Is !! a safe way to convert to bool in C++?

[This question is related to but not the same as this one.]

If I try to use values of certain types as boolean expressions, I get a warning. Rather than suppress the warning, I sometimes use the ternary operator (?:) to convert to a bool. Using two not operators (!!) seems to do the same thing.

Here's what I mean:

typedef long T;       // similar warning with void * or double
T t = 0;
bool b = t;           // performance warning: forcing 'long' value to 'bool'
b = t ? true : false; // ok
b = !!t;              // any different?

So, does the double-not technique really do the same thing? Is it any more or less safe than the ternary technique? Is this technique equally safe with non-integral types (e.g., with void * or double for T)?

I'm not asking if !!t is good style. I am asking if it is semantically different than t ? true : false.

The argument of the ! operator and the first argument of the ternary operator are both implicitly converted to bool, so !! and ?: are IMO silly redundant decorations of the cast. I vote for

b = (t != 0);

No implicit conversions.

Alternatively, you can do this: bool b = (t != 0)

Careful!

• A boolean is about truth and falseness.
• An integer is about whole numbers.

Those are very distinct concepts:

• Truth and falseness is about deciding stuff.
• Numbers are about counting stuff.

When bridging those concepts, it should be done explicitly. I like Dima's version best:

b = (t != 0);

That code clearly says: Compare two numbers and store the truth-value in a boolean.

All valid techniques, all will generate the same code.

Personally, I just disable the warning so I can use the cleanest syntax. Casting to a bool is not something I'm worried about doing accidentally.