Is My Understanding of the & Operator Correct?

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koranged
367
koranged  

I am a complete beginner, just trying to make sure I understand each section of the manual before progressing to the next.

My understanding of the & operator is as follows:

& compares the binary digits of two integers and returns a new integer. If the digits from Integer1 correspond with a '1' wherever the digits from Integer2 also have a '1' then the statement can be executed. Otherwise the statement cannot be executed. 

For Example:

----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

YES;int = 1 (00000001)

NO;int = 2  (00000010)

showPopup(button:int)

if(button & YES)

if(button & NO)

----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

In the above example, my understanding is that if the user clicks the button 'YES' then 'button' will return the binary digits of 00000001. If so then the '1' in 'button' would line up with the '1' in 'YES' and the statement could be executed. 

Along the same lines, the following statement - 'if(button & NO)' - would not be executed because the binary digits of the button clicked were 00000001, in which the '1' would not line up with the '1' in the 'NO' statement (00000010).

----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Is my understanding correct? Or have I got this totally wrong?

Anthony Garot
1844
Anthony Garot  

You aren't totally wrong, but your example would be better if you set NO = 0. Then you have conventional boolean logic, which is appropriate for a push button.

Thus, you have:

YES = 1 = true = 00000001

NO = 0 = false = 00000000

- - - - - - - - - - - - - - - - - - -

As for bitwise "and," that is using the "&" operator, the following are all true because the last binary digit (bit) is 1.

1 & 1 (00000001 & 00000001) = 1 (00000001)

3 & 1 (00000011 & 00000001) = 1 (00000001)

5 & 1 (00000101 & 00000001) = 1 (00000001)

7 & 1 (00000111 & 00000001) = 1 (00000001)

This Wikipedia article describes bitwise masking in more detail.

https://en.wikipedia.org/wiki/Mask_(computing)#Masking_bits_to_0

koranged
367
koranged  
Anthony Garot:

You aren't totally wrong, but your example would be better if you set NO = 0. Then you have conventional boolean logic, which is appropriate for a push button.

Thus, you have:

YES = 1 = true = 00000001

NO = 0 = false = 00000000

- - - - - - - - - - - - - - - - - - -

As for bitwise "and," that is using the "&" operator, the following are all true because the last binary digit (bit) is 1.

1 & 1 (00000001 & 00000001) = 1 (00000001)

3 & 1 (00000011 & 00000001) = 1 (00000001)

5 & 1 (00000101 & 00000001) = 1 (00000001)

7 & 1 (00000111 & 00000001) = 1 (00000001)

This Wikipedia article describes bitwise masking in more detail.

https://en.wikipedia.org/wiki/Mask_(computing)#Masking_bits_to_0

Hi Anthony,

I was under the impression that bitwise operators were not of boolean type?

I am aware that they are similar but my understanding was that a boolean will simply return 'true' or 'false' whereas the '&' operator will create a new integer with the value of '1' wherever the two compared integers also had a '1'. 

Is this not correct?

Marco vd Heijden
Moderator
8170
Marco vd Heijden  

Operation

Desciption

Execution Order

()

[]

.

Function Call

Referencing to an array element

Referencing to a structure element

From left to right

!

~

++

--

(type)

sizeof

Logical negation

Bitwise negation (complement)

Sign changing

Increment by one

Decrement by one

Typecasting

Determining size in bytes

Right to left

*

/

%

Multiplication

Division

Module division

From left to right

+

Addition

Subtraction

From left to right

<<

>>

Left shift

Right shift

From left to right

<

<=

>

>=

Less than

Less than or equal

Greater than

Greater than or equal

From left to right

==

!=

Equal

Not equal

From left to right

&

Bitwise AND operation

From left to right

^

Bitwise exclusive OR

From left to right

|

Bitwise OR operation

From left to right

&&

Logical AND operation

From left to right

||

Logical OR operation

From left to right

?:

Conditional Operator

Right to left

=

*=

/=

%=

+=

-=

<<=

>>=

&=

^=

|=

Assignment

Multiplication with assignment

Division with assignment

Module with assignment

Addition with assignment

Subtraction with assignment

Left shift with assignment

Right shift with assignment

Bitwise AND with assignment

Exclusive OR with assignment

Bitwise OR with assignment

Right to left

,

Comma

From left to right


Bitwise AND Operation

The bitwise AND operation of binary-coded x and y representations. The value of the expression contains a 1 (TRUE) in all digits where both x and y contain non-zero, and it contains 0 (FALSE) in all other digits.

b = ((x & y) != 0);

Example:

   char a='a',b='b'; 
//--- AND operation 
   char c=a&b; 
   Print("a = ",a,"  b = ",b); 
   Print("a & b = ",c); 
// The result will be: 
// a = 97   b = 98 
// a & b = 96


koranged
367
koranged  
Marco vd Heijden:

Operation

Desciption

Execution Order

()

[]

.

Function Call

Referencing to an array element

Referencing to a structure element

From left to right

!

~

++

--

(type)

sizeof

Logical negation

Bitwise negation (complement)

Sign changing

Increment by one

Decrement by one

Typecasting

Determining size in bytes

Right to left

*

/

%

Multiplication

Division

Module division

From left to right

+

Addition

Subtraction

From left to right

<<

>>

Left shift

Right shift

From left to right

<

<=

>

>=

Less than

Less than or equal

Greater than

Greater than or equal

From left to right

==

!=

Equal

Not equal

From left to right

&

Bitwise AND operation

From left to right

^

Bitwise exclusive OR

From left to right

|

Bitwise OR operation

From left to right

&&

Logical AND operation

From left to right

||

Logical OR operation

From left to right

?:

Conditional Operator

Right to left

=

*=

/=

%=

+=

-=

<<=

>>=

&=

^=

|=

Assignment

Multiplication with assignment

Division with assignment

Module with assignment

Addition with assignment

Subtraction with assignment

Left shift with assignment

Right shift with assignment

Bitwise AND with assignment

Exclusive OR with assignment

Bitwise OR with assignment

Right to left

,

Comma

From left to right


Bitwise AND Operation

The bitwise AND operation of binary-coded x and y representations. The value of the expression contains a 1 (TRUE) in all digits where both x and y contain non-zero, and it contains 0 (FALSE) in all other digits.

b = ((x & y) != 0);

Example:

   char a='a',b='b'; 
//--- AND operation 
   char c=a&b; 
   Print("a = ",a,"  b = ",b); 
   Print("a & b = ",c); 
// The result will be: 
// a = 97   b = 98 
// a & b = 96


Hi Marco,

That is just copy and pasted straight from the manual which I am already reading from mate.

Not really sure what the point in that was lol...

Anthony Garot
1844
Anthony Garot  
koranged:

I was under the impression that bitwise operators were not of boolean type?

You are mixing your metaphor. Operators are not types and types are not operators.

Bitwise operators act upon types of: boolean, int, long, char, etc. (but not on floats or doubles)

But bitwise operators use "boolean logic." That all sounds confusing, doesn't it?

Think in terms of what's really going on. There are NAND and NOR gates that operate on high/low voltage values.

a boolean will simply return 'true' or 'false'

A boolean variable holds a 'true' or 'false'. These are, in binary, 1 and 0, but since MQL booleans are "a whole number 1 byte large," technically these are 00000001 and 00000000. But you can mix types (so long as not float/double). The following is perfectly acceptable:

bool b = 1;
int c = 3;
Print (b&c);

My point, above, wasn't with your understanding of bitwise arithmetic. My point is that you would never use integer values of 1 and 2 to denote YES and NO states of a button. It simply isn't done. That's the whole point of the boolean type.

Now a tri-state button, yes, you might use something like:

OFF = 0 = 00000000
YES = 1 = 00000001
N/A = 2 = 00000011

Then you would check the state doing something like:

if ( state & 2 == 2 )
{
    Print("Is N/A");
}
koranged
367
koranged  
Anthony Garot:

You are mixing your metaphor. Operators are not types and types are not operators.

Bitwise operators act upon types of: boolean, int, long, char, etc. (but not on floats or doubles)

But bitwise operators use "boolean logic." That all sounds confusing, doesn't it?

Think in terms of what's really going on. There are NAND and NOR gates that operate on high/low voltage values.

A boolean variable holds a 'true' or 'false'. These are, in binary, 1 and 0, but since MQL booleans are "a whole number 1 byte large," technically these are 00000001 and 00000000. But you can mix types (so long as not float/double). The following is perfectly acceptable:


My point, above, wasn't with your understanding of bitwise arithmetic. My point is that you would never use integer values of 1 and 2 to denote YES and NO states of a button. It simply isn't done. That's the whole point of the boolean type.

Now a tri-state button, yes, you might use something like:

OFF = 0 = 00000000
YES = 1 = 00000001
N/A = 2 = 00000011

Then you would check the state doing something like:

Hi Anthony, my apologies I have only just noticed this reply. 

Okay I understand the principle of using int value 0 and 1 rather than 1 and 2, that makes sense. 

The confusion I have is more over the wording of 'if(button & YES)' or to use your recent example - 'if ( state & 2 == 2 )' 

What is 'button'?

In this code 'button' is never assigned with a value so how can the binary digits be compared with another integer if there are no binary digits?

Is it the case that 'button' is in fact representing what button the user pressed? And for example if the button pressed was 'YES' then this will return binary digits '00000001' which would obviously correspond with the binary digits of 'YES', therefore making the condition 'if(button & YES)'  true. 

Is this correct or have I just gone off on another tangent haha?

Either way thanks for your help bud. 

Marco vd Heijden
Moderator
8170
Marco vd Heijden  
koranged:

Hi Marco,

That is just copy and pasted straight from the manual which I am already reading from mate.

Not really sure what the point in that was lol...

To illustrate the difference between logical and bitwise.

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