Consider the following C function:

void swap (int a, int b)
{
   int temp;
   temp = a;
   a = b;
   b = temp;
}

In order to exchange the values of two variables x and y.

Consider the following C program segment:

char p[20];
char *s = "string";
int length = strlen(s);
for(i=0; i < length; i++)
  p[i] = s[length-i];
printf("%s",p);

The output of the program is

The following finite state machine accepts all those binary strings in which the number of $$1's$$ and $$0's$$ are respectively

Which of the following grammar rules violate the requirements of an operator grammar ? $$P,$$ $$Q, R$$ are non-terminals and $$r, s, t$$ are terminals. $$$\eqalign{ & 1)\,\,\,P \to Q\,R\,\,\,\,\,2)\,\,\,P \to Q\,s\,R \cr & 3)\,\,\,P \to c\,\,\,\,\,\,\,\,\,\,\,4)P \to Q\,t\,R\,r \cr} $$$