hi the article was good.
where do constant variable stored in ???

Very nice article.. It helped me lot. Thanks

Very nice article.. It helped me lot. Thanks

Hi Narendra,
I like your explaination.
I have few doubts if you can explain it: What do you mean by “Read from program file by exec” And “initialized to zero by exec” ?

hay hi
can anyone explain if i want to create matrix of size greater than the physical memory is it possible?and yes then how?and show with example.
please help me..

Hi ,In Example 1 ,the size of BSS segment is 8bytes .Can you explain how BSS has 8bytes?

Thanks a lot. I highly appreciate your work.

Thanks … the best article available for emory layout on Net…

very nice explanation Mr. Narendra, also making your points substantiated by the real memory stats is marvellous. Please continue this beautiful practice to teach others through learning by doing.

very good.. hats off

in the above program if i add a structure and assign value to its members, it is not reflected in the memory layout. Can anybody explain ?

Very Good Explaination…. !!! Simple and well explained

Excellent!

Nice explanation…

But i get a different result(as follows) , am i missing something ?

int a[5];
main()
{}
——————————————-
bash-3.2$ gcc -c 1.c

bash-3.2$ size 1.o

text data bss dec hex filename

62 0 0 62 3e 1.o

———————————————
Why the data and bss segment shows zero here ?

Where is the memory-mapping-segment, where shared memory, memory mapped I/O,… is hold???

it’s very useful, thank you for providing.

Excellent information. Adding small finding the static/global variables that are initialized to non zero will reside in data segment(Initialized data segment). The other static/global variables not initialized or initialized to zero will reside in BSS.

Nice Explanation indeed 

Thanks for explanation.

nice explanation

awesome,excellent explanation sir than k u so much for posting this.If you dont mind can you post any article about virtual destructors and why cant we have virtual constructors in c++..

awesome explanatation

sir i want to know that what is the maximum size of a heap for any process

Good tutorial. Output would be different depends on compiler you use, but the thought introduced here is excellent.

may I know the above examples are on which IDE . I want to check it . I am using eclipse. please help me…….

in my program the starting address and size are as follows:

code end: 90a00000
data start:90b00000 size 3mb
bss start: 91200000 size 1mb

but when i am declaring a global variable..its address is:910bead0
can anyone tell me what is going on?

Here in the above Programming example. when the global and static variables are intialized then it increased memory size in data segment (fine).

but why it did not reduce the size in unintialized data segment.

please reply ASAP

In which segment static variable declared in function goes,and why?

We say, global variables are stored on heap (BSS) but here we can see that initialized global variable goes into DS. What shall we say about this ? Is it still on the heap or has it gone into stack.

hello sir,
why integer value is increase with compiler(32 bit, 64 bit) , other wise all are some (float,char etc.) ?

thanks .. even the foll tutorial is real nice

http://www.ualberta.ca/CNS/RESEARCH/LinuxClusters/mem.html

I had compiled a 3 line program

 

int main(){

return 0;

}

 

and got the following:
gaurav@pruthi-linux:~$ size a.out
text data bss dec hex filename
1053 276 4 1333 535 a.out

Since there is no data, no include files why its showing bss value as 4 and data as 276.

interesting article though it would be good if the underlying assumptions are made explicit
a) that the underlying machine is x86 and OS as Linux.

Not every OS has a per process heap like Linux. viz Vxworks has a common system vide heap for all processes.

Also note that bss is not initialized to zero by kernel rather its the preparatory code that gets called before invocation of main in every c Program. true it gets invoked in the context of execve but technically its not kernel that is clearing this rather the preamble code of __main.

The Stack section too deserves some mention. for ARM, the return address and input parameters to a subroutine are in registers itself namely R13-LR, r14 as SP and r0-r3 for input parameters. its different on CISC machines though.

 

/* Paste your code here (You may delete these lines if not writing code) */

 

Thanks for the information. Simply awesome!

Also, can someone explain whether stack always grow downwards ??

Really an awsm explanation.
Can you please also explain that where are the pointers stored ??

 

/* Paste your code here (You may delete these lines if not writing code) */

 

Nice Article
I have one doubt.
data size in the BSS is getting rounding off to meet address (x86 or x64) architecture
but in (initialized)data segment size is not getting rounding off

Can you explain why?

Eg: my machine is x64 (64 bit )
if i have a int variable in BSS it is taking 8 bytes

Before:

#include

int main()
{
return 0;
}

text data bss dec hex filename
1044 496 16 1556 614 a.out

After:

#include

int a;
int main()
{
return 0;
}

text data bss dec hex filename
1044 496 24 1564 61c a.out

As u can see 8 bytes increased even-though size of int is 4 bytes in my machine

#include

int a=9; //since initialized it goes to Data segment
int main()
{
return 0;
}

text data bss dec hex filename
1044 500 16 1560 618 a.out

but if i have int in data segment it is taking only 4 bytes as u can see data segment changed from 496 to 500

Y rounding off in BSS according to machine architecture (64bit-8 bytes) and Y not in data segment…..?
Can you explain…

 

/* Paste your code here (You may delete these lines if not writing code) */

 

How a structure will be stored in memory in c. For example
struct node
{
int a;
char *p;
}var;

is there a way to do the same on windows

 

/* Paste your code here (You may delete these lines if not writing code) */

 

Why the bss size is 8 bytes by default?.

 

/* Paste your code here (You may delete these lines if not writing code) */

 

very informative code 

In which part of the memory locally declared constant variable and globally declared constant variable will store?. Whether both the variable will store in data segment(initialized read only area )?

How to see total stack and heap size?

thank you,for giving such an information…

Nice article, very informative.

Nice and informative for beginners and advanced .

nice post…figures very helpful…thnx 

nice post…very helpful

 

/* Paste your code here (You may delete these lines if not writing code) */

 

Awesome …

hey. I tried the exact same code in a gcc compiler. I am not getting any difference in the bss value for int global;(outside main) and for int local;(inside the main). Nor am I getting any difference in bss for int global=0 and int local =0. Is this example meant to be compiler dependent?

char* string = “hello world” ;
const int a=5;

according to the post both string and a are stored in initialized read-only area.
so when we write
a=6 // modifaication of a gives compilation error and

string[0]=’m’ // or *string=’m’ gives segementation fault why.

super , awesome

char s[] = “hello world” will not be stored on data segement if it’s local to function. It will be stored on stack.
In case of global, yes it will be on data segment.

Good content. Adding further, the following example increases TEXT section size

 

/* TEXT - As good as integer literal */

static const int read_only_global = 10;

/* Initialized DATA */

int read_write_global = 100;

int main(int argc, char *argv[], char *evn[])

{

	/* BSS */

	int static localized_global;

	return 0;

}

 

   text	   data	    bss	    dec	    hex	filename

    804	    256	     12	   1072	    430	mem_layout