regarding variable storage

[madhuti]

char *s = "Hello"; // 1
where does this "Hello" will store in memory region, in data segment or stack..
As *s is a pointer of auto variable and we are not allocating memory for that. But we directly assigning value.
Can anyone clarify this.

AND

what about volatile key word. where it will sit in the memory.

[gerard4143]

If you try compiling your program with -S switch it will produce an assembler file so you can see where and what the compiler is doing with your variables. Check out the code examples below:

test.c C code example

#include <stdio.h>
#include <stdlib.h>

__volatile__ int myint = 1234;

void myfunc(void)
{
	char *s = "hello, world!\n";
	fprintf(stdout, "%s\n", s);
}

int main(int argc, char**argv)
{
	myfunc();
	fprintf(stdout, "myint->%d\n", myint);
	exit(EXIT_SUCCESS);
}

Assembler file generated by gcc -S test.c(I only copied the section that mattered)
test.s

	.file	"test.c"
.globl myint
	.data
	.align 4
	.type	myint, @object
	.size	myint, 4
myint:
	.long	1234
	.section	.rodata
.LC0:
	.string	"hello, world!\n"
.LC1:
	.string	"%s\n"
	.text

You can see that char *s is in the rodata(read only data) section of the code. As for the second question, the volatile keyword informs the compiler that the variable may change unexpectedly so don't perform any optimization on it....Gerard4143