What's this =! operator?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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!

Answer 5

Operators in 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.

1. It could be typo mistake as I've noticed that =! operators is in if statement and someone is trying to type != instead of =! because != is the comparison operator which returns true or false.
2. Possibly, the developer was trying to assign the boolean negation of b to a 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

Answer 6

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)

Answer 7

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.

Answer 8

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.

Answer 9

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.

Answer 10

There could be three reasons for this:

1. It could be a mistyping of the != operator, meaning not equal to. Example:

   if (a != b) {
       // a is not equal to b
   }
   

2. It could be a mistyping a == !b, meaning a is equal to not b, 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)
   }
   

3. It could be trying to assign a to the inverse of b. Example:

   bool a = !b; // Sets a to the opposite of b
   

Answer 11

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) .

Answer 12

This can be taken as a is assigned to !b.

Answer 13

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.