-5
main()    
{

   int k = 5;

    if(++k <5 && k++/5 || ++k<=8);  // how to compiler compile this statement 
  print f("%d",k);
}

// Here answer is 7 but why ?

Grijesh Chauhan
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ankit dixit
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  • Note: short circuit operators are sequence points. – Paul R Aug 18 '13 at 15:13
  • **Read**: [Behaviour of && in C programming language](http://stackoverflow.com/questions/18229657/behaviour-of-in-c-programming-language/18229763#18229763) – Grijesh Chauhan Aug 18 '13 at 15:14
  • how to statement is executed inside "if()" how to : ++k <5 && k++/5 || ++k<=8 this executed ? – ankit dixit Aug 18 '13 at 15:15
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    @ankitdixit: You need to understand short circuit behavior of `&&` and `||` in C and C++, so feel free to search and read that up. – Alok Save Aug 18 '13 at 15:16
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    You are missing the return type (`int`) for `main`. The `if` is unrelated to the `printf`; it only controls the empty-statement `;`. – aschepler Aug 18 '13 at 15:17
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    When you ask why the output is the way it is, it's also good to explain what you expect. And `main()` is illegal in C++, you need to spell out the return type `int`. – jrok Aug 18 '13 at 15:19

4 Answers4

6

++k < 5 evaluates to false (6 < 5 = false), so the RHS of the && operator is not evaluated (as the result is already known to be false). ++k <= 8 is then evaluated (7 <= 8 = true), so the result of the complete expression is true, and k has been incremented twice, making its final value 7.

Note that && and || are short circuit boolean operators - if the result of the expression can be determined by the left hand argument then the right hand argument is not evaluated.

Note also that, unlike most operators, short circuit operators define sequence points within an expression, which makes it legitimate in the example above to modify k more than once in the same expression (in general this is not permitted without intervening sequence points and results in Undefined Behaviour).

Paul R
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4

Unlike many questions like this, it appears to me that your code actually has defined behavior.

Both && and || impose sequence points. More specifically, they first evaluate their left operand. Then there's a sequence point1. Then (if and only if necessary to determine the result) they evaluate their right operand.

It's probably also worth mentioning that due to the relative precedence of && and ||, the expression: if(++k <5 && k++/5 || ++k<=8) is equivalent to: if((++k <5 && k++/5) || ++k<=8).

So, let's take things one step at a time:

int k = 5;

if(++k <5 &&

So, first it evaluates this much. This increments k, so its value becomes 6. Then it compares to see if that's less than 5, which obviously produces false.

k++/5

Since the previous result was false, this operand is not evaluated (because false && <anything> still always produces false as the result).

|| ++k<=8);

So, when execution gets to here, it has false || <something>. Since the result of false | x is x, it needs to evaluate the right operand to get the correct result. Therefore, it evaluates ++k again, so k is incremented to 7. It then compares the result to 8, and finds that (obviously) 7 is less than or equal to 8 -- therefore, the null statement following the if is executed (though, being a null statement, it does nothing).

So, after the if statement, k has been incremented twice, so it's 7.

  1. In C++11, the phrase "sequence point" has been replaced with phrases like "sequenced before", as in: "If the second expression is evaluated, every value computation and side effect associated with the first expression is sequenced before every value computation and side effect associated with the second expression." Ultimately, the effect is the same though.
Jerry Coffin
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2

In following:

for && "something" is evaluated when first condition is True

for || "something" is evaluated when first condition is False

( (++k <5)      &&          (k++/5) )      ||      (++k<=8)

( 6 < 5         AND        something )     OR   something

( False         AND        something )     OR   something
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
                False                      OR   7 < 8

                False                      OR   True
                ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^                
                                          True

So k comes out to be 7

P0W
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1

Initially k is assigned 5, in your declaration, then in following if condition: 

   ++k < 5 && k++ / 5 || ++k <= 8
//  ^
// evaluates

Increments k to 6 then its and LHS of && operand evaluates false. 

6 < 5  && k++ / 5   ||  ++k <= 8     
// ^      ^^^^^^^
//   0    not evaluates

Then because of Short-Circuit behavior of && operator k++ / 5 will not evaluates.

Short-Circuit behavior of && operator is:
0 && any_expression == 0, so any_expression not need to evaluate.

So above step 2 conditional expression becomes: 

0 || ++k <= 8     
//   ^^^^^^^^   
//  evaluates

Increments k to 7 then its:

0 || 7 <= 8

because you have ; after if, so no matter whether if condition will evaluates True or False printf() will be called with k = 7.

Grijesh Chauhan
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