What's this =! operator?
C++COperatorsC++ Problem Overview
I was surprised by this code:
if (a =! b) { // let it be on false
...
}
But a
is never assigned by a value. What's this operator about?
C++ Solutions
Solution 1 - C++
That's two operators, =
and !
, not one. It might be an obfuscated way of writing
a = !b;
if (a) {
// whatever
}
setting a
to the logical inverse of b
, and testing whether the result is true (or, equivalently, whether b
was false).
Or it might be a mistyping of a != b
.
Solution 2 - C++
Long ago, when dinosaurs roamed the earth and C ran on 5th edition UNIX on PDP-11s, =!
was the 'not equals' operator. This usage was deprecated by the creation of Standard C, so now it means 'assign the logical inverse', as in a = !b
. This is a good argument for always surrounding binary operators with spaces, just to make it clear to the humans reading the code what the compiler is thinking.
I'm a bit surprised nobody else mentioned this, but then again I may be the only SO user to have ever touched a C compiler that old.
Solution 3 - C++
a
is assigned the boolean negation of b
in that line. It is just a misformatted
if( a = !b ) {
... and an evil hidden assignment inside a condition.
Solution 4 - C++
a =! b
is just a funny way of putting
a = !b
i.e. the assignment of not b
to a
.
The value of the expression is a
after the assignment.
With the code below you can see that the value of the expression a = !b
is !false
(i.e. true
), and you can then see the assignment has taken place by checking the value of a
, which is also true
.
#include <iostream>
int main()
{
bool a = false;
bool b = false;
if(a)
printf("a is true!\n");
else
printf("a is false!\n");
if(a = !b)
printf("expression is true!\n");
else
printf("expression is false!\n");
if(a)
printf("a is true!\n");
else
printf("a is false!\n");
}
Result:
a is false!
expression is true!
a is true!
Solution 5 - C++
According to C/C++ operators list there is no operator such as =!
. However, there is an operator !=
(Not equal to, Comparison operators/relational operator)
There are two possibilities.
- It could be typo mistake as I've noticed that
=!
operators is inif
statement and someone is trying to type!=
instead of=!
because!=
is the comparison operator which returns true or false. - Possibly, the developer was trying to assign the boolean negation of
b
toa
and he/she has done a typo mistake and forgot to put a space after equal sign. This is how the compiler interprets it, anyways. According to Operator precedence in c++:
- Operator Logical NOT (
!
) precedence is 3 and Associativity is Right-to-left - Operator Direct assignment (=) precedence is 16 and Associativity is Right-to-left
Solution 6 - C++
They are two different operators: the =
(assignment) operator together with the !
operator. It can basically be translated to an assignment of a
to the negated value of b
.
if (a = !b)
But, what the user, probably, meant to write was the !=
operator:
if (a != b)
Solution 7 - C++
That is not a single operator, it is however, a great way to obfuscate code.
If it were written a=!b
instead, the white space might not lead you to believe that it was a single operator.
Compilers have warnings for assignment in a conditional expression unless you wrap the entire statement in a set of parenthesis, and this is a perfect example of when this warning would be useful.
Both of these statements are functionally identical, but one generates a warning and the other does not:
if (a =! b) // Generates a warning with `-Wparentheses` (gcc)
if ((a =! b)) // No such warning
> -Wparentheses >>Warn if parentheses are omitted in certain contexts, such as when there is an assignment in a context where a truth value is expected, or when operators are nested whose precedence people often get confused about.
This, of course, assumes that you are a responsible programmer and actually read the warnings your compiler spits out.
Using white space in a more reasonable way, the statement is actually:
if (a = !b) // Assign A the value of (logical) NOT B and then test the truth
The compiler warning mentioned above is actually useful in the case where the person who wrote this code accidentally transposed !
and =
. However, from the cryptic inline comments in your original question, a = !b
is probably the author's intention.
Solution 8 - C++
C++ does not have an =!
operator, which immediately means that this is an =
operator followed by a !
operator. So, you simply have a = !b
in the if
condition.
Solution 9 - C++
This is all about clarity of code:
It should be written as: if (a = !b)
if (a = !b)
is the same as saying if a
is assigned to !b
. So there are technically 2 separate operators being used, =
which is an assignment operation, and !
which is a logical expression.
Just put a space between =
and !
it solves this confusion.
Solution 10 - C++
There could be three reasons for this:
-
It could be a mistyping of the
!=
operator, meaning not equal to. Example:if (a != b) { // a is not equal to b }
-
It could be a mistyping
a == !b
, meaninga
is equal to notb
, which would most commonly be used with booleans. Example:if (a == !b) { // The boolean a is equal to not b (a is not equal to b) }
-
It could be trying to assign
a
to the inverse ofb
. Example:bool a = !b; // Sets a to the opposite of b
Solution 11 - C++
I think this is one of the writing or printing mistake. Because if we can not compile this code as it holds a compiletime error. And here this is inside if clause, the return type must be boolean. So how can it possible. The answer is simply a mistake or as Mike said obfuscated way of writing.
And only possible if both variables a and b are boolean which will be compiled as a is equal to (not)b i.e (a = !b) .
Solution 12 - C++
This can be taken as a
is assigned to !b
.
Solution 13 - C++
I am trying to answer your question and, in agreement with Mike's answer, I am just adding up.
You can also understand it by a=!!b
.
The code a=!b
returns 0
to a
and a=!!b
returns 1
to a
.
The =!
is not a single operator but the mix of these two.
These types of question are asked to misguide students to check their coding skills.