double negation to check against not NULL

Question

I have been seeing people writing code like

SomeType c=....

    if(!!c)
    {
     ....
    }

in what circumstance, would it be difference from

if (c)
{
   .....
}

When `c` isn't a `bool` (with a `true` or `false` value at least). — chris, Jul 09 '14 at 15:38
[Operator overloading](http://en.cppreference.com/w/cpp/language/operators). — BoBTFish, Jul 09 '14 at 15:38
so this means it is acutally same as if(c == true), which can also do the conversion of c to a bool — Peaceful, Jul 09 '14 at 15:39
@BoBTFish, Yeah, I didn't pay attention to the context too much. It would differ if the numerical value was used. — chris, Jul 09 '14 at 15:40
If there's ever a discrepancy between those two forms, that's a sign of bad coding. — David G, Jul 09 '14 at 15:40
Note that this is one of the cases where an expression is "contextually converted to bool", so `if(X)` is considered the same as `if((bool)X)`. (This works even if `X` has an **`explicit`** conversion to `bool`). — BoBTFish, Jul 09 '14 at 15:40
@Max, As long as a conversion from `SomeType` to `bool` exists. — chris, Jul 09 '14 at 15:40
@KKKoo0 No, for integers `(c == true)` checks whether `(c == 1)` (the `true` is promoted to an integer, rather than `c` to being demoted to a boolean). — , Jul 09 '14 at 15:43
@chris indeed, this can be [used to normalize to 0 or 1](http://stackoverflow.com/a/20916491/1708801). — Shafik Yaghmour, Jul 09 '14 at 15:44
Before C++11, one might define `operator!` to give some kind of "validness" test, but not define `operator bool` since that could cause confusing implicit conversions to numeric types. In that case, you'd need `!!` to express "valid". (These days, `explicit operator bool` does the job nicely.) — Mike Seymour, Jul 09 '14 at 16:08

Drew Dormann · Accepted Answer · 2014-07-09T15:55:59.893

4

This practice originates from the C language. Before C had a boolean type.

When c is 0, !!c is also 0.

When c is any other value, !!c is always 1.

This converts c into a true 2-state boolean. Testing expressions like ( !!bool1 == !!bool2 ) or ( !!bool3 == TRUE ) will give the expected result, even if these values are different bitwise representations of "true".

edited Jul 09 '14 at 15:55

answered Jul 09 '14 at 15:42

Drew Dormann

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Is it necessary in C++ any more? – Neil Kirk Jul 09 '14 at 15:51
1

@NeilKirk assuming `operator!` isn't overloaded to do something bizarre for the given type, it is not necessary. It's functionally equivalent to any cast to `bool`. – Drew Dormann Jul 09 '14 at 15:54

score 0 · Answer 2 · answered Jul 09 '14 at 16:30

I've found that the following generates a warning in Microsoft Visual C++:

int i = GetSomeValue();
bool b = (bool) i;

warning C4800: 'int' : forcing value to bool 'true' or 'false' (performance warning)

My intent is clear but the compiler insists on generating the warning. If I use the double negation then the warning goes away.

int i = GetSomeValue();
bool b = !!i;

double negation to check against not NULL

2 Answers2