C Programming - Pointers - Discussion

str has "%s" stored in it so the printf will work like,
printf(Str,"K\n"); => printf("%s","K\n"); => will print K as it's string and \n is for newline.

I think that "%s", "%d", . Etc are pointers to string, integer, etc respectively themselves. That is why we use them in printf as (pointer to the variable of the type of its content, then the variable name).

So when we write str="%s" we are assigning a pointer to another pointer.

Your explanation is good for understanding. Thank you @Megha.

Can anyone explain me this:

If,
int *str="abcd"
Then first character will be 'a' not '"'(Double quotes).

Then, if we initialize as,
str = "%s";
printf(str, "K\n");

Then printf requires "--"(double quotes).

And if we replace str value in printf function, then it will write as,
printf(%s,"K\n");

Got my point everybody? Please reply me.

If we try to print only str it will show (null) as output and in the code, there is no need of str.
It is only taking printf(" ");
Whatever you write inside the quotes is only printing that's it.

%s is the format specifier type of string.

We can declare it anywhere in the local scope.

Change %s with %c, %x. You will see that str is replace by different output.

Its like
char str[]="good";
printf("%s",str);
its jst like replacing str with "good"
it becomes printf("%s","good");
output will be:
good

int i;
int main()
{
int i;
i=78;
void *vptr;
vptr = &i;
printf("%d",& *vptr);

\\as this gives garbage value because type casting should be done.(*(int*)vptr)); and in calling function we have given **q.we know that *q=&vptr and **q=value of vptr so vptr value is 0 because it is initialised with void data type.and size to store 0 we get the size to store 0 because any pointer by default gets int. So we get the size and 0 are stored.

So why this not possible when I replace char data type with int?