Is My Understanding of the & Operator Correct?
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
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?
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'; |
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'; |
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...
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 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"); }
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.
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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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)
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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?