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Determine Output:
void main()
{
      static char *s[] = {"black", "white", "yellow", "violet"};
      char **ptr[] = {s+3, s+2, s+1, s}, ***p;
      p = ptr;
      ++p; 
      printf("%s",*--*++p + 3); 
}

A. te

B. ow

C. et

D. ck

Answer: Option D

Solution(By Examveda Team)

In this problem we have an array of char pointers "s" pointing to start of 4 strings. Then we have ptr which is a pointer to a pointer of type char and a variable p which is a pointer to a pointer to a pointer of type char. p holds the initial value of ptr, i.e. p = s+3. The next statement increment value in p by 1 , thus now value of p = s+2. In the printf statement the expression is evaluated as follows: *++p causes gets value s+1 then the pre decrement is executed and we get s+1-1 = s which is pointing to the word "black". The indirection operator now gets the value from the array of s and adds 3 to the starting address. The string is printed starting from this position. Thus, the output is 'ck'.


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