sizeof

[Michael DeWulf]

Why does the code in example 1 output 100 and the code in example 2 output
4? Rather, I understand example 2's output, but why doesn't example 1
also output 4?

-Mike

----Example 1----
#include <iostream>

using namespace std;

int main()
{
char c[100];
cout << sizeof(c) << endl;
return 0;
}

----Example 2----
#include <iostream>

using namespace std;

int main()
{
char * c = "what's up?";
cout << sizeof(c) << endl;
return 0;
}

 

[Frederick Gotham]

Michael DeWulf posted:

int main()
{
char c[100];
cout << sizeof(c) << endl;
return 0;
}

This should print 100 on every implementation, as sizeof(char) is exactly 1
on every implementation. Therefore, sizeof( char[100] ) should be equal to
100 * 1.

#include <iostream>

using namespace std;

int main()
{
char * c = "what's up?";
cout << sizeof(c) << endl;
return 0;
}

In your former example, "c" is an array". In _this_ example, "c" is a
pointer. If the bell is ringing in your head yet, I suggest you read a good
book which explains the nature of arrays and pointers in C/C++.

 

[Jim Langston]

Why does the code in example 1 output 100 and the code in example 2 output
4? Rather, I understand example 2's output, but why doesn't example 1
also output 4?

-Mike

----Example 1----
#include <iostream>

using namespace std;

int main()
{
char c[100];
cout << sizeof(c) << endl;
return 0;
}

c is an array with 100 elements.

----Example 2----
#include <iostream>

using namespace std;

int main()
{
char * c = "what's up?";
cout << sizeof(c) << endl;
return 0;
}

c is a char pointer that takes 4 bytes of memory.

 

[Martin =?iso-8859-1?Q?J=F8rgensen?=]

c is a char pointer that takes 4 bytes of memory.

And it always does so, independent of what *it points to* (in this example
an arbitrary char-array). So if the char-array was bigger or smaller,
the pointer would still take 4 bytes of memory...


Best regards
Martin Jørgensen

 

[Pete Becker]

Why does the code in example 1 output 100 and the code in example 2 output
4? Rather, I understand example 2's output, but why doesn't example 1
also output 4?

char c[100];

c is an array of 100 char, so its size is 100. This is one of a couple
contexts where the name of an array does not decay to a pointer.

--

-- Pete

Author of "The Standard C++ Library Extensions: a Tutorial and
Reference." For more information about this book, see
www.petebecker.com/tr1book.

 

[Michael DeWulf]

On Sat, 16 Sep 2006, Pete Becker wrote:

Ok, thanks. That is what I was looking for. I was thinking it would
decay to a pointer.

Why does the code in example 1 output 100 and the code in example 2 output
4? Rather, I understand example 2's output, but why doesn't example 1 also
output 4?

char c[100];

c is an array of 100 char, so its size is 100. This is one of a couple
contexts where the name of an array does not decay to a pointer.

 

[Default User]

Ok, thanks. That is what I was looking for. I was thinking it would
decay to a pointer.

Please don't top-post. Message rearranged. See the FAQ for more details.


The major time the name of an array name is not converted in when used
with the address-of operator (&). So if you did &c, it would not be a
pointer to pointer to char, but a pointer to array 100 of char.



Brian (WUSTL grad)

 

[Ron Natalie]

And it always does so, independent of what *it points to* (in this example
an arbitrary char-array). So if the char-array was bigger or smaller,
the pointer would still take 4 bytes of memory...

Furthermore, it always points to a single char. If that single
char might be followed by others is no concern of the pointer.