This question already has answers here:
Is short-circuiting logical operators mandated? And evaluation order?
(7 answers)
Closed 8 years ago.
I'm getting it difficult to understand how the following programs work, kindly help me in understanding.
int x=2,y=0;
(i)
if(x++ && y++)
cout<<x<<y;
Output:
(ii)
if(y++ || x++)
cout<<x<<" "<<y;
Output: 3 1
(iii)
if(x++||y++)
cout<<x<<" "<<y;
Output: 3 0
Kindly explain me how the program is working and also, what makes the difference between (ii) and (iii).
You are looking at a "C++ puzzle" which is using two languages tricks.
The first is that postincrement uses the value of the variable first, then increments the variable.
So x++ has the value 2 in the expression and then x becomes 3
y++ has the value 0 in the expression, and then y becomes 1
&& is and
|| is or
both operators are short-circuiting.
Look at && first.
In order for and to be true, both sides must be true (non-zero)
Since it is x++ && y++, first the x++ happens, and since it is non-zero (true)
the y++ has to happen to determine whether the result is true or not.
Therefore x is 3 and y is 1.
The second case is the same. y is zero, but in order to determine if the OR is true,
the if statement executes the second half of the expression.
The third case, since it is in the opposite order, never executes y++ because
x++ || y++
the first half being x++, it is already true, so the compiler does not bother to execute the second half of the test.
int x=2,y=0;
Condition in if() is evaluated according to the rules:
operator && evaluates left operand first and if the value is logically false then it avoids evaluating the right operand. Typical use is for example if (x > 0 && k/x < limit) ... that avoids division by zero problems.
operator || evaluates left operand first and if the value is logically true then it avoids evaluating the right operand. For example if (overwrite_files || confirm("File existing, overwrite?")) ... will not ask confirmation when the flag overwrite_files is set.
To understand the following it is also important to note that x++ is postincrementation what means that in if( x++) x has the old value but just in the next line (or even same line but after if() was tested) x is incremented.
(i)
if(x++ && y++) // x=2,y=0;
// 2 so y++ is evaluated
cout<<x<<y; // skipped, but now x=3,y=1;
Output:
(ii)
if(y++ || x++) // x=2,y=0;
//0 so x++ is evaluated
cout<<x<<" "<<y; // x=3,y=1;
Output: 3 1
(iii)
if(x++||y++) // x=2,y=0;
//2 so y++ is not evaluated
cout<<x<<" "<<y; // x= 3,y=0;
"IF" argument evaluation order?
In C++ (and a few other languages) && and || are your logical operators in conditions. && is the logical AND operator and || is the logical OR operator.
When using these in a condition, the result of the condition is dependent on what is on either side of these operators and is read as if you were just saying the words.
Example:
if(thisNum > 0 && thisNum < 10) is read as "if thisNum is greater-than zero AND less-than 10, the condition will be true." Therefore, any number between 0 and 10 will make the condition true.
if(thisNum > 0 || thisNum < 10) is read as "if thisNum is greater-than zero OR less-than 10, the condition will be true." Therefore, ANY number at all will make this statement true because only one part of this statement has to be true to evaluate to TRUE.
A more detailed explanation on how these work is this:
OR (||) - If EITHER or BOTH sides of the operator is true, the result will be true.
AND (&&) - If BOTH and ONLY BOTH sides of the operator are true, the result will be true. Otherwise, it will be false.
There are many more explanations, including truth tables around the internet if you do a Google search to assist in understanding these.
As for your examples.
if(x++ && y++)
cout<<x<<y;
You are doing an if statement that is comparing the the values of x and y. The important thing to remember is that you are NOT comparing the values together. In C++, and non-zero value will evaluate to TRUE while a zero value will result in FALSE. So, here you are checking to see if x AND y are TRUE and then increasing their values by one AFTER the comparison. Since x = 2 and y = 0, we have one TRUE value and one FALSE value and since we have a && (AND) operator between them we know the result of the condition is FALSE (since both aren't TRUE) and the output is skipped.
if(y++ || x++)
cout<<x<<" "<<y;
In this example, we are doing the same thing EXCEPT we are doing a logical OR operation instead of AND (since we have || and not &&). This means, as I said above, that only one part of the condition has to be TRUE to result in TRUE. Since our values are still x = 2 and y = 0, we have one TRUE value and one FALSE value. Since part is TRUE, the whole condition is TRUE and the output occurs.
The reason the result is 3 1 and not 2 0 is because of the ++ operators after the variables. When the ++ operators are after the variable, it will add one AFTER the rest of the line has occurred and done the actions it needs. So it does evaluates the condition THEN increases the values by one.
if(x++||y++)
cout<<x<<" "<<y;
This example is EXACTLY the same, the only difference is that the values have been swapped. And since it is an OR (||) operation, only one part has to be TRUE. And as before, we have one TRUE value (since x is non-zero) and one FALSE value (since y is zero). Therefore, we have a TRUE result, meaning the output is NOT skipped and we get the answer 3 1 again (since the ++ operators came AFTER the variables).
Now, speaking of the ++ operator. What if in the first example it had been:
if(++x && ++y)
cout<<x<<y;
We would get a DIFFERENT result in this situation. Since the ++ operator comes BEFORE the variables, the first thing that happens is that the values are increased by one AND THEN the condition is evaluated. Since our new values would be x = 3 (TRUE) and y = 1 (also TRUE), we can conclude that we would get the same result as the other two examples since BOTH values are TRUE (which, since we have an AND (&&) operator, BOTH variables have to be TRUE to result in TRUE).
Hope this helps.
EDIT:
One of the other answers made an excellent point about the second and third examples. Result wise, there is no difference. HOWEVER, in terms of how it is evaluated there is a difference.
C++ automatically does something called "short-circuit evaluation". This means, it will skip any evaluations if it knows the result is already going to be TRUE or FALSE. This means that with an OR operation, if the first part is TRUE it will just skip the second part since it already knows the result HAS TO BE TRUE. (Remember, with OR only one part of the comparison has to be TRUE.) So this means that IF the first part is FALSE then the program HAS TO evaluate the second part too, since it determines whether the result will be TRUE or FALSE.
With an AND operation, it will do the opposite. If the first part is FALSE then the program knows the result has to be FALSE and will skip the rest. However, if the first part is TRUE then the program HAS TO evaluate the rest to determine if the result is TRUE or FALSE.
In terms of your examples, this means the second one HAS TO evaluation both sides since the first part results in FALSE.
In the third example, the second part is skipped because the first part is TRUE, meaning the whole condition is TRUE.
For or(||) operator, if the expression of left side is true, it will ignore the expression of right side.
For and(&&) operator, if the expression of left side is false, it will ignore the expression of right side.
This is the answer of your question what makes the difference between (ii) and (iii).
In the output (ii), it is:
as the time where compiler is evaluating y, y is equal to 0 so the statement is false it s evaluating the other member x which is equal to 2 so the statement is true. And then it operates the increment operator.
In the output (iii), it is:
as the time where compiler is evaluating x, x is equal to 2 so the statement is true as the operator of the condition is || it won't evaluate the other expression so the y++ is ignored. Only x will be incremented.
Related
OK so obviously this question might sound dumb for more experienced people, but, for the following lines the result I get is 0:
int x = 2,
y = -2;
cout << (x++ - y && (--x + y));
I understand it means that either one of these two expressions equals 0, but how? As far as I understand, this should be (3 && -1)?
Also, a little subquestion: when does x++ exactly take effect? On the next occurance of x within the same expression, after the left-shift operator within the same line, or in the next statement?
Thank you!
As far as I understand, this should be (3 && -1)?
you understand wrong:
first left side is fully evaluated, as it is necessary for short circuit evaluation with logical and (details can be found here)
x++ - y == 4 // as result of x++ == 2 so (2-(-2)), after that x == 3
result is true so right side is evaluated:
--x + y == 0 // as result of --x == 2 so (2+(-2)), after that x == 2
result on the right is false so result of and is false as well which printed by std::ostream as 0
Note: short circuit evaluation of logical or and and operations make such code valid (making them sequenced) but you better avoid such questionable expressions. For example simple replacing logical and to binary would make it UB.
Here is a part of code
int main()
{
int x=5,y=10;
if(x=!y)
{
cout<<"h";
}
else
{
cout<<"p";
}
getch();
}
The output was p, please explain, how the code works and the meaning of x=!y.
Looks like a typo that produces valid code. Expanding it helps--
if (x = (!y))
Since y is 10, !y == 0, and assignments themselves produce a value. In particular the value of x = 0 is 0, so the test evaluates to 0 and that's why you get the result.
But this is a crazy thing to write in this context, presumably what was, or what should have been intended was
if (x != y)
I.e., not-equals.
x=!y is an assignment.
x is being assigned the value of !y expression, which is a logical "NOT" operation. This operation returns true if the operand is zero, or false otherwise. The value true becomes 1 when assigned back to int; false becomes zero.
In C and C++ it is OK to use assignment expressions inside if conditionals and other control statements, such as for and while loops. The value being assigned is used to evaluate the condition, and the assignment itself is performed as a side effect. In this case, the condition is !y.
int x = 12;
false && x++;
After this snippet executes, x still has the value 12. If the ++ operator has higher precedence than &&, why isn't x incremented? Is there an explanation for this is or is it one of those things that you just accept? I am sorry if this is answered somewhere else. I did not come across it in my search.
Thank you for your time.
It checks for the false which is actually false for x=12 so it skips the incrementation of x as 13. if you swap the places as x++ && false then it would do x=13 and then skip the false leaving x with an incrementation of 1.
while ( (i=t-i%10 ? i/10 : !printf("%d\n",j)) || (i=++j<0?-j:j)<101 );
I came across this on codegolf
Please explain the usage of ? and : and why is there no statement following the while loop? As in why is there a ; after the parenthesis.
There is a boolean operation going on inside the parentheses of the while loop:
while (boolean);
Since the ternary operator is a boolean operator, it's perfectly legal.
So what's this doing? Looks like modular arithmetic, printing going on over a range up to 101.
I'll agree that it's cryptic and obscure. It looks more like a code obfuscation runner up. But it appears to be compilable and runnable. Did you try it? What did it do?
The ?: is a ternary operator.
An expression of form <A> ? <B> : <C> evaluates to:
If <A> is true, then it evaluates to <B>
If <A> is false, then it evaluates to <C>
The ; after the while loop indicates an empty instruction. It is equivalent to writing
while (<condition>) {}
The code you posted seems like being obfuscated.
Please explain the usage of ? and :
That's the conditional operator. a ? b : c evaluates a and converts it to a boolean value. Then it evaluates b if its true, or c if its false, and the overall value of the expression is the result of evaluating b or c.
So the first sub-expression:
assigns t-i%10 to i. The result of that expression is the new value of i.
if i is not zero, the result of the expression is i/10
otherwise, print j, and the result of the expression is zero (since printf returns a non-zero count of characters printed, which ! converts to zero).
Then the second sub-expression, after ||, is only evaluated if the result of the first expression was zero. I'll leave you to figure out what that does.
why is there no statement following the while loop?
There's an empty statement, ;, so the loop body does nothing. All the action happens in the side effects of the conditional expression. This is a common technique when the purpose of the code is to baffle the reader; but please don't do this sort of thing when writing code that anyone you care about might need to maintain.
This is the Conditional Operator (also called ternary operator).
It is a one-line syntax to do the same as if (?) condition doA else (:) doB;
In your example:
(i=t-i%10 ? i/10 : !printf("%d\n",j)
Is equivalent to
if (i=t-i%10)
i/10;
else
!printf("%d\n",j);
?: is the short hand notation for if then else
(i=t-i%10 ? i/10 : !printf("%d\n",j)<br>
equals to
if( i= t-i%10 )
then { i/10 }
else { !printf("%d\n",j) }
Your while loop will run when the statement before the || is true OR the statement after the || is true.
notice that your code does not make any sense.
while ( (i=t-i%10 ? i/10 : !printf("%d\n",j)) || (i=++j<0?-j:j)<101 );
in the most human-readable i can do it for u, it's equivalent to:
while (i < 101)
{
i = (t - i) % 10;
if (i > 0)
{
i = i / 10;
}
else
{
printf("%d\n",j);
}
i = ++j;
if (i < 0)
{
i = i - j;
}
else
{
i = j;
}
}
Greetings.
I am the proud perpetrator of that code. Here goes the full version:
main()
{
int t=getchar()-48,i=100,j=-i;
while ((i=t-i%10?i/10:!printf("%d\n",j)) || (i=++j<0?-j:j)<101 );
}
It is my submission to a programming challenge or "code golf" where you are asked to create the tinniest program that would accept a digit as a parameter and print all the numbers in the range -100 to 100 that include the given digit. Using strings or regular expressions is forbidden.
Here's the link to the challenge.
The point is that it is doing all the work into a single statement that evaluates to a boolean. In fact, this is the result of merging two different while loops into a single one. It is equivalent to the following code:
main()
{
int i,t=getchar()-'0',j=-100;
do
{
i = j<0? -j : j;
do
{
if (t == i%10)
{
printf("%d\n",j);
break;
}
}
while(i/=10);
}
while (j++<100);
}
Now lets dissect that loop a little.
First, the initialisation.
int t=getchar()-48,i=100,j=-i;
A character will be read from the standard input. You are supposed to type a number between 0 and 9. 48 is the value for the zero character ('0'), so t will end up holding an integer between 0 and 9.
i and j will be 100 and -100. j will be run from -100 to 100 (inclusive) and i will always hold the absolute value of j.
Now the loop:
while ((i=t-i%10?i/10:!printf("%d\n",j)) || (i=++j<0?-j:j)<101 );
Let's read it as
while ( A || B ) /* do nothing */ ;
with A equals to (i=t-i%10?i/10:!printf("%d\n",j)) and B equals to (i=++j<0?-j:j)<101
The point is that A is evaluated as a boolean. If true, B won't be evaluated at all and the loop will execute again. If false, B will be evaluated and in turn, if B is true we'll repeat again and once B is false, the loop will be exited.
So A is the inner loop and B the outer loop. Let's dissect them
(i=t-i%10?i/10:!printf("%d\n",j))
It's a ternary operator in the form i = CONDITION? X : Y; It means that first CONDITION will be evaluated. If true, i will be set to the value of X; otherwise i will be set to Y.
Here CONDITION (t-i%10) can be read as t - (i%10). This will evaluate to true if i modulo 10 is different than t, and false if i%10 and t are the same value.
If different, it's equivalent to i = i / 10;
If same, the operation will be i = !printf("%d\n",j)
If you think about it hard enough, you'll see that it's just a loop that checks if any of the decimal digits in the integer in i is equal to t.
The loop will keep going until exhausting all digits of i (i/10 will be zero) or the printf statement is run. Printf returns the number of digits printed, which should always be more than zero, so !printf(...) shall always evaluate to false, also terminating the loop.
Now for the B part (outer loop), it will just increment j until it reaches 101, and set i to the absolute value of j in the way.
Hope I made any sense.
Yes, I found this thread by searching for my code in google because I couldn't find the challenge post.
This question already has answers here:
Is Short Circuit Evaluation guaranteed In C++ as it is in Java?
(2 answers)
Closed 9 years ago.
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 ?
++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).
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.
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.
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
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.