CODEWITHSUNDEEP
cexpressionoperator-precedence
user1369975
user1369975

Reputation: 420

Evaluation of the following expression

The following code snippet:

int i=-3,j=2,k=0,m;

m=++i && ++j || ++k;

can be evaluated using two concepts,I believe:

1.Since ++ operator has greater precedence than the logical operators,so first all increment operators will be evaluted,then && having higher precedence than || will be computed.In this process,k will be incremented.

2.First && operator will be evaluated.For this ++ i and ++j will be computed.Since the result of the && operator is 1,no need to evaluate the ++k.So k will not be incremented.

When I try it on a system, the result proves reasoning 2 to be correct and 1 to be wrong. Why is it so?

Upvotes: 0

Views: 477

Answers (3)

John Bode
John Bode

Reputation: 123458

The && and || operators force left-to-right evaluation. So i++ is evaluated first. If the result of the expression is not 0, then the expression j++ is evaluated. If the result of i++ && j++ is not 1, then k++ is evaluated.

The && and || operators both introduce sequence points, so the side effects of the ++ operators are applied before the next expression is evaluated. Note that this is not true in general; in most circumstances, the order in which expressions are evaluated and the order in which side effects are applied is unspecified.

Upvotes: 0

Taylor Brandstetter
Taylor Brandstetter

Reputation: 3623

Oli is right... You're confusing precedence with evaluation order.

Precedence means that the expression is interpreted as:

m = ((((++i) && (++j)) || (++k));

As opposed to, say:

m = (++(i && ++(j || (++k)))

Precedence doesn't change the fact that the LHS of the || operator will always be evaluated before the RHS.

Upvotes: 2

Floris
Floris

Reputation: 46365

In attempting to be efficient, evaluation of an OR statement (executed from left to right) stops when the LHS is true. There is no need to start evaluating the RHS - there is no concept of "precedence" except within the same group of an expression (when it matters to the value of the expression whether you first do A or B. Example: 5 + 3 * 2 should evaluate to 11. But in evaluating ( 5 + 6 > 3 * 2) it doesn't matter whether you do the addition before the multiplication - it doesn't change the result of the comparison. And in practice this gets evaluated left-to-right. Thus you get the result you observed.

See also this earlier answer

Upvotes: 0

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