() has the highest priority, why is it short-circuited？
int a = 1, b = 0;
(--a)&&(b++);
Why is (b++) still short-circuited?
I don't find the term "short-circuit" particularly helpful.
It is better to say that && and || have left-to-right order of evaluation. The C standard (6.5.13) actually explains this well:
the && operator guarantees left-to-right evaluation ...
If the first operand compares equal to 0, the second
operand is not evaluated.
And that's it. The left operand --a evaluates to 0.
What is the difference between operator precedence and order of evaluation?
In this case, all that matters is order of evaluation. Operator precedence only guarantees that the expression is parsed as expected - which operands that "glue" to which operator.
The expression (--a) uses the prefix operator and evaluates to 0. This is false and the 2nd expression is not evaluated due to short-circuit as you noted.
To understand consider the following expression
a * b + c * d
If to assume that the operands of the operator + are evaluated from left to right then at first the expression a * b will be evaluated and then the expression c * d.
The assumption relative to the order of evaluations of operands of the operator + is not valid. But it is valid for the operator &&.
So left operand of the expression
(--a)&&(b++);
is evaluated first. And if its value is not equal to 0 then the second operand is evaluated.
From the C Standard (6.5.13 Logical AND operator)
4 Unlike the bitwise binary & operator, the && operator guarantees
left-to-right evaluation; if the second operand is evaluated, there
is a sequence point between the evaluations of the first and second
operands. If the first operand compares equal to 0, the second operand
is not evaluated.
You are right that you can override operator precedence using parenthesis ().
For example, the expression
a && b || c
is equivalent to
(a && b) || c,
because && has a higher precedence than ||. You can override this precedence by changing it to
a && (b || c)
so that now the || has precedence over &&.
However, operator procedence doesn't change the behavior of the individual operators. In particular, both operators will still use short-circuiting. So, even if || has higher precedence than &&, the && operator will still evaluate its left-hand operand before its right-hand operand.
In both of my examples above, the && operator will still evalulate a before evaluating b or (b || c). Operator precedence will only affect whether the right-hand operand of the && operator is the expression b or (b || c).
Therefore, in your example, by writing (--a)&&(b++) instead of --a&&b++, you are merely ensuring that the -- and ++ operators have precedence over the && operator. This is not necessary, because he C language specifies that these operators already have precedence.
In the hypothetical scenario that the C language had instead specified that the && operator had precedence over the -- and ++ operators, then the expression --a&&b++ would be interpreted as --(a&&b)++, and it would be necessary to write (--a)&&(b++) to prevent this interpretation. But this is not the case.
Related
I'm using C language, for the below code:
#include <stdio.h>
int main(int argc, char const *argv[])
{
int num1=0;
int res = ++num1 && num1++;
printf("%d\t%d\n",num1,res);
}
In the above code I get output as 2 1. I think the output should be 2 0.
Please correct me if wrong, to solve this statement, the num1++(0) would be executed first due to highest precedence and then ++num1(2) would be executed and then at last && will be preformed because it has the lowest precedence.
Please comment how is this statement getting executed.
In some of the tutorials I find that postfix ++ and prefix ++ have the same precedence, but if that is true then according to the associativity rule again num1++ should be executed first(right to left) which should again lead to answer as 2 0.
In the expression used as an initializer
int res = ++num1 && num1++;
there is a sequence point for the operator &&.
From the C Standard (6.5.13 Logical AND operator)
3 The && operator shall yield 1 if both of its operands compare
unequal to 0; otherwise, it yields 0. The result has type int.
and
4 Unlike the bitwise binary & operator, the && operator guarantees
left-to-right evaluation; if the second operand is evaluated, there is
a sequence point between the evaluations of the first and second
operands. If the first operand compares equal to 0, the second operand
is not evaluated.
At first the left operand of the operator is evaluated and as a result num1 will be equal to 1 due to the unary (prefix) increment operator. As the sub-expression is not equal to 0 then the second operand is evaluated. Its value is the value before incrementing that is 1. As this second operand is also unequal to 0 then the whole expression is evaluated to the logical true and its value is 1 (see the first quote from the C Standard).
This value 1 is assigned to the variable res while the variable num1 after the postfix increment will be equal to 2.
So you will have that after this declaration res is equal to 1 and num1 is equal to 2.
Lots of misconceptions here. First of all, operator precedence states the order of parsing, not the order of execution. There are two related but different terms, operator precedence and order of evaluation.
See What is the difference between operator precedence and order of evaluation?.
Once you understand order of evaluation, the && operator specifically comes with well-defined sequencing, which isn't normally the case for C operators. It guarantees a left-to-right order of evaluation. C17 6.5.14/4:
Unlike the bitwise | operator, the || operator guarantees left-to-right evaluation; if the
second operand is evaluated, there is a sequence point between the evaluations of the first
and second operands. If the first operand compares unequal to 0, the second operand is
not evaluated.
Normally, you wouldn't be able to wild and crazy things with the ++ operator mixed with other operators, but the above && rule makes it possible in this specific case.
See Why can't we mix increment operators like i++ with other operators? It explains sequencing/sequence points.
In some of the tutorials I find that postfix ++ and prefix ++ have the same precedence,
They don't, prefix ++ takes precedence over postfix (and other unary operators). So associativity does not apply.
Please correct me if wrong, to solve this statement, the num1++(0) would be executed first due to highest precedence and then ++num1(2) would be executed and then at last && will be preformed because it has the lowest precedence.
Precedence only controls which operators are grouped with which operands - it does not affect the order in which expressions are evaluated.
The &&, ||, ?:, and comma operator all force left-to-right evaluation - the left operand is fully evaluated (and any side effects applied) before the right operand. && and || both short circuit - for &&, the right operand will be evaluated only if the left operand is non-zero.
The unary (prefix) ++ operator yields the current value of the operand plus 1, so the result of ++num1 is 1. As a side effect the value in num1 is incremented. Since this result is non-zero, num1++ is also evaluated. The postfix ++ operator yields the current value of the operand, so the result of num1++ is 1. As a side effect the value in num1 is incremented.
The result of an && expression is 1 if both operands are non-zero, 0 otherwise.
It's roughly equivalent to writing
tmp = num1 + 1;
num1 = num1 + 1;
res = 0;
if ( tmp != 0 )
{
if ( num1 != 0 )
{
res = 1;
}
}
num1 = num1 + 1;
So the result of ++num1 && num1++ is 1, and the value stored in num1 at the end of it is 2.
In some of the tutorials I find that postfix ++ and prefix ++ have the same precedence,
That is very wrong and you should stop using those tutorials immediately. Postfix operators have higher precedence than unary operators - *a++ is parsed as *(a++), ++a[i] is parsed as ++(a[i]), etc. An expression like ++i++ would be parsed as ++(i++), but you can't write such an expression in C - the result of i++ isn't an lvalue and cannot be the operand of a unary ++ like that.
In the following code, I am getting 10 | 1 | 1 as a result. But according to precedence rules shouldn't 'and' operator must be evaluated first?(and yield c=9) Like : d = a || (--b)&&(--c) since 'and' has higher precedence. ( or shortcutting breaks precedence rules ?)Thanks in advance.
#include <stdio.h>
#include <stdlib.h>
int main(){
int a,b,c,d;
a =1;
b = c = 10;
d = a|| --b&&--c;
printf("%d\n",c);
printf("%d\n",a);
printf("%d\n",d);
return 0;
}
Precedence and order of evaluation are two different things. From Logical OR
documentation (emphasis mine):
There is a sequence point after the evaluation of lhs. If the result of lhs compares unequal to zero, then rhs is not evaluated at all (so-called short-circuit evaluation).
In case of exp1 || exp2, exp1 is always evaluated first as there is a sequence point after it and if exp1 is non-zero then exp2 is not evaluated.
Precedence only determines which operands are grouped with which operators - it does not control the order in which expressions are evaluated.
In your example, it means the expression is parsed as
a || (––b && ––c)
Both || and && force left-to-right evaluation1. Both introduce a sequence point (IOW, the left hand operand will be evaluated and all side effects will be applied before the right hand operand is evaluated).
Both operators short-circuit - if the left operand of || evaluates to non-zero, then the result of the expression is 1 (true) regardless of the value of the right operand, so the right operand isn’t evaluated at all. If the left operand of && is 0, then the result of the expression is 0 (false) regardless of the value of the right operand, so the right operand isn’t evaluated at all.
In your expression, a is evaluated first. It has a non-zero value (1), so ––b && ––c is not evaluated.
Along with the ?: and comma operators. All other operators (arithmetic, equality, subscript, etc.) do not force a particular order of evaluation.
If they do lie in different row then shouldn't in Exp1 || Exp2 && Exp3, (Exp2 && Exp3) be evaluated first since && has higher precedence than ||?
And, If they do lie in same row then shouldn't order of execution be First (Exp1 || Exp2) and then && Exp3 since both are left associative?
Exp1 || Exp2 && Exp3 is grouped as Exp1 || (Exp2 && Exp3), due to operator precedence.
Note that (Exp2 && Exp3) is only actually evaluated if Exp1 evaluates to 0, and then Expr3 is only evaluated if Exp2 evalutes to 1. This is called short-circuiting.
Their being in a different "row" is not relevant.
Operator precedence in C is "as indicated by the syntax", meaning in plain English that each sub chapter of the standard chapter 6.5 is one operator group. The precedence is: first group, highest precedence:
6.5 Expressions
6.5.1 Primary expressions
6.5.2 Postfix operators
6.5.3 Unary operators
6.5.4 Cast operators
6.5.5 Multiplicative operators
6.5.6 Additive operators
6.5.7 Bitwise shift operators
6.5.8 Relational operators
6.5.9 Equality operators
6.5.10 Bitwise AND operator
6.5.11 Bitwise exclusive OR operator
6.5.12 Bitwise inclusive OR operator
6.5.13 Logical AND operator
6.5.14 Logical OR operator
6.5.15 Conditional operator
6.5.16 Assignment operators
6.5.17 Comma operator
As you can see, Logical AND is a separate group from Logical OR. It appears earlier in the standard so it has higher precedence.
If they had been "in the same row", that is if they would have the same precedence, then the operator associativity would determine how the expression should be parsed. a || b || c must be parsed as (a || b) || c because the associativity of this operator group is left-to-right.
Note that operator precedence only mandates which operands that belong with which operators. It doesn't specify order of evaluation (execution), which is a different but related term.
In your case, operator precedence specifies that Exp1 || Exp2 && Exp3 is equivalent to
Exp1 || (Exp2 && Exp3)
The order of evaluation of the || operator specifically is a special case, it must evaluate Exp1 completely first, and if it is false, and only then, evaluate the right-hand operator.
Usually in C, the order of evaluation of operands is not specified though. Take for example Exp1 | Exp2, where no special rule of order of evaluation exists. Then either Exp1 or Exp2 could be evaluated first.
This question already has answers here:
Is short-circuiting logical operators mandated? And evaluation order?
(7 answers)
Closed 6 years ago.
I always wondered about something and couldn`t find an answer somewhere else. If I have this piece of code:
if ((cond1) &&(cond2) && (cond 3) && (cond 4))
{
// do something
}
Let`s say the first condition is false, then my program will verify the other conditions too, or just skip verifying them?
But if I have
if ((cond1) ||(cond2) || (cond 3) || (cond 4))
{
// do something
}
and cond 1 is true, will my program go instantly on the if part or continue to verify the other conditions too?
Quoting C11 standard, chapter §6.5.13, Logical AND operator (emphasis mine)
Unlike the bitwise binary & operator, the && operator guarantees left-to-right evaluation;
if the second operand is evaluated, there is a sequence point between the evaluations of
the first and second operands. If the first operand compares equal to 0, the second
operand is not evaluated.
So, if the first condition (LHS operand) evaluates to false, the later conditions, i.e., RHS operand of the && is not evaluated.
Similarly (Ironically, rather), for Logical "OR" operator,
Unlike the bitwise | operator, the || operator guarantees left-to-right evaluation; if the
second operand is evaluated, there is a sequence point between the evaluations of the first
and second operands. If the first operand compares unequal to 0, the second operand is
not evaluated.
In C both && and || "sort-circuit", meaning if evaluation of the left operand is enough to determine the outcome then the right operand is not evaluated.
I have executed the following code in Code::Blocks 10.05 on Windows 7.
int a=0,b=0,c;
c=a++&&b++;
printf("\na=%d\nb=%d\nc=%d\n\n",a,b,c);
The output I obtained is given below,
a=1
b=0
c=0
This makes perfect sense because of short circuit evaluation.
The expression a++ is post increment and 0 is returned to the logical and (&&). Hence the part b++ is not evaluated since both 0 && 0 and
0 && 1 evaluates to 0.
But here arises my doubt. The precedence value of operators clearly states that ++ is having higher precedence over &&. So my understanding was like this, both a++ and b++ are evaluated and then && only checks the result of expression a++ to come to a decision. But this has not happened only a++ is evaluated here.
What is the reason for this behavior? Does && being a sequence point has something to do with this behavior? If so why we say that && is having lower precedence than ++?
You are confused about precedence and order of evaluation.
Precedence defines how the operators are grouped, i.e
c = a++ && b++;
is equivalent to:
c = ((a++) && (b++));
Order of evaluation defines how the expression is evaluated, the short circuit of && means a++ is evaluated first, if it's zero, the end; if it's not zero, b++ is then evaluated.
As another example:
c = (a++) + (b++);
Is a++ evaluated before b++? The answer is we don't know. Most operators don't define the order of evaluation. && is one of the few operators that do define. (The rest are ||, , and ?:)
There are two concepts here - order of precedence and order of evaluation. Order of precedence will have an impact only if an expression (or sub-expression) is evaluated.
In general, the order of evaluation is not sequenced. Given an operator, its operands can be evaluated in any order. The arguments of a function can be evaluated in any order.
From the C++ Standard:
1.9 Program execution
15 Except where noted, evaluations of operands of individual operators and of subexpressions of individual expressions are unsequenced.
and
8.3.6 Default arguments
9 Default arguments are evaluated each time the function is called. The order of evaluation of function arguments is unspecified.
For the logical AND operator, &&, the C++11 standard says:
5.14 Logical AND operator
1 The && operator groups left-to-right. The operands are both contextually converted to type bool (Clause 4). The result is true if both operands are true and false otherwise. Unlike &, && guarantees left-to-right evaluation: the second operand is not evaluated if the first operand is false.
Similar exception is specified for the logical OR operator, ||.
Since b++ is not evaluated due to short circuiting of the expression because of && operator, the order of precedence of the operators has no significance in this particular case.