Can't reset QFlag in simple class

mrjj

Hi
Its store OR combination of the enums. (in your case zero and 1 )
The code is

 Q_DECL_RELAXED_CONSTEXPR inline QFlags &setFlag(Enum flag, bool on = true) noexcept
    {
        return on ? (*this |= flag) : (*this &= ~Int(flag));
    }

becomes
value =0
value |= 0
value |= 1
value |= 0
...

So it seems you are expecting it to behave differently?

Please_Help_me_D

@Christian-Ehrlicher yes, I set a breakpoint and line by line watch the status of variable flag. And in other my application the behavior is the same and I can see this not only with debugger but also how the application runs.

@mrjj thank you for explanation but this code exceeds my knowledge on C/C++ operators :)
I wanted that QFlag would be a trigger wich has two positions: Yes and No. I could replace it with bool variable but I want to get experience with QFlags (or C++ flags).
So inside my class I need to be able to change the satus of flag manually. I thought if I write flag.setFlag(Yes); then flag.testFlag(Yes); should return true and in opposite: flag.setFlag(No); flag.testFlag(No); should return true
But this doesn't work like I expected...

mrjj

@Please_Help_me_D
Hi
yes a bool would act as a toggle but QFlags is
actually more like a list of bools.
You can then set one or more of these bools.

so say you had something to describe a person

 enum MyFlag {
        Clever, // default
        Handsome,
       Rich
    };

then
flag.setFlag( Clever );
flag.setFlag( Handsome );

Then you set the flags for the person to be Clever and handsome.

For your exmaple to make it flip/flop you would have to do
flag.setFlag(Yes ,true ); // turn on
flag.setFlag(Yes, false); // turn off
flag.setFlag(No); // turn ON

Setting No alone , will not turn off Yes.
It means that No is on also. (or off)

Hope this makes better sense then :)

Please_Help_me_D

@mrjj thank, now I understood but the problem is in this way they also don't work.
I just tried to rewrite MyClass here it is:

#include <QFlags>

class MyClass
{
public:
    enum MyFlag {
        Yes, // default
        No
    };
    Q_DECLARE_FLAGS(MyFlags, MyFlag)

    MyClass(){
        flag.setFlag(Yes, true);
        bool a = flag.testFlag(Yes); // a == true

        flag.setFlag(Yes, false);
        bool b = flag.testFlag(Yes); // b == true

        flag.setFlag(Yes, true);
        bool c = flag.testFlag(Yes); // c == true
    }

private:
    MyFlags flag;
};

Why all the three bool vars are true? Is that ok?

mrjj

Hi
I think its due to using a value of zero for one of the enums.

try with

enum MyFlag {
        Yes = 1, // default
        No =2
    };

then it looks as expected
alt text

Please_Help_me_D

@mrjj huh now it works :)
thank you!

SGaist

Hi,

It is shown in and explained in the QFlags that each item of your flags need to have a different value (1 -> 2 -> 4 -> 8 -> etc.) that can be ORed otherwise you cannot establish which was set. You can do that using a binary format for your number, it will make things clearer.

Please_Help_me_D

@SGaist thank you!
But it is always difficult for me to read documentation about something new that I don't have any idea. It becomes much easier when I already have some representation about studied topic.
As example I've read both this and that but nothing helped me to solve my task. I'm like a slowpoke :)
But now when I know that QFlag is something like a list of boolean variables I can match documentation with my needs. I always search "how to solve my problem" in Internet and only then I disturb people from communities.
This is just my excuses so you would not think that I don't do anything to solve my problem by myself :)

I really appreciate forum's people help because you save a lot of time and nerves and honestly saying I would have never learned programming (Matlab and C/C++) if there were not such communities.

mrjj

@Please_Help_me_D

Hi
Yes its like a list
but to make it take up far less space, it uses bits instead of a bool.
So basically QFlags helps you get/and set the bits.

You know for programming a value can also be shown as binary 0 and 1 ?

When you OR something, then just one of the bit has to be set then it becomes set in the result too.

0 1 0 1 OR
0 1 1 0
-------
0 1 1 1

with AND operator both bits must be set to be set in the result.

Thats how QFlags works interinally

if we look at

 Q_DECL_RELAXED_CONSTEXPR inline QFlags &setFlag(Enum flag, bool on = true) noexcept
    {
        return on ? (*this |= flag) : (*this &= ~Int(flag));
    }

then it says

if (on is true )
value = value OR flag ( in your case yes )
else if on is false ( to turn off )
value = value AND (negative) flag

the last part is taking out the bits that belong to the value for the flag.

So AND / OR is not that difficult to understand at a high level as it simply
manipulate the values as bits.

Please_Help_me_D

@mrjj thank you!
I understood something but some interesting things I don't. Here what I get.
Yes I understand byte/bit

This:

on ? (*this |= flag) : (*this &= ~Int(flag));

is equal to:

if ( on == true ){
    *this = (*this || flag);
} else {
    *this = (*this &~ Int(flag));
}

right?
If so then to understand the mathematics I need to understand what in bit representation is: *this, flag, Int(flag)
In my case flag is equal to 1 right ( enum MyFlag { Yes = 1} )? in bit representation 0001
But what is in *this before assigning anything (operator =) and what function Int() does? simply converts flag to integer? as I know this type of casting doesn't change bits of the number it simply tells "these bits should be read as int"
Something like that:)

||________________
By the way how to turn on notifications when somebody answers me or writes in topic created by me?

mrjj

@Please_Help_me_D

Yes odd syntax is called Conditional ternary operator:
but is just a one liner if else.

(condition) ? (if_true) : (if_false)

The *this de reference the this pointer.
so its what it points to.
This is done to trigger the operator = which QFlags have overloaded.
so its a way to assign the value to it self using =

the Int() is a type cast like (Int)flag;
But its a bit more if we look at source

#if defined(Q_CC_MSVC) || defined(Q_CLANG_QDOC)
    // see above for MSVC
    // the definition below is too complex for qdoc
    typedef int Int;
#else
    typedef typename std::conditional<
            std::is_unsigned<typename std::underlying_type<Enum>::type>::value,
            unsigned int,
            signed int
        >::type Int;
#endif

so when not MSVC compiler it also checks the type

And if you think that looks strange, you are right :)

Its c++ templates and while really useful, its also
often ugly and unreadable. So ignore it until later in your training.

Please_Help_me_D

@mrjj all the things that depend on the compiler features are far outside of my current possibilities :)
So I think I begin to understand the magic behind QFlag.
returned value is of type bool and bool is not a single bit but rather a byte (maybe few bytes). So any numbers written in Yes except zero in this expression:

*this || flag

returnes some byte (bytes) where even a single (or more) bits is not zero. A bool which has even a single (or more) non zero bit should return true right?
And I believe that behind this dark veil:

*this &~ Int(flag)

are hidden all zeroed bits :)

By the way is it necessary to assign numbers to enum definitions (Yes, No etc) that are equal the power of 2 (1, 2, 4, 8, 16 ...)? as in many examples

mrjj

@Please_Help_me_D

Ok that's good :)
maybe you are ready for
https://www.youtube.com/watch?v=TKEaakjKrMc
Bitwise AND / OR

Actually the dark veil:
is using AND and BINARY NOT
to remove the bits for a flag

Yes, you must use power of 2 values as else some of the values will have
overlapping bit representation and things get funny when you combine certain flags.

Please_Help_me_D

@mrjj thank you
Video helped me to remember what XOR. Also now it I understand how to easilly transform binary data representation to decimals :)