a question about obtaining address of variable .......

purpleburple · 07-04-2002, 11:24 PM

Hi. New to C and, linux <-- (sorta). In my C book (one of them anyways) I do the following example

#include <stdio.h>

int main(void)
{
char c;

printf("The address of c = 0x%p\n", &c);

return 0;
}

the problem is the ouput is

The address of c = 0x0xbfffa50

the example says to put the 0x in front of '%p' but it looks like it's printing another '0x'
and the output is supposed to be '0xbfffa50'

Im following the example except I think the author is using a windows machine
Am I doing something wrong here?

thanks

Config · 07-05-2002, 02:43 AM

Weird... if you output a pointer, the 0x gets added anyway. I have done this on Windows this way. And from what you're saying, Linux does this as well.
Except if... what happens if you ommit the & before the Variable c in the printf statement? I know this refers to the address of this variable, but may be, if you have a %p in the printf, it does that already.
Otherwise, you can send an E-Mail to the author

abi_sh · 07-05-2002, 10:32 AM

y r u using & before the variable name its used only when u have to accept the values into the variable.u can use this following code rather
main()
{
char c;
printf("the address of c =%u",c);
return 0;
}

llama_meme · 07-06-2002, 05:28 AM

You must pass '&c' to printf, not just 'c'. Remember, printf is not a builtin command - it has no way of obtaining the address of a variable that you pass it by value. Clearly the prepending of 0x is just a little formatting niceity provided by printf when it prints an address, it's not really a problem, is it?

Alex