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

An alternative way of invoking the main function would be

Code:

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

This will hopefully make it more clear what argv[] actually is. It's an array of strings. argv[0] is the name of the program and then argv[1] through to argv[argc-1] are the actual command line arguments. You'd print them this way:

Code:

printf( "%s\n", argv[1] );

Your problem with ppch is similar. Declaring it as double pointers makes it look like you want an array of strings (which is an array of arrays of chars).

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

Originally Posted by kunalnandi

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.

Why are you initialising ppch with "Hello World"? "Hello World" is a char[12] rvalue, which means that it is converted to a char*, which clearly is not a char**.

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.