walrus operator of python language ( := )
work:- assign the value & also return that value.
language like swift at value assign it return nothing.
how to implement walrus operator kind a thing in swift language ?
I think it done by make function, pass address of variable & value.
assign value in that address & return value.
Is this work or any other way for this?
Joakim is correct.
Swift doesn't have a unary operator like C.
In C, you could do:
b = 10;
while (b>0) {
print(b--);
}
In Swift, there isn't a unary ++ or -- operator, so you would do:
var b = 10
while (b > 0) {
print b
b -= 1
}
but, really, in Swift, you'd do this instead
for b in (0...10).reversed() {
print b
}
See Reverse Range in Swift
Related
I am currently programming in BCPL for an OS course and wanted to write a simple is_digit() function for validation in a program of mine.
A code snippet of my current code follows:
let is_digit(n) be {
if ((n >= '0') /\ (n <= '9')) then
resultis true;
}
I am aware that BCPL has no notion of types, but how would I be able to accomplish this sort of thing in the language?
Passing in a number yields a false result instead of the expected true.
is_digit() is a function returning a value, rather than a routine, so should use = VALOF rather than BE. Otherwise, the code is OK.
let is_digit(n) = valof {
.....
resultis true
}
Functions that return values should be using valof rather than be, the latter (a routine rather than a function) can be called as a function but the return value you get back from it will be undefined(a).
In addition, you should ensure you return a valid value for every code path. At the moment, a non-digit will not execute a RESULTIS statement, and I'm not entirely certain what happens in that case (so best to be safe).
That means something like this is what you're after, keeping in mind there can be implementation variations, such as & and /\ for and, or {...} and $(...$) for the block delimiters - I've used the ones documented in Martin's latest manual:
LET is_digit(n) = VALOF {
RESULTIS (n >= '0') & (n <= '9')
}
(a) Since Martin Richards is still doing stuff with BCPL, this manual may help in any future questions (or see his home page for a large selection of goodies).
I was thinking of making an integer power function in Swift based on this StackOverflow answer:
func **<T : IntegerType>(var base: T, var exponent: T) -> T {
var result: T = 1
assert(exponent >= 0, "Exponent cannot be negative")
while exponent > 0 {
if exponent & 1 != 0 {
result *= base
}
exponent = exponent >> 1
base *= base
}
return result
}
I figured I could use generics to implement the function so that it would work with any integer type.
Unfortunately, I get an error when I attempt to use exponent >> 1:
Binary operator '>>' cannot be applied to two 'T' operands
Checking the function definitions for >>, I see that there is one for each of the ten integer types, but no other ones are defined. I was surprised therefore that all the other operators were working, such as &, but I noticed that & was actually defined to work on all types which conform to BitwiseOperationsType, which IntegerType appears to conform to.
Is there a reason why the >> and << operators are not implemented for BitwiseOperationsType?
I'm very confused even after looking through similar questions of what the(!) operator does when it is prefixed on a variable or other object in if statements, functions, etc?
Example:
mutating func add(value: T)
{
if !contains(items, value)
{
items.append(value)
}
}
the exclamation mark ! is used for two purposes. When you see it appear at the beginning of an object, such as in !contains(items, values), it means "NOT". For example...
let x = 10
let y = 5
if x == y {
print("x is equal to y")
} else if x != y {
print("x is NOT equal to y")
}
The above code will print => "x is NOT equal to y" .
The logical NOT (!) operator can be used to reverse boolean values. For example...
var falseBoolValue = false
falseBoolValue = !falseBoolValue
print(falseBoolValue)
The above code will print => "true"
In addition to the usage as the logical NOT operator, the exclamation mark is also used to implicitly unwrap optional values. Whenever you see the exclamation mark appear at the end of an object name, such as in someVariable!, it is being used to implicitly unwrap an optional value. Read about optionals to gain a better understanding of how ! is used with optional values.
it is NOT prefix there. Meaning your if looks for "items NOT containing value".
I know the difference between i++ and ++i in Swift. As the official document said, it is better to use ++i to increment i.
But I wonder why I get a syntax error using i++ in the for loop.
The code looks like this:
for var i = 0; i < 10; i++{
println("hello")
}
However, it is OK to use either i++ or ++i in other cases. Is there any restrictions in for loop?
The error says that:
Operator is not a known binary operator
The cause is very simple: you need to add a blank between the operator and the opening curly brace:
i++ {
^
without that, the compiler takes ++{ as a binary operator, with i and print("hello") as its arguments
The problem doesn't happen with the prefixed version of the increment operator because the i variable makes a clear separation between the ++ operator and the curly brace (letters and numbers cannot be used to define operators).
I have a problem with If statement in OpenScad.
I have 4 variables
a=20;
b=14;
w=1;
c=16;
I want to check witch number is bigger a or b.
And after depending who is smaller to take the value of smaller variable(in our case b < a) and to make a simple operation with c variable ( c=b-w).
I tried like this but it doesn't work.
a=20;
b=14;
w=1;
c=16;
if(a>b)
{
c=b-w;
}
if (a<b)
{
c=a-w;
}
if (a==b)
{
c=a-w;
}
It seems logic, but in openscad as I understood you can't change the value of variable inside a If statement. What trick can I use in order to get my goal.
Thank you!
in the 3. leg you confused the assignment-operator „=“ with the equal-operator „==“ (correct if (a==b)).
in your 3. leg you do the same as in the 2., so you could handle both as an „else“-leg.
Correct: assignment is not allowed in if-statement. In openscad you can use the ? operator instead:
c = a > b ? b-w : a-w;
After = follows the condition. The statement after the ? becomes the value if the condition is true, and the statement after the : becomes the value if the condition is false. Nested conditions are possible, e.g. your conditions:
c = a > b ? b-w : (a < b ? a-w : a-w);
More information in the documentation.
OpenSCAD's variable assignment is different. You can only assign variables inside a bracket. So c = b - w will only be assigned inside the if bracket. Outside if this bracket it will still be 16. Don't ask me why. You can read more in the Documentation of OpenSCAD.
c = min(c,min(a,b)/2-w);
this also solve the problem )