CODEWITHSUNDEEP
c
Ashwyn
Ashwyn

Reputation: 669

order of evaluation of || and && in c

If the precedence of && is greater than that of ||, shouldn't this code evaluate --b && ++c first, and thus the output should be 1 2 4 11. But here it seems to be short circuited to give 1 2 5 10. Please help!

int x;
int a=1,b=5,c=10;
x=a++||--b&&++c;
printf("%d %d %d %d\n",x,a,b,c);
return 0;

Upvotes: 5

Views: 1893

Answers (5)

Seth Carnegie
Seth Carnegie

Reputation: 75130

The precedence of && is higher, which means it binds tighter to the things on the left and right of it than ||. So that expression is equivalent to

a++ || (--b && ++c)

|| only evaluates the thing on the right if the expression on the left evaluates to non-0. Since a is 1, only a++ will be evaluated, and b will not be decremented and c will not be incremented.

Upvotes: 2

Sushant Sharma
Sushant Sharma

Reputation: 447

There are two concepts at work here

  1. Operator associativity
  2. Compiler optimization Short Circuiting

In C, || operator is left associative. As a result, a++ will be evaluated first. Since the left side is TRUE, compiler optimization short-circuiting will make sure that the right side of the || is not evaluated because it will not change the result of the expression.

Upvotes: 1

Kevin
Kevin

Reputation: 56059

Yes, && has higher precedence, but that only determines the grouping of the operands, not the order of evaluation. The base operation here is ||, which guarantees its right side is not evaluated if the left is true, regardless of what operations are on the right-hand side.

Upvotes: 2

Manlio
Manlio

Reputation: 10865

Lazy evaluation.

--b && ++c is not evaluated at all.

Upvotes: 0

Luchian Grigore
Luchian Grigore

Reputation: 258608

shouldn't this code evaluate --b && ++c first

No Operator precedence doesn't affect evaluation order. It just means that

a++||--b&&++c

is equilvalent to

a++||(--b&&++c)

so it's still a++ that is evaluated first, and thus short-circuits the statement.

Upvotes: 13

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