Issue with pointer to character array C++

• The declaration char buffer[9] = "12345678"; creates an array of 9 chars.
  Here, buffer is the address of its first element but not of the array.

• char* pBuffer = buffer; is a correct expression as pBuffer is a pointer to char and can address the first element.

• But the expression char* pBuffer = &buffer is wrong, because pBuffer can't address an array. (error in your code, &buffer address of array as explained below)

Difference between buffer and &buffer

&buffer means address of array. The values of buffer and &buffer are really the same, but semantically both are different. One is an address of a char, while the other is an address of an array of 9 chars.

buffer[9] = "12345678";


+----+----+----+---+---+----+----+----+---+----+ 
| '1'| '2' |'3'|'4'|'5'| '6'| '7'|'8' | 0 |  ...........
+----+----+----+---+---+----+----+----+---+---+----+  
 201   202  203 204 205 206  207   208 209 210  211
  ^     ^                                         
  |     |                                         
(buffer) (buffer + 1)                                         
|                                         |
|-----------------------------------------|--------
|201                                      | 210
  ^                                          ^
  |                                          |
(&buffer)                                 (&buffer + 1)     

_{I used decimal numbers for address instead of hexadecimal}

Even though the value of buffer is 201 and the value of &buffer is 201, their meaning is different:

• buffer: first element's address—its type is char*.
• &buffer: complete char array's address—its type is char(*)[9].

Additionally, to observe the difference, add 1 :

buffer + 1 gives 202 that is the address of the second array element '2' but
&buffer + 1 gives 210 which is the address of the next array.

On My System, I write following code:

int main(){
   char buffer[9] = "12345678";
   char (*pBuffer)[9] =  &buffer; 

   printf("\n %p, %p\n",buffer, buffer+1);
   printf("\n %p, %p\n",(&buffer), (&buffer+1));
}  

And the output is as below:

0xbfdc0343, 0xbfdc0344

0xbfdc0343, 0xbfdc034c

_[ANSWER]

That's the reason Error is:

error: cannot convert 'char ()[9]' to 'char' in initialization

You are trying to assign 'char (*)[9]' type value to char*.

• char (*ptr2)[9]; Here ptr2 is pointer to an array of 9 chars, And this time
  ptr2=&buffer is a valid expression.

How to correct your code?

As in Nate Chandler's answer:

char buffer[9] = "12345678";
char* pBuffer =   buffer;   

or another approach,

char buffer[9] = "12345678";
char (*pBuffer)[9] =  &buffer;       

Which you choose depends on what you need.

array decays to pointer, you only need

char* pBuffer = buffer;

Also checkout: std::decay

   std::cout << "int[] -> int*; // " << std::boolalpha
              << std::is_same<int *, std::decay<int[]>::type>::value;

Output:

int[] -> int*; // true

The error message is very clear; &buffer is of the type char (*)[9], i.e., a pointer to an array of char with 9 elements (yes, arrays are fully fledged types).

That is not the same as a char*. However, arrays degrade to pointers to the first element when needed, so...

char *pbuffer = buffer;

Name of array indicates it's base address, so if you just write buffer, it means address of buffer[0]. If you are using &buffer, you need to take it in char** not in char *.

So,

char* pBuffer = buffer;

Would be correct as suggested.

because the compiler can't implicitly convert char()[] to char. So you need explicit type conversion

char buffer[9] = "12345678";
char* pBuffer = (char*)&buffer;