Re: [Dev-C++] Very simple Q 'm sure...

[Jason H. <jas...@bt...>]

The * is the pointer operator. The program below is pretty bad, because it creates a pointer but
doesn't initialise it to point at anything, and when you get input to it you are writing to
unallocated memory, which could cause a memory access violation.

Books dedicate whole chapters to explaining what a pointer is and how to use them, so I won't here.

char *PName;
or
char* PName;

These two are the same statement really, declaring a pointer to a character or the start of an array
of characters. You should know that a null terminated array of characters in C/C++ represents a C
style string.

char PName;

This just declares a character. Essentialy allocating space for a single character (a byte) which is
why you get only one character from it.

A modified version of your program would be:

#include <iostream>

int main(int argc, char* argv[])
{
using namespace std;
char namebuf[32];
char *PName = namebuf;
cout << "Type passenger name\t";
cin >> PName;
cout << PName;
return 0;
}

Which won't cause any memory errors as long as the name doesn't go over 31 characters (remember 1
char for the null terminator).

Lookup pointers and arrays in your C++ reference of choice.

Jason.

----- Original Message -----
From: "Daniel Glenfield" <sp...@ho...>
To: <dev...@li...>
Sent: Saturday, April 06, 2002 4:37 PM
Subject: [Dev-C++] Very simple Q 'm sure...


Look at the declaration for PName. It is declared as char *PName. The line
cout<<PName; outputs the entire name.
HOWEVER if I declare the variable as char PName (with no *) only the first
letter is outputted.
Can anyone tell me what the * is doing?

Thanks,
Daniel



#include <iostream>

#pragma warning (disable : 4101)

int main(int argc, char* argv[])
{
using namespace std;
char *PName;
cout << "Type passenger name\t";
cin >> PName;
cout << PName;
return 0;
}



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