions/8225776/why-does-sizeofx-not-increment-x) so nothing bad would happen, and it's actually a good way of passing the size to `malloc()` for example, for many reasons. For example, if you change the type of `name` you don't need to change *every* `malloc()` and also because it helps you allocate the correct size since it's always correct. — Iharob Al Asimi, May 18 '16 at 17:00
Questions are required to contain all relevant information **in the question itself**! Not a a link, nor as an image. — too honest for this site, May 18 '16 at 18:31

score 3 · Accepted Answer · answered May 18 '16 at 16:50

First the -> operator is evaluated and then the resulting pointer which is pointing to what name points to is dereferenced with *. So it's the same as

row->name[0]

and IMHO this is a better way of expressing the same (althought sometimes using the indirection operator * is clearer).

score 0 · Answer 2 · answered May 18 '16 at 17:55

The -> component selection operator has higher precedence than the unary * operator, so *row->name is parsed as *(row->name); you're dereferencing the row->name member.

The same is true for all other postfix¹ operators:

*arr[i] == *(arr[i])
*foo()  == *(foo())
*a.p    == *(a.p)
*p++    == *(p++)
*q--    == *(q--)

^{Component selection operators `.` and `->` are grouped with postfix operators even though they look like infix operators.}

C - how do I read * and -> together?

2 Answers2