[SOLVED] c++ pointers

tarunchawla · 09-07-2011, 02:49 AM

I have two questions:

In c++ pointer is of 4 bytes,then how can we access the whole 4gb Ram in linux,I mean how it is mapped to the physical address.

Second is when I write a statement like this:
char *a="tarun";
as "tarun" is temporary ,then how memory is allocated to 'a'.It runs fine when I cout<<a; it prints tarun,just clear me the facts.

dwhitney67 · 09-07-2011, 05:25 AM

Quote:

Originally Posted by tarunchawla

I have two questions:

In c++ pointer is of 4 bytes,then how can we access the whole 4gb Ram in linux,I mean how it is mapped to the physical address.

Second is when I write a statement like this:
char *a="tarun";
as "tarun" is temporary ,then how memory is allocated to 'a'.It runs fine when I cout<<a; it prints tarun,just clear me the facts.

A pointer occupies 4-bytes on a 32-bit architecture, but 8-bytes on a 64-bit architecture. The kernel takes care of mapping virtual memory to physical memory. AFAIK, when running a user application, physical memory is never addressed directly; the app uses virtual memory. Here's some more info on this topic: http://en.wikipedia.org/wiki/Virtual_memory

As for your second question, the hard-coded string "tarun" is not temporary. It resides in the program space (memory) for the entirety of the life-time of the program. The variable 'a' merely points to this location in memory. And since it appears that you are developing in C++, you can avoid compiler warnings by declaring 'a' as a const char* type. The const is necessary because the "tarun" string cannot be modified.

ta0kira · 09-07-2011, 06:22 AM

Quote:

Originally Posted by tarunchawla

Second is when I write a statement like this:
char *a="tarun";
as "tarun" is temporary ,then how memory is allocated to 'a'.It runs fine when I cout<<a; it prints tarun,just clear me the facts.

That is similar to the following:

Code:

static const char a_static[] = "tarun";
char *a = a_static;

The problem with what you wrote is that a points to to a non-const char, which means the string might inadvertently be changed. That would be "undefined behavior", possibly having negative side-effects. You should either use const char *a = "tarun"; or char a[] = "tarun";. The second makes a copy on the stack.
Kevin Barry