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C | Operators | Question 12

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  • Difficulty Level : Medium
  • Last Updated : 12 Oct, 2021
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int main() 
  char *s[] = { "knowledge","is","power"}; 
  char **p; 
  p = s; 
  printf("%s ", ++*p); 
  printf("%s ", *p++); 
  printf("%s ", ++*p); 
  return 0; 

(A) is power
(B) nowledge nowledge s
(C) is ower
(D) nowledge knowledge is

Answer: (B)

Explanation: Let us consider the expression ++*p in first printf(). Since precedence of prefix ++ and * is same, associativity comes into picture. *p is evaluated first because both prefix ++ and * are right to left associative. When we increment *p by 1, it starts pointing to second character of “knowledge”. Therefore, the first printf statement prints “nowledge”.
Let us consider the expression *p++ in second printf() . Since precedence of postfix ++ is higher than *, p++ is evaluated first. And since it’s a postfix ++, old value of p is used in this expression. Therefore, second printf statement prints “nowledge”.
In third printf statement, the new value of p (updated by second printf) is used, and third printf() prints “s”.

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