Thread: Question about pointers

First I would want to know why,if I declare the following:

Code:

char a[]="abcd";

and then print:

Code:

printf("%s\n",a);

it;s ok and even

Code:

printf("%c\n",a[1]);

is good
But when I replace the %c into %s so I get:

Code:

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

that's an error
I mean , why can;t it print bcd?
is it some logic point I missed or simply syntax?

Another question:
I have teh following declarations:

Code:

	char *c[]={"door","me","pointers","balul"};
	char **cp[]={c+3,c+2,c+1,c};
	char ***cpp=cp;

And the following printing lines:

Code:

    printf("%s\n",**++cpp);
	printf("%s\n",++**cpp);
	printf("%s\n",*cpp[-2]+3);
	printf("%d\n",***cpp);
	printf("%d\n",**cp);

Well I understand the 1st and 2ed lines, it;s the 3rd and 4th I dun quite get...
Why the 3rd line gives trash
And the 4rth gives "inters"?

What you could do is:

Code:

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

But it's redundant really, since a[1] is effectively a de-reference of a at index 1. So, a more common sense way to do, is what laserlight suggested.

Thank you I understood that part

But what about the other I can;t figure that out.

Thanks for the help :P.
after realizing that it was ++cpp...somehow I missed that ._. (I didn;t keep the change but always followed the initiate value...)
All became clear.

Okies now all is clear as ice ^^ thanks for the help!