Does anyone have some code that performs a bitwise XOR operation on two numbers?

Thanks in advance

Thanks in advance

If you were able to put the bin in a string with DCS7 commands, you could do something like:

**Code:**

```
:" "→str3
```

:For(A,1,length(str1

:expr(sub(str1,A,1→B

:expr(sub(str2,A,1→C

:If B xor C

:Then

:str3+"1"→str3

:Else

:str3+"0"→str3

:End:End

:Sub(str3,2,length(str3)-1→str3

I fixed it! It furs out I was using the wrong variable in the code that called this program, here's the source in case anyone would like to see it:

**Code:**

Variables:

N = 1st number

Z = 2nd number

R = Output

S = Starting bit to calculate from

E = Ending bit to calculate to

```
8->F
```

While F<S or F<E

F*2->F

End

0->R

For(A,1,F)

int(2fPart(.5N/2^(A-1->D

int(2fPart(.5Z/2^(A-1->L

If A<=E and A>=S and (D xor L):R+(2^(A-1))->R

End

Variables:

N = 1st number

Z = 2nd number

R = Output

S = Starting bit to calculate from

E = Ending bit to calculate to

I came up with the following technique, but I know Weregoose is about to ninja this post with the code he posted on IRC after you logged off last night. My version performs A xor B -> C. Shameless fact: I ripped off the complex number concept from Weregoose.

**Code:**

I am sure I can shrink the code that figures out how many bits to compute. Anyway, here's how that works:

1) Computes the number of binary digits that will need to be generated: ~min(int(~log(A)/log(2)),int(~log(B)/log(2. This uses a pair of change of base operations plus a quirk of the int() command to get the ceiling of the two numbers of digits, and effectively gets the max of the two.

2) Computes a complex number for each bit indicating the value of the bit in A and B: fPart((A+Bi)/(2^X)).

3) Make sure that it's either 1+0i or 0+1i: 1=abs(int(2...))

4) Give each resulting bit its proper place value: .5(2^X)

5) Generate a list of place-valued bits and sum them: sum(seq(

`sum(seq(.5(2^X)(1=abs(int(2fPart((A+Bi)/(2^X))))),X,1,~min(int(~log(A)/log(2)),int(~log(B)/log(2`

I am sure I can shrink the code that figures out how many bits to compute. Anyway, here's how that works:

1) Computes the number of binary digits that will need to be generated: ~min(int(~log(A)/log(2)),int(~log(B)/log(2. This uses a pair of change of base operations plus a quirk of the int() command to get the ceiling of the two numbers of digits, and effectively gets the max of the two.

2) Computes a complex number for each bit indicating the value of the bit in A and B: fPart((A+Bi)/(2^X)).

3) Make sure that it's either 1+0i or 0+1i: 1=abs(int(2...))

4) Give each resulting bit its proper place value: .5(2^X)

5) Generate a list of place-valued bits and sum them: sum(seq(

Wow, thanks! I'll post my other bitwise operation code (AND, Shift, OR), I'm sure you'll come up with a one liner for those

Let's see how similar these end up being...

**Code:**

If you want to retool this into another bitwise operation, go to the "=" and substitute it with "<" for AND, or "≤" for OR.

[EDIT]

Three fewer bytes and noticeably faster:

**Code:**

`sum(seq(2^Xabs(1=abs(int(2fPart(.5/2^X(A+Bi))))),X,0,log(2)⁻¹log(max({A,B}`

If you want to retool this into another bitwise operation, go to the "=" and substitute it with "<" for AND, or "≤" for OR.

[EDIT]

Three fewer bytes and noticeably faster:

```
:int(log(2)⁻¹log(max({A,B
```

:2^cumSum(binomcdf(Ans,0

:sum(Ans.5(1=abs(int(2fPart(Ans⁻¹(A+Bi

Wow, that is fantastic!

I have one question though, what is the ⁻¹ symbol?

the emulator won't accept it.

EDIT: Found it on catalog

I have one question though, what is the ⁻¹ symbol?

the emulator won't accept it.

EDIT: Found it on catalog

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