char array in C++ - expression must be a modifiable lvalue

[Shervan360]

Hello,

Could you please explain why we have an erorr?

#include <iostream>
using namespace  std;

int main()
{
	char s[80];
	cin.getline(s, 80);

	while (*s != '\0')
	{
		cout.put(char(tolower(*s)));
		s++; // error
	}
	return 0;
}

.

[Ganado]

Incrementing (++ operator) an array makes no sense. An array is not a pointer, although an array will degrade into a pointer if you are using it by value.

Copy the array's pointer to another variable first, e.g.

char* ptr = s;

// ...

ptr++;

[Shervan360]

Thank you, but the name of an array is the pointer of the first element. isn't it?

[mzimmers]

No, not really, though it seems a lot of schools teach that these days.

An (unindexed) array name will point to the address of the first element, but an array is not a pointer, and shouldn't be treated as one. Your program would benefit from a char pointer that is initialized to the first element of the array, and advanced within your while loop.

[jonnin]

Thank you, but the name of an array is the pointer of the first element. isn't it?

the c++ language ensures that the compiler converts the name of an array to a pointer for you no questions asked. So syntactically, yes, it may as well be, for a rather limited scope of things you can do with it. Its more like a heavily constantified pointer that you can't new, delete, assign, move, or much else -- all it gives you is a memory address where you can look around.

More technically and precisely, no, it isn't. Consider this: you can use [] on a pointer. Does that make a pointer an array? (spoiler: no).

Think of it as just a shortcut to avoid having to type &array[0].

[Ganado]

^good analogy, it's very much like avoiding "&array[0]", because just like how arr++ doesn't work, neither would (&array[0])++ (trying to increment a temporary value).