Let us say I have an if condition which checks for one of the events a, b, c,... to be true as shown:
if(a || b || c || .... || z)
do something
My question is that let us say that the condition c turns out to be true. Then will the program evaluate the conditions d to z or will it proceed to execute the "do something" instructions?
Using || will evaluate each condition left to right until it hits a value that's true. It will then stop the check.
if(a || b || c || .... || z)
If you use |, then every condition will be evaluated regardless of the result.
if(a | b | c | .... | z)
Let's say we have int a = b = c = 1.
If I do ++a || ++b;, a will be 2 after the statement, as an OR-clause only needs one of the condition needs to be true and b will remain 1, as the compiler doesn't continue to check it, as the expression is true already.
If we have --a || ++b however, a evaluates to 0, which will be false in terms of logic. So it continues to check the other condition: b will be 2, which is logically true.
So what do you need to remember?
the compiler evaluates those conditions from left to right
if one of the OR conditions is true, the compiler doesn't even continue to evaluate the other condition, is it will be true no matter if the other condition is true or false
AND conditions need both conditions to be true, so if the first conditions is false, the whole expression is guaranteed to be false, so the compiler doesn't continue to evaluate the other condition. If the first one is true however, the compiler has to check if the other condition is true aswell.
All this is called
short circuit evaluation
For more, click here.
This is called short circuit evaluation.
For example java checks it from left to right; means if c is true it will not check rest of the portion.
But this is not true for all languages.
For example csh performs short circuit evaluation from right to left.
Related
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.
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.
I didn't think these if's would compile but they do:
if (a>>b&&c&&d)
if (month==1,2,3,5,7,9,10)
The first I'm clueless about. In the second statement is the comma supposed to be an (||) or operator ?
Syntax wise was it always this way or was it introduced some time ago ?
I'm using Visual Studio 2010.
if (a>>b && c && d)
it is equal to
if ((a>>b) && c && d)
if the result of a shifted right b times evaluates to a bool, c and d also evaluates to bool respectively, then all these booleans will be AND-ed to each other.
In your context, the all expressions within commas will be evaluated and then the last expression will be passed to if expression:
if (month==1,2,3,5,7,9,10) -> is equal to
if (2,3,5,7,9,10) -> is equal to
if (3,5,7,9,10) -> is equal to
if (5,7,9,10) -> is equal to
if (7,9,10) -> is equal to
if (9,10) -> is equal to
if (10)
which is always true.
It's not suppose to be || or &&. If you want OR or AND write it like below:
if (month==1 || month==2 || month==3 || ....)
or
if (month==1 && month==2 && month==3 && ....)
// Also month can not simultaneously be equal to more than one value!
// then, it's equal to
if (false)
The first if statement would be evaluated like:
if(((a >> b) && c) && d)
Essentially bitshift a by b bits and then logical and with c and then with d
The second is the comma operator which will evaluate the first term and throw it away, then the second, and so on and return the result of the final term. So in our case the statement is equivalent to:
if(10)
which is always true.
I had a question in my test paper in which we had to compare the values of int type variables. The first thought that came to my mind was that it was missing the && operator but i am not sure.
int a=2, b=2, c=2;
if(a==b==c)
{
printf("hello");
}
I have a doubt, will the above statement will execute or not in c or c++? Can i have the reason as well.
Thank You
It will execute but with what I believe unexpected results to you.
One of the == will evaluate to a boolean value, which will then be converted to an int and then the second comparison will be performed, comparing an int to either 1 or 0.
The correct statement is a==b && b==c.
For example:
3 == 3 == 3
evaluates to
true == 3
1 == 3
false
a==b==c
is equivalent to
(a == b) == c
The result of a == b is 1 (if true) or 0 (if false), so it will probably not achieve what you expect.
Use a == b && b == c to check if the value of the three objects are equal.
a == b == c is a comparison between c and result of a==b (1 or 0) operation.
use a==b&&b==c.
the condition a==b==c is equivalent to (a==b)==c which will provide the required result iff c==1, else the code will fail.
what is the correct syntax for checking a varable value and then setting a varable in the same condition also checking that new set varables var, all in one if statement?
so basically something like
if(this->somevar > 0 && this->newvar = this->GetNewVar(this->somevar),this->newvar > 0)
i know that is not the correct syntax or at least its not working for me anyway, hence the topic, i am using that as an example, so if this->somevar is null or 0, i don't want it to execute the next condition which is && this->newvar = this->GetNewVar(this->somevar,this->newvar but instead skip the statement and ignore that part.
what is the correct syntax for something like this?
&& is an operator with short circuit evaluation, right part is not executed if left part is true.
But why don't you simply write:
if(this->somevar > 0)
{
this->newvar = this->GetNewVar(this->somevar);
if (this->newvar > 0)
{
...
This will certainly makes things clearer ...
the logical AND && operator is short-circuited if this->somevar evaluates to zero, meaning the rest of your if expression would not be evaluated in that situation
The expression after the comma is not necessary. Also, there is one thing missing, parentheses arround the assignment:
if(this->somevar > 0 && (this->newvar = this->GetNewVar(this->somevar)) > 0)
Without the parentheses you may end up setting this->newvar to the value of the boolean expression
this->GetNewVar(this->somevar),this->newvar > 0, which will be evaluated to a boolean result (true/false which, in turn, may be converted to 0 or 1 or -1 depending on the compiler, when cast to the type of this->newvar).
I think only the bit after the comma is evaluated for the if condition. The expression on the left of the comma is ignored.
int main() {
if( false, true) { cout << " got to if( false, true ) "; }
if ( true, false ) { cout << "got to if( true, false ) "; }
}
to answer your question, you can put anything on the left of the comma and do whatever you like, as long as the expression you want to evaluate is the last expression in the list.
so if ( exp1, exp2, exp3 , exp4 ) dowhatever(); only gets run if exp4 is true. You should really run exp1 to exp3 outside the if condition for readability.