This code is giving me datatype mismatch error.
const double dd = 0.225;
af::array aa = af::constant(111, 10, f32) + dd;
af::array bb = af::constant(111, 10, f64) + dd;
af::eval(aa, bb);
Both aa and bb should be f64 because both add a double value. There shouldn't be a datatype mismatch error.
I got their response on github.com/arrayfire/arrayfire/issues/2426, it remains as f32 after adding the double scalar on purpose. Anyway, it is actually a good thing for me to write program that supports precisions based on the caller's input. So, it is actually a good feature I want to keep. Marking this as resolved.
Related
I have a 4 byte hexadecimal value that I have a script to print out, But I want to now take that value then subtract the value C8 from it 37 times and save them as different variables each time, But the problem is I don't know how to do hexadecimal calculations in lua, If anyone can link me to any documentation on how to do this then that would be much appreciated.
You can make a hexadecimal literal in Lua by prefixing it with 0x, as stated in the reference manual. I found this by googling "lua hex"; such searches usually get good results.
"Hexadecimal numbers" aren't anything special, hexadecimal is just a way to represent numbers, same as decimal or binary. You can do 1000-0xC8 and you'll get the decimal number 800.
Code to convert:
function convertHex()
local decValue = readInteger(0x123456);
hexValue = decValue
end
function hexSubtract()
for i = 1,37 do
local value = 0xC8
hexValue = hexValue - 0xC8
result = hexValue
if i == 37 then
print(result) --Prints dec value
print(string.format('%X',result)); --Prints hex value
end
end
end
Replace 0x123456 with your address, use those functions like this convertHex(),hexSubtract()
I've done most of my development in C# and am just learning F#. Here's what I want to do in C#:
string AddChars(char char1, char char2) => char1.ToString() + char2.ToString();
EDIT: added ToString() method to the C# example.
I want to write the same method in F# and I don't know how to do it other than this:
let addChars char1 char2 = Char.ToString(char1) + Char.ToString(char2)
Is there a way to add concatenate these chars into a string without converting both into strings first?
Sidenote:
I also have considered making a char array and converting that into a string, but that seems similarly wasteful.
let addChars (char1:char) (char2: char) = string([|char1; char2|])
As I said in my comment, your C# code is not going to do what you want ( i.e. concatenate the characters into a string). In C#, adding a char and a char will result in an int. The reason for this is because the char type doesn't define a + operator, so C# reverts to the nearest compatable type that does, which just happens to be int. (Source)
So to accomplish this behavior, you will need to do something similar to what you are already trying to do in F#:
char a = 'a';
char b = 'b';
// This is the wrong way to concatenate chars, because the
// chars will be treated as ints and the result will be 195.
Console.WriteLine(a + b);
// These are the correct ways to concatenate characters into
// a single string. The result of all of these will be "ab".
// The third way is the recommended way as it is concise and
// involves creating the fewest temporary objects.
Console.WriteLine(a.ToString() + b.ToString());
Console.WriteLine(Char.ToString(a) + Char.ToString(b));
Console.WriteLine(new String(new[] { a, b }));
(See https://dotnetfiddle.net/aEh1FI)
F# is the same way in that concatenating two or more chars doesn't result in a String. Unlike C#, it results instead in another char, but the process is the same - the char values are treated like int and added together, and the result is the char representation of the sum.
So really, the way to concatenate chars into a String in F# is what you already have, and is the direct translation of the C# equivalent:
let a = 'a'
let b = 'b'
// This is still the wrong way (prints 'Ã')
printfn "%O" (a + b)
// These are still the right ways (prints "ab")
printfn "%O" (a.ToString() + b.ToString())
printfn "%O" (Char.ToString(a) + Char.ToString(b))
printfn "%O" (String [| a;b |]) // This is still the best way
(See https://dotnetfiddle.net/ALwI3V)
The reason the "String from char array" approach is the best way is two-fold. First, it is the most concise, since you can see that that approach offers the shortest line of code in both languages (and the difference only increases as you add more and more chars together). And second, only one temporary object is created (the array) before the final String, whereas the other two methods involve making two separate temporary String objects to feed into the final result.
(Also, I'm not sure if it works this way as the String constructors are hidden in external sources, but I imagine that the array passed into the constructor would be used as the String's backing data, so it wouldn't end up getting wasted at all.) Strings are immutable, but using the passed array directly as the created String's backing data could result in a situation where a reference to the array could be held elsewhere in the program and jeopardize the String's immutability, so this speculation wouldn't fly in practice. (Credit: #CaringDev)
Another option you could do in F# that could be more idiomatic is to use the sprintf function to combine the two characters (Credit: #rmunn):
let a = 'a'
let b = 'b'
let s = sprintf "%c%c" a b
printfn "%O" s
// Prints "ab"
(See https://dotnetfiddle.net/Pp9Tee)
A note of warning about this method, however, is that it is almost certainly going to be much slower than any of the other three methods listed above. That's because instead of processing array or String data directly, sprintf is going to be performing more advanced formatting logic on the output. (I'm not in a position where I could benchmark this myself at the moment, but plugged into #TomasPetricek's benckmarking code below, I wouldn't be surprised if you got performance hits of 10x or more.)
This might not be a big deal as for a single conversion it will still be far faster than any end-user could possibly notice, but be careful if this is going to be used in any performance-critical code.
The answer by #Abion47 already lists all the possible sensible methods I can think of. If you are interested in performance, then you can run a quick experiment using the F# Interactive #time feature:
#time
open System
open System.Text
let a = 'a'
let b = 'b'
Comparing the three methods, the one with String [| a; b |] turns out to be about twice as fast as the methods involving ToString. In practice, that's probably not a big deal unless you are doing millions of such operations (as my experiment does), but it's an interesting fact to know:
// 432ms, 468ms, 472ms
for i in 0 .. 10000000 do
let s = a.ToString() + b.ToString()
ignore s
// 396ms 440ms, 458ms
for i in 0 .. 10000000 do
let s = Char.ToString(a) + Char.ToString(b)
ignore s
// 201ms, 171ms, 170ms
for i in 0 .. 10000000 do
let s = String [| a;b |]
ignore s
I'm trying to pattern match a binary against this
<<_:(A * ?N + A + B)/binary,T:1/binary,_/binary>>
However it seems erlang throws an error saying that variable T is unbound. Just a quick explanation: I want to ignore a certain number of bytes and then read a byte and then ignore the remaining bytes. How can I achieve this?
In bit syntax we can't use runtime expressions as bit size.
We can use only constants, compile time expressions like _:(4*8)/binary and variables: _:Var/binary.
In your case, solution is to bind A * ?N + A + B to variable first.
IgnoredBytes = A * ?N + A + B,
<<_:IgnoredBytes/binary,T:1/binary,_/binary>> = SomeBinary,
T.
It's Better explained in answer from [erlang-questions]
I have a record declared as
T3DVector = record
X,Y,Z: Integer;
end;
One variable V of type T3DVector holds:
V.X= -25052
V.Y= 34165
V.Z= 37730
I then try to the following line. D is declared as Double.
D:= (V.X*V.X) + (V.Y*V.Y) + (V.Z*V.Z);
The return value is: -1076564467 (0xFFFFFFFFBFD4EE0D)
The following code should be equivalent:
D:= (V.X*V.X);
D:= D + (V.Y*V.Y);
D:= D + (V.Z*V.Z);
But this,however, returns 3218402829 (0x00000000BFD4EE0D), which actually is the correct value.
By looking at the high bits, I thought this was an overflow problem. When I turned on overflow checking, the first line halted with the exception "Integer overflow". This is even more confusing to me because D is Double, and I am only storing values into D
Can anyone clarify please ?
The target of an assignment statement has no bearing on how the right side is evaluated. On the right side, all you have are of type Integer, so the expression is evaluated as that type.
If you want the right side evaluated as some other type, then at least one operand must have that type. You witnessed this when you broke the statement into multiple steps, incorporating D into the right side of the expression. The value of V.Y * V.Y is still evaluated as type Integer, but the result is promoted to have type Double so that it matches the type of the other operand in the addition term (D).
The fact that D is a double doesn't affect the type of X, Y and Z. Those are Integers, and are apparently not large enough to store the squares of such large numbers, and their multiplication is what overflows. Why don't you declare them as doubles, too?
I have trouble with integer division in Dart as it gives me error: 'Breaking on exception: type 'double' is not a subtype of type 'int' of 'c'.'
Here's the following code:
int a = 500;
int b = 250;
int c;
c = a / b; // <-- Gives warning in Dart Editor, and throws an error in runtime.
As you see, I was expecting that the result should be 2, or say, even if division of 'a' or 'b' would have a result of a float/double value, it should be converted directly to integer value, instead of throwing error like that.
I have a workaround by using .round()/.ceil()/.floor(), but this won't suffice as in my program, this little operation is critical as it is called thousands of times in one game update (or you can say in requestAnimationFrame).
I have not found any other solution to this yet, any idea? Thanks.
Dart version: 1.0.0_r30798
That is because Dart uses double to represent all numbers in dart2js. You can get interesting results, if you play with that:
Code:
int a = 1;
a is int;
a is double;
Result:
true
true
Actually, it is recommended to use type num when it comes to numbers, unless you have strong reasons to make it int (in for loop, for example). If you want to keep using int, use truncating division like this:
int a = 500;
int b = 250;
int c;
c = a ~/ b;
Otherwise, I would recommend to utilize num type.
Integer division is
c = a ~/ b;
you could also use
c = (a / b).floor();
c = (a / b).ceil();
if you want to define how fractions should be handled.
Short Answer
Use c = a ~/ b.
Long Answer
According to the docs, int are numbers without a decimal point, while double are numbers with a decimal point.
Both double and int are subtypes of num.
When two integers are divided using the / operator, the result is evaluated into a double. And the c variable was initialized as an integer. There are at least two things you can do:
Use c = a ~/ b.
The ~/ operator returns an int.
Use var c;. This creates a dynamic variable that can be assigned to any type, including a double and int and String etc.
Truncating division operator
You can use the truncating division operator ~/ to get an integer result from a division operation:
4 ~/ 2; // 2 (int)
Division operator
The regular division operator / will always return a double value at runtime (see the docs):
for (var i = 4; i == 4; i = 3) {
i / 2; // 2 (double)
}
Runtime versus compile time
You might have noticed that I wrote a loop for the second example (for the regular division operator) instead of 4 / 2.
The reason for this is the following:
When an expression can be evaluated at compile time, it will be simplified at that stage and also be typed accordingly. The compiler would simply convert 4 / 2 to 2 at compile time, which is then obviously an int. The loop prevents the compiler from evaluating the expression.
As long as your division happens at runtime (i.e. with variables that cannot be predicted at compile time), the return types of the / (double) and ~/ (int) operators will be the types you will see for your expressions at runtime.
See this fun example for further reference.
Conclusion
Generally speaking, the regular division operator / always returns a double value and truncate divide can be used to get an int result instead.
Compiler optimization might, however, cause some funky results :)