Assigning char array to pointer

I must be confused here because I would think this says assign the address of a 1d array to the pointer; however, when I try to compile this, I get an error: error: cannot convert char (*)[4]' to 'char*' in assignment p = &input_line; It looks like I should be doing this:

const char input[] = "test?"; int quest_count = 0; const char *i = input; while(*i)

This doesn't make sense to me because you are assigning a char array to a char pointer which stores an address.

This doesn't make sense to me because you are assigning a char array to a char pointer It makes sense because arrays decay to pointers.

p = &input_line; Tries to assign a "pointer to a char array", to a "pointer to char ". Such types are incompatible. Did you mean to write p = input_line; ?

Use neither pointers nor arrays in C++. There is no reason to, as there are better alternatives nowadays.

2 Answers 2

The error in this line of code:

Can be resolved by changing it with:

p = input_line; 

That's because your're assigning the memory direction of the array. Its value is a pointer to a char pointer. That's why the error is raised. Remember, the operator & gives to you the variables memory direction.

A pointer stores a memory direction of an object. An array is a sequence of objects by a certain type that are located in consecutives reserved amount of space in memory.

Each index of an array is a number composed of the digits from 0 to 9. Each element of an array is an object that you can take the address of, like a pointer to an object memory location. In an array the objects are located in consecutive memory locations. When you assign an array to a pointer, you're assigning the pointer to the arrays first element, it's array[0] .

When you increase by 1 the pointer value, the pointer will point to the next object in a memory location. So, arrays and pointers have similar behavior. If you assign to a pointer an array and then you increase by 1 the pointer value, it will point now to the object in the array.

This is not only for char type, it's for every type in C++. In this page you can get more information about pointers and array. You must note that the pointer and the array must contain or point to the same variable type.

int* ptr; int a[5]; ptr = &a[2]; // &a[2] is the address of third element of a[5]. 

An output example from the example in this page is:

Displaying address using arrays:

&arr[0] = 0x7fff5fbff880 &arr[1] = 0x7fff5fbff884 &arr[2] = 0x7fff5fbff888 &arr[3] = 0x7fff5fbff88c &arr[4] = 0x7fff5fbff890 

Displaying address using pointers:

 ptr + 0 = 0x7fff5fbff880 ptr + 1 = 0x7fff5fbff884 ptr + 2 = 0x7fff5fbff888 ptr + 3 = 0x7fff5fbff88c ptr + 4 = 0x7fff5fbff890 

As you can note in the output example, both are pointing to the same memory location, so you can access the objects from both methods.

Formally, in the C++ 11 standard is mentioned that:

Array-to-pointer conversion:

An lvalue or rvalue of type “ array of N T ” or “ array of unknown bound of T ” can be converted to a prvalue of type “pointer to T ”. The result is a pointer to the first element of the array .

You can see those pages for more information about this theme: