pointer to char

[cuteanu]

why is it that when a pointer is pointed to an int, it displays the address while when a pointer is pointed to char there is no output, in the following program:
int *pptr;
char *cptr;
cout<<pptr; //o/p shows the hexadecimal address
cout<<cptr; // no o/p, it compiles but the output screen is blank

Please help, this has caused a hindrance in understanding pointers

Please also refer simple tutorials to understand pointers

[Savage]

why is it that when a pointer is pointed to an int, it displays the address while when a pointer is pointed to char there is no output, in the following program:
int *pptr;
char *cptr;
cout<<pptr; //o/p shows the hexadecimal address
cout<<cptr; // no o/p, it compiles but the output screen is blank

Please help, this has caused a hindrance in understanding pointers

Please also refer simple tutorials to understand pointers

Valid hexadecimal address of char is containing only 0(and x ofcourse),so cout don't print it out.If address of a pointer to char contain value differant then 0 it will print out error:cannot convert int to char*.U can test this by using printf instead.

Savage

[weaknessforcats]

The problem is the << operator.

This is really a function call so something like:

ostream& operator<<(ostream& os, char);
 

This function is programmed to display the letter and not the numeric value of the char.

Do this instead:

char c = 'A';
int temp = c;
cout << temp;
 

Now the << operator is a call to this function:
CODE]
ostream& operator<<(ostream& os, int);
[/code]
This function is programmed to show the numeric value of the int.

Remember, the << operator means FORMATTED OUTPUT. The formatting is accomplished by an appropriate overload of the << operator.

Avoid printf(). This function cannot be used to output to anythign other than the screen. AND is only works with the buiklt-in types of C. However, a << function will output to whereever the os object says to. That means without changing your code you can output to the screen, the disc, or maybe even send a packet across a network.