Thread: Segmentation fault char**

Find the below program,

Code:

#include <stdio.h>

int main(int argc, char** argv)
{
  char** ppch = "Hello World";

  printf( "%s\n", *argv );           
  printf( "%s\n", *ppch );           // Segmentation fault

  return 0;
}

why this is giving Segmentation fault while printing *ppcch? and why not in *argv? as both are pointer to pointer to char they should behave in same way.

but if I try to print the value value of ppch using

Code:

printf( "%s\n", ppch )

It is working fine but if I try this same thing with

Code:

printf( "%s\n", *argv );

It will print some junk values.

can any one give me the logical reason why this is happening?

--
Regards
Kunal

Thanks for the reply TheBigH, but your reply is not the answer of my question.

there should be a rule not to post example code that emits compiler warnings. or at least the submitter must acknowledge there are warnings being ignored on purpose.