Interesting example indeed!
Replace the declaration of s to
char s[] = "hello";
and see "Hello" being printed :-)
The point is: in your program is is only a pointer. When you pass s
as a parameter to printf, the compiler assumes that only s is being
used so the (effective) assignment
*s = 'H'
is deleted as dead code when optimization is enabled.
If you do not optimize the program, you get a segmentation fault (rightly so).
When s is converted from pointer to array, the assignment s[0] is not
a dead assignment because the array base address is passed to the
printf function.
Hope that helps.
Uday.
----------------------------------------------------------------------
Dr. Uday Khedker
Professor
Department of Computer Science & Engg.
IIT Bombay, Powai, Mumbai 400 076, India.
email : [email protected]
homepage: http://www.cse.iitb.ac.in/~uday
phone : Office - 91 (22) 2572 2545 x 7717, 91 (22) 2576 7717 (Direct)
Res. - 91 (22) 2572 2545 x 8717, 91 (22) 2576 8717 (Direct)
----------------------------------------------------------------------
Godmar Back wrote, On Tuesday 14 September 2010 11:08 PM:
int
main()
{
char * s = (char *)"hello"; // read-only
printf("%s\n", s);
s[0] = 'H'; // gcc -O elides this
printf("%s\n", s);
return 0;
}
Could someone briefly provide justification/rationale for this decision?
