Precedence & Associativity Questions

#include<stdio.h> int main() {   char *s[] = { "knowledge","is","power"};   char **p;   p = s;   printf("%s ", ++*p);   printf("%s ", *p++);   printf("%s ", ++*p);   getchar();   return 0; }

Output: nowledge nowledge s

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 psotfix ++, 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”.

Question 2: int main() {   int x, y = 5, z = 5;   x = y==z;   printf("%d", x);   getchar();   return 0; }

The crux of the question lies in the statement x = y==z. The operator == is executed before = because precedence of comparison operators (<=, >= and ==) is higher than assignment operator =.

The result of a comparison operator is either 0 or 1 based on the comparison result. Since y is equal to z, value of the expression y == z becomes 1 and the value is assigned to x via the assignment operator.

Question 3:

int main() {   char arr[]  = "geeksforgeeks";   char *ptr  = arr;   while(*ptr != '\0')       ++*ptr++;   printf("%s %s", arr, ptr);   getchar();   return 0; }

Output:  hffltgpshfflt
Explanation:
The crust of this question lies in expression ++*ptr++.
If one knows the precedence and associativity of the operators then there is nothing much left. Below is the precedence of operators.

Postfixx ++            left-to-right
   Prefix  ++             right-to-left
   Dereference *          right-to-left

Therefore the expression ++*ptr++ has following effect
Value of *ptr is incremented
Value of ptr is incremented

Question 4:

int main()
{
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
return 0;
}
Output:0 0 1 3 1
Explanation:Logical operations always give a result of 1 or 0 . And also the logical
AND  (&&)  operator  has  higher  priority over   the  logical  OR  (||)  operator.  So  the
expression  ‘i++ && j++ && k++’ is executed first. The result of this expression is 0
(-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR
operator always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the
value of m is 1. The values of other variables are also incremented by 1.

Question 5:

int main()
{
int i=10;
i=!i>14;
Printf ("i=%d",i);
}
Output:i=0
Explanation:In the expression !i>14 , NOT (!) operator has more precedence than ‘
>’ symbol.   !  is a unary logical operator. !i (!10) is 0 (not of true is false).   0>14 is
false (zero).