C ++：char * vs char（*）[]

Question

I'm new with C++ and came to this problem, here is my code:

shared_ptr<char[]>var(new char[20]);

char *varptr = *(var.get());

So I'm creating smart pointer of char array.

The problem that I'm having is that during compilation it gives me error saying:

cannot convert argument 1 from 'char *' to 'char(*)[]'

As the declaration of shared_ptr::get says T* get() const noexcept; it returns the pointer of template, in my case pointer to char[] . Dereferencing it should give me the char[] and assigning it to char* should be ok.

It seems like I'm missing something and can't figure out what.

What is the difference between char* and char(*)[] ? Isn't char(*)[] just a pointer of char array? Shouldn't assigning it to char* be ok?

Answer 1

char(*)[] is a pointer to an array (of unknown size) of char , it can point to an array.

char* is a pointer to a char , it can point to an element of an array.

Semantically the two types are very different things.

To get a char * from a char(*)[] , you need to get a pointer to an specific element of the array (typically the first element):

char* varptr = &(*var.get())[0];

And if all you want is a string, then use std::string instead.

---

To explain the difference between pointer to element, and pointer to array, consider the following array:

char a[3];

A pointer to the first element ( a[0] ) can be expressed as &a[0] , which is what plain a decays to. A pointer to the whole array is &a .

If we show it how it's laid out in memory, with some "pointers" added, it's like this:

+------+------+------+-----
| a[0] | a[1] | a[2] | ...
+------+------+------+-----
^      ^      ^      ^
|      |      |      |
a      a+1    a+2    |
|                    |
&a                   (&a)+1

Even through the pointers a (which is equal to &a[0] ) and &a point to the same location, they are two different pointers and doing pointer arithmetic on them will do different things.