Question: If we have a `char *hello` - and the string is `"hello"`

and i do 

```
char *ptr;
ptr = hello;

```

then ptr will be pointing at `'h'`, correct?

Now I have just done an assignmnet in this and completed it using the following terms

if i wanted to move the pointer to the next chatachter i would just do `ptr++`. If i wanted to use the value of the pointer for some check, i would use `if(*ptr == '\0')...`

When i was doing the assignmnets our teacher gave us some pre built methods, and they used stuff like 

```
*string++ = *s++;

```

ok, so why would we want to do `*string` (which gets a value) - and combine it with `++`

I hope i make sense in explaining what is not clear. Its just I managed to do the whole assignment with ptr++ to move to next element or \*ptr to check its value

Answer: 
> 
> so why would we want to do \*string(which gets a value) - and combine it with ++
> 
> 
> 

When `*string` is on the left hand side of the equals, it doesn't get the value, it **sets** the value.

The statement `*string++ = *s++;` is equivalent to:

```
*string = *s;
s++;
string++;

```

This is because `x++` is the [postfix increment operator](http://msdn.microsoft.com/en-us/library/e1e3921c.aspx) (as opposed to `++x` which is the prefix increment operator). The pointer x is updated but the value that `x` was originally pointing to is used in the expression.

Personally I'd say that the one-liner is more confusing to read and you should generally try to avoid complex expressions with side-effects. But in this case it's a fairly standard idiom in C, so you might as well get used to it.
> 
> then ptr will be pointing at h, correct?
> 
> 
> 

~~Yes.~~ No. See [@paxdiablo's answer](https://stackoverflow.com/questions/2305805/stuck-with-syntax-in-c-pointers/2305813#2305813).

> 
> `*string++ = *s++;`
> 
> 
> 

This statement can be viewed as

```
*string = *s;
string ++;
s ++;

```

That means:

1. Copy the character in `s` to the location pointed by `string`.
2. Move both `s` and `string` to the next character.