Why doesn't a character array return a memory address when called directly?

Question

I understand from here that the name of an array is the address of the first element in the array, so this makes sense to me:

int nbrs[] = {1,2};
cout << nbrs << endl;   // Outputs: 0x28ac60

However, why is the entire C-string returned here and not the address of ltrs?

char ltrs[] = "foo";
cout << ltrs << endl;   // Outputs: foo

Because it's been overloaded to print the string for convenience. — Mysticial
– Mysticial, Commented Aug 3, 2012 at 22:18
A name of an array is not the address of the first item. They're very different. However, an array will sometimes pretend to be a pointer to the first item. — Mooing Duck
– Mooing Duck, Commented Aug 3, 2012 at 22:22
You're writing C++ rather than C. I would recommend you find a good C++ tutorial instead. — Roddy
– Roddy, Commented Aug 3, 2012 at 22:24

Jerry Coffin · Accepted Answer · 2012-08-03 22:18:30Z

8

Because iostreams have an overload for char * that prints out what the pointer refers to, up to the first byte that contains a \0.

If you want to print out the address, cast to void * first.

answered Aug 3, 2012 at 22:18

Jerry Coffin

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mathematician1975 · Accepted Answer · 2012-08-03 22:18:52Z

4

cout has operator<<() overloaded for char* arrays so that it outputs every element of the array until it reaches a null character rather than outputting the address of the pointer

answered Aug 3, 2012 at 22:18

mathematician1975

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pb2q · Accepted Answer · 2012-08-03 22:26:06Z

2

cout, and generally, C++ streams, can handle C strings in a special way. cout operators <<, >> are overloaded to handle a number of different things, and this is one of them.

edited Aug 3, 2012 at 22:26

answered Aug 3, 2012 at 22:18

pb2q

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