Pointers and array of char[]

[strokebow]

	char name1[50];
	char *pname;

	cout << endl << i << ". Please enter your full name: ";
	cin.getline(name1, 50);

	pname = name1;	// The address of the pointer is now the address of the first element of the array

	system("pause");
	cout << "\n*pname is " << *pname << " and " << "pname is " << pname << endl;
	system("pause");

Hi,

I cant understand for the life of me why pname becomes the full length of the string.

Can someone explain this code for me please?

Thanks

[matsp]

pname holds the address of the first member of your array, so the memory represented by pname is the same as name1. The length of name1 is defined by the 50 you give, and if you enter a string into that memory, strlen(pname) and strlen(name1) will give the same result (actually, if you where to stop inside strlen() and look at the string argument, it would be the same for both of those cases).

A pointer is really just a variable that holds the memory location of something - in this case name1.

Note also that the comment is slightly misleading. It should say "the address in the pointer is now the address of the first element ...", not "the address of the pointer". The address OF the pointer is still the same as when the variable is first introduced a few lines up - 52 more than the start of name, most likely (52 because it's rounded to an even 4 - it could also be 56 if the machine has 64-bit (8 byte) pointers).

--
Mats

[anon]

Another thing is that cin and cout simply have a special behaviour for arguments of char* type. If you tried with an array of a different type, getline would not compile, and cout would print the address of the pointer, not the contents pointed to.

So if you want to see the address of the char*, you can cast it to a void*, so that cout would treat it like any other pointer:

cout << "\n*pname is " << *pname << " and " << "pname is " << (void*)pname << endl;

[nucleon]

So your question is why does *pname output a single char while pname outputs the entire string? pname is a pointer to char, so *pname is a char, thus you get a char. Being a pointer to char, pname is considered by cout to be a C-style string (the char array must be zero-terminated), so it outputs the whole string. As anon mentioned, casting to another pointer type will output the address.