About struct and pointer in C++

Question

I would like to create a linked list.

For the following code:

struct sampleStruct
{
    int a;
    sampleStruct *next = NULL;
};
sampleStruct *sample = new sampleStruct;

What is the difference between sample.next and sample->next?

see http://stackoverflow.com/questions/1238613/what-is-the-difference-between-the-dot-operator-and-in-c — Vincent van der Weele, Apr 07 '13 at 11:25
The first one is when you have a struct (or a class) the second is when you have a pointer to a struct (or class). — john, Apr 07 '13 at 11:25
I understand now, thank you for your help :) I will do some research before asking next time, I have lost so many marks (cry) — redcap, Apr 07 '13 at 15:59

score 2 · Accepted Answer · answered Apr 07 '13 at 11:50

Okay to explain it in a more complete way. The most other guys had already wrote that you have to use '->' when ever you have a pointer. But you could also do this with '.', to do this you must respect the priority of the operators. We need the '*' to get the content of the pointer, but this has a lower priority than the '.', so you must write it into brackets to give it a higher priority, so when you want to do this with a '.' you have to wrote:

(*sample).next

and you see this is a complex syntax, to do this in a more easy way the '->' was introduced. With it you could write code in a more comfortable way.

So this is is equal to the example and it looks much better.

sample->next

score 1 · Answer 2 · answered Apr 07 '13 at 11:29

Since sample is a pointer, there can't be any way to access a data member through . rather than the indirection operator ->. For example, this won't compile:

sample.next; // error: member reference type 'sampleStruct *' is a pointer;
             //        maybe you meant to use '->'?

The error actually speaks for itself.

About struct and pointer in C++

2 Answers2