This question already has answers here:
OR and AND operation in C
(2 answers)
Closed 7 years ago.
I am struck on a basic problem with operators. Here is the problem
int i=3, j=2, k=1, m;
m = ++i || ++j && ++k;
printf("%d %d %d %d", i, j, k, m);
As i worked on this and evaluated the ans as
4 3 2 1
But C is not happy with my ans and evaluates the above as
4 2 1 1
But i dont understand how. Why ++j is doesn't updates the value of j. And also ++k doesn't updates the value of k.
Anybody here please explain this how it worked.
m = ++i || ++j && ++k;
The && operator has higher precedence than ||, so this is equivalent to:
m = ++i || (++j && ++k);
Both && and || are short-circuit operators; if the left operand determines the result, the right operand is not evaluated. Since ++i (which yields 4) is "truthy", the right operand (++j && ++k) is not evaluated, and j and k are not incremented. So i becomes 4, and j and k retain their initial values of 2 and 1, respectively.
The result of the || operator is either 0 (for false) or 1 (for true); since the condition is true, the value 1 is assigned to m.
Result:
4 2 1 1
In C, || is a short-circuit operator. This means when evaluating the expression a || b, it will only evaluate b if a is false. If a is true, then the expression will always be true regardless of b.
Since ++i gives a truthy value, ++j && ++k will not be evaluated.
Related
So, I was writing some code and I was getting an unexpected output in one part of my program which disrupted the entire system.
I managed to extract and simplify the problem to a basic logical expression. Let's say:
int i = 1, j = 1, k = 0;
printf("%d\n", ++i || ++j && k);
printf("%d, %d, %d\n", i, j, k);
return 0;
The output for this program is:
1
2 1 0
I am thinking that the value of j was not incremented to 2 due to the short circuit nature of the || operator. However I am confused how the value of the first "%d" is 1. Shouldn't the value of k be non-zero for the && statement to return 1? Or isn't this statement executed at all since ++i || ++j is not 0 and thus returns a 1?
I know that && is a logical and, and expr1 && expr2 has the value 1 if values of expr1 && expr2 are both non-zero.
Any clarifications would be appreciated and please excuse the basic nature of this question.
&& has higher precedence than ||.
(See https://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B )
so
++i || ++j && k
is
++i || (++j && k)
and || shortcircuits if the first operator is truthy, as per 6.5.14p4 .
If you're on gcc or clang and compile your code with -Wall, the compiler will nudge you to put those parentheses there. It's probably a good idea to heed that advice, as some people get confused by the precedences (I hear).
From the C Standard (6.5.14 Logical OR operator)
3 The || operator shall yield 1 if either of its operands compare
unequal to 0; otherwise, it yields 0. The result has type int.
This expression
++i || ++j && k
is equivalent to
++i || ( ++j && k )
and according to the quote from the Standard the expression returns integer value 1 because ++i is not equal to zero. The subexpression ( ++j && k ) is not evaluated.
++i || ++j && k is evaluate to 1 (true) because i equals 2 (++1), (++j && k) isn't evaluated because short circuit.
Operators precedence. && has higher precedence than ||.
Your expression is the same as: ++i || (++j && k)
++i is TRUE, the parenthesis is not evaluated anymore.
This question already has answers here:
Short circuit behavior of logical expressions in C in this example
(1 answer)
Operator Precedence vs Order of Evaluation
(6 answers)
Closed 7 years ago.
In the program shown below, prefix should be evaluated first because it has higher precedence, But answer is -2, 2, 0, 1 and it is explained in book "as LHS of || is true RHS is not evaluated."
Why is it so? All the increments should performed first and then logical should be checked because of precedence.
#include<stdio.h>
int main()
{
int i=-3, j=2, k=0, m;
m = ++i || ++j && ++k;
printf("%d, %d, %d, %d\n", i, j, k, m);
return 0;
}
Don't get confused with Precedence and Order of evaluation.
The order of evaluation of logical OR || is left to right.
So if left = true then left || right will never execute right. In your code exactly same happened.
As you know, any non zero value treated as true in C, hence, ++i or -2 is true. So,
m = ++i || ++j && ++k;
m = true || bla bla bla; //right not even checked!
m = true
m = 1
And you get the output as expected.
The logical && and || operators fully evaluate the LHS before doing any evaluation of the RHS.
In the code shown, since ++i is -2, the LHS of the || evaluates to true (1) and the RHS is not evaluated. Therefore, neither j nor k is incremented. The printed result follows: m was assigned 1, i became -2, and j stayed as 2 and k stayed as 0.
The only remaining issue is that && binds tighter than ||, so:
a || b && c
is equivalent to:
a || (b && c)
so if a evaluates to true (non-zero), then neither b nor c is evaluated.
Operator precedence is a completely different thing. Order of evaluation is determined by side effects and sequence points.
see this manual on order of evaluation.
This is My Code:
int main()
{
int i=2,j=2,c=2,d;
d= ++i || ++j && c;
printf("I is %d\n",i);
printf("J is %d\n",j);
printf("C is %d\n",c);
printf("D is %d\n",d);
return 0;
}
The output of the following code is:
i is 3
j is 2
c is 2
d is 1
Now my question is, if ++i is 3 then why ++j is 2?
What is the difference between ++i & i++? Also I want to know that how come d is 1 ?
You asked:
Now my question is if ++i is 3 then why ++j is 2?
++j is never executed since ++i evaluates to true.
The expression ++i || ++j && c is equivalent to ++i || (++j && c) due to operator precedence.
The run time evaluates the expression ++i || ++j && c from left to right.
The first term, ++i evaluates to 3, with the side effect that the value of i is 3. Since it is non-zero, the run time does not need to evaluate anything after that. Since ++j is never evaluated, the value of j is not changed.
What is the difference between ++i & i++?
The value of the expression ++i is i+1 with the side effect that the value of i is incremented by 1.
The value of the expression i++ is i with the side effect that the value of i is incremented by 1.
int i = 2;
int x = ++i; // x == 3, i == 3 at the end of execution of the statement
int y = i++; // y == 3, i == 4 at the end of execution of the statement
I want to know that howcome D is 1
The value of d is set to 1 since the boolean value of the expression is true.
++i is pre increment.
i++ is post increment.
Suppose here i = 2
In Post In crement
printf("%d",i++); // it will print output 2 then increment value of i to 3
In Pre Increment
printf("%d",++i); //first it will increment value of i to 3, then print output 3
d= ++i || ++j && c;
Condition will be performed in this manner ++i || (++j && c);
As i=2, after ++i it will become 3.
In OR if first condition is true it will skip second condition. means ++j && c will not be performed.
Result of ++i || ++j && c is 1
so d = 1.
printf("I is %d\n",i); // i = 3;
printf("J is %d\n",j); // j = 2;
printf("C is %d\n",c); // c = 2;
printf("D is %d\n",d); // d = 1;
Short Circuit evaluation
Reason for the evaluation is short-circuiting of the boolean operators && and ||.
For &&, if the left-hand side expression is false, the combined result is false (the right-hand side expression is never evaluated).
For ||, if the left-hand side expression is true, the combined result is true (the right-hand side expression is never evaluated).
This is mostly important when the right-hand expression has side effects, such as altering values(or calling some functions with side-effects), for example in your case:
d= ++i || ++j && c;
Note that the combined conditional expression would increment i by 1 since the conditional expression on the left of || is true, the conditional expression on the right-hand side i.e. ++j && c is never evaluated.
Thus, the total expression will lead eventually to:
d = 1 || ++j && c;
which explains why d=1?
Now if you change the value of i=-1 then the expression on the right of the || will be executed and the j will be incremented as per your expectation
Now for the performance effects of i++ vs ++i check here :)
This question already has answers here:
Why isn't "k" incremented in the statement "m = ++i && ++j || ++k" when "++i&&++j" evaluates to true? [duplicate]
(5 answers)
Closed 9 years ago.
#include<stdio.h>
int main()
{
int i=-3,j=2,k=0,m;
m= ++i && ++j || ++k;
printf("%d %d %d %d",i,j,k,m);
return 0;
}
Could someone tell how does is the m evaluated. I am confused as how k remain '0'
m= ++i && ++j || ++k;
First, ++i && ++j takes effect, i become -2 and j become 3, which makes ++i && ++j non-zero, so m will be evaluated to 1 (true) because of short-circuit. All the later expression(++k) won't be executed.
m= ( (++i) && (++j) ) || ++k;
| |
| |
( T (-2) AND T(3) ) OR something
^^^^^^^^^^^^^^^^^
T OR something
^^^^^^^^^^^^^^^^^^^^^^^^^
= T
= 0
Changing j=-1
m= ( (++i) && (++j) ) || ++k;
| |
| |
( T (-2) AND T(0) ) OR something
^^^^^^^^^^^^^^^^^
F OR T (0) //++k evaluated
^^^^^^^^^^^^^^^^^^^^^^^^^
= T
= 0
Initialized Values:
int i=-3,j=2,k=0,m;
Original expression:
m= ++i && ++j || ++k;
With implicit operator precedence:
m= (++i && ++j) || ++k;
The expression evaluates left to right. What you need to understand is how Short Circuit Logic works.
Evaluation Walkthrough:
i gets incremented (-3 --> -2)
The logical AND operator is evaluated (&&). It checks if the left hand argument i evaluates to true (i!=0). If the left hand side evaluates to false, the and operator short circuits the evaluation and returns false (0). This doesn't happen because i==-2, so the right hand side (++j) is also evaluated.
j get incremented (2 --> 3)
Logical AND finishes its right hand side evaluation i!=0 && j!=0. It returns true (!0).
The logical OR operator is evaluated (||). It checks if the left hand argument (++i && ++j) evaluates to true (!0). If the left hand side evaluates to true, the and operator short circuits the evaluation and returns true (!0). This happens because ++i && ++j already evaluated true. Consequent k++ never executes!
Logical OR (||) assigns the evaluated value (!0) to m. m!=0 after the expression evaluates.
This is because of short circuting. If you know about how an OR expression is evaluated, you will get the hang of this expression. 1 OR Anything is always 1
int i=-3,j=2,k=0,m;
m= ++i && ++j || ++k;
`-2 && +3 is +1 therefore +1 OR anything is 1.` Hence C skips evaluation of k.
the answer would be
-2 3 0 1
Explanation :
There are two basic facts that work here
FACT 1 :
Whenever there is an expression with PRE-INCREMENT operator, All these operators would be operated on the corresponding variables first, and then the whole expression would be evaluated with updated values of variables.
FACT 2
whenever there is an expression having OR operator, If first operator is true(or non zero in some cases), second operator wouldn't be executed.
for example
a=2;
b=2;
c=3;
boolean k = c || a+b ;
The values of variable k would be true and the expression "a+b" wouldn't be even considered as c is non zero.
In your case variable i and j would be incremented first and the expression would be
-2 && 3 || ++k
Note : variable k wouldn't be incremented here because in case of OR logical operator Left operand would be solved completely first.
Left operand would would be then TRUE because both -2 and 3 are non zero .
so the whole expression would be
TRUE || ++k
This wouldn't be evaluated further because no matter what, the result always be TRUE .
as we are assigning this result (TRUE) to an integer variable, it would be 1.
Hence the answer is ....
This question already has answers here:
Is short-circuiting logical operators mandated? And evaluation order?
(7 answers)
Closed 5 years ago.
Why is k not getting incremented whereas,i and j are getting incremented in the same expression.And i also want to know what is the output of the program.I am getting the output as
-2 3 1 0
#include <stdio.h>
void main()
{
int i=-3, j=2, m, k=0;
m=++i && ++j || ++k;
printf("%d %d %d %d", i, j, m, k);
}
The logical or, || short-circuits, and after
++i && ++j
the value of the entire expression is determined, so the right operand of the || isn't evaluated.
m=++i && ++j || ++k;
is parenthesized m = (++i && ++j) || ++k; since the && has higher precedence than the ||.
The short-circuiting of the logical operators means that the right operand is only evaluated when the evaluation of the left has not yet determined the final result, for || that means the right operand is only evaluated if the left evaluated to 0, and for &&, the right operand is only evaluated if the left evaluated to a nonzero value.
So first ++i && ++j is evaluated, and for that, first ++i is evaluated. i had the value -3 before, so ++i evaluates to -2, which is not 0, hence the ++j is evaluated too. j had the value 2 before, so ++j evaluates to 3, which is again nonzero, and thus ++i && ++j evaluates to 1 (true). Since the left operand of the || is not zero, its result is already determined (to be 1), and the right operand isn't evaluated, thus k remains unchanged and m is set to 1.
If the item on the left of an || condition evaluates to true, there is no point evaluating the right hand side since the OR condition is already satisfied. That is why the ++k is not being evaluated
These operators are known to be short-circuited operators. So, if the expression ++i && ++j is true, it does not evaluate k (we know the value of the expression regardless to the value of k).
It has to do with order precedence. Anytime a logical OR is executed, it will stop if the first operand is true, which in this case is j.
The logical operator || works like this: if first condition is true then the second one is not evaluated.
So first (++i && ++j) checks. It returns true, so after it || is not evaluated. Thus, the value of k is not increased.
#include<stdio.h>
int add(int);
void main()
{
int i=3,k,l;
clrscr();
k= add(++i);
l=add(i++);
printf("i=%d k= %d l=%d",i,k,l);
getch();
}
int add(int ii)
{
++ii;
return(ii);
}
/*th eoutput is
5
5
5 can anyone expalin how? */
#include"stdio.h"
#include"conio.h"
main()
{
int rmv=10,vivek=10;
clrscr();
rmv++;//this post incremant means incremant next step 10+1
//rmv+1 and rmv++ both are same
printf("1.rmv++=%d\n",rmv);//value is 10
printf("2.rmv++=%d\n",rmv++);//the value is increment 11
++vivek;//this pre increment means first increment the value so,1+10=11
printf("1.\t++vivek=%d\n",vivek);////++vivek and 1+vivek both are same
printf("2.\t ++vivek=%d",++vivek");
getch();
}