C Programming - Pointers - Discussion

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.

int main()
{
int *str;
str = "%d";
printf(str, "5\n");
return 0;
}

What will be the output?

So, can we assign a format specifier to the variable ? If so, the above solution will be correct explain me?

@Nithiya.

The output will be garbage value.

@Shahin.

How it will give garbage value as an output?

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

%s is the format specifier type of string.

We can declare it anywhere in the local scope.

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

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.

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.