I am learning Kotlin and I came across string template, and find it very amazing!
Soon I discovered that a number can be converted to string with the help of it.
I know that toString() is the recommended way. But I want to understand, is the below method advisable to use and what good is the efficiency?
fun main() {
val integer: Int = 19
val string1: String = "$integer"
val string2: String = integer.toString()
println(string1)
println(string2)
}
Output:
19
19
Check This SO thread to understand how string interpolation is actually implemented in kotlin:
Bottom line you'll have a StringBuilder that is created every time when you perform an interpolation. This alone makes this method less efficient then a "simple and straight-to-the-point" integer.toString()
An additional reason is code clarity I would say: the usage of string interpolation for conversion from integer to string looks non-intuitive - so use the right tool for the right job, you know :)
Related
I'm new to Groovy. When I want convert some integer number to hex string, I have tried codes like this:
theNumber.toString(16)
as what I did in JavaScript. (Groovy is just like yet another script language looks similar to Java, right?)
But the code above not work as my expected. When the number is very large, the result is correct; but most of the time, it just return 16.
println(256.toString(16)) // 16
println(36893488147419103232.toString(16)) // 20000000000000000
I'm confused why Groovy behavior such strange. Could anyone help me to explain this? And, what is the best way to convert integer number to hex string?
Thanks.
Java is not JavaScript.
Groovy is a language built for the Java platform.
Java code also works directly with Groovy. So you can use .toHexString()
Integer.toHexString(256)
Long.toHexString(28562)
For numbers larger than the maximum value of long (9223372036854775807) the BigInteger datatype can be used.
String bigInt = new BigInteger("36893488147419103232").toString(16);
What you are calling is the static toString(int) from e.g. Integer. docs:
public static String toString(int i)
Returns a String object representing the specified integer. The argument is converted to signed decimal representation and returned as a string, exactly as if the argument and radix 10 were given as arguments to the toString(int, int) method.
E.g.:
groovy:000> Integer.toString(16)
===> 16
So what you want is:
groovy:000> Integer.toString(256,16)
===> 100
I'm starting to like the Swift string formatting since it uses variable names in the string rather than ambiguous formatting tags like "%#"
I want to load a large string from a file that has Swift-style formatting in it (like this)
Now is the time for all good \(who) to come to babble incoherently.
Then I want to feed the contents of that String variable into a statement that lest me replace
\(who)
with the contents of the constant/variable who at runtime.
The code below works with a string constant as the formatting string.
let who = "programmers"
let aString = "Now is the time for all good \(who) to come to babble incoherently."
That code does formatting of a quoted string that appears in-line in my code.
Instead I want something like the code
let formatString = "Now is the time for all good %# to come to babble incoherently."
aString = String(format: formatString, who)
But where I can pass in a Swift-style format string in a constant/variable I read from a file.
Is that possible? I didn't have any luck searching for it since I wasn't exactly sure what search terms to use.
I can always use C-style string formatting and the String class' initWithFormat method if I have to...
I don't think there's a way to do this. String interpolation is implemented via conforming to the StringInterpolationConvertible protocol, and presumably you're hoping to tap into that in the same way you can tap into the methods required by StringLiteralConvertible, a la:
let someString = toString(42)
// this is the method String implements to conform to StringLiteralConvertible
let anotherString = String(stringLiteral: someString)
// anotherString will be "42"
print(anotherString)
Unfortunately, you can't do quite the same trick with StringInterpolationConvertible. Seeing how the protocol works may help:
struct MyString: Printable {
let actualString: String
var description: String { return actualString }
}
extension MyString: StringInterpolationConvertible {
// first, this will get called for each "segment"
init<T>(stringInterpolationSegment expr: T) {
println("Processing segment: " + toString(expr))
actualString = toString(expr)
}
// here is a type-specific override for Int, that coverts
// small numbers into words:
init(stringInterpolationSegment expr: Int) {
if (0..<4).contains(expr) {
println("Embigening \(expr)")
let numbers = ["zeo","one","two","three"]
actualString = numbers[expr]
}
else {
println("Processing segment: " + toString(expr))
actualString = toString(expr)
}
}
// finally, this gets called with an array of all of the
// converted segments
init(stringInterpolation strings: MyString...) {
// strings will be a bunch of MyString objects
actualString = "".join(strings.map { $0.actualString })
}
}
let number = 3
let aString: MyString = "Then shalt thou count to \(number), no more, no less."
println(aString)
// prints "Then shalt thou count to three, no more, no less."
So, while you can call String.init(stringInterpolation:) and String.init(stringInterpolationSegment:) directly yourself if you want (just try String(stringInterpolationSegment: 3.141) and String(stringInterpolation: "blah", "blah")), this doesn't really help you much. What you really need is a facade function that coordinates the calls to them. And unless there's a handy pre-existing function in the standard library that does exactly that which I've missed, I think you're out of luck. I suspect it's built into the compiler.
You could maybe write your own to achieve your goal, but a lot of effort since you'd have to break up the string you want to interpolate manually into bits and handle it yourself, calling the segment init in a loop. Also you'll hit problems with calling the combining function, since you can't splat an array into a variadic function call.
I don't think so. The compiler needs to be able to resolve the interpolated variable at compile time.
I'm not a Swift programmer, specifically, but I think you can workaround it to something pretty close to what you want using a Dictionary and standard string-replacing and splitting methods:
var replacement = [String: String]()
replacement["who"] = "programmers"
Having that, you can try to find the occurrences of "\(", reading what is next and prior to a ")", (this post can help with the split part, this one, with the replacing part), finding it in the dictionary, and reconstructing your string from the pieces you get.
this one works like a charm:
let who = "programmers"
let formatString = "Now is the time for all good %# to come to babble incoherently."
let aString = String(format: formatString, who)
When doing format string interpolation in Sweden I get a comma instead of a dot when creating strings with decimal numbers:
scala> val a = 5.010
a: Double = 5.01
scala> val a = 5.0101
a: Double = 5.0101
scala> f"$a%.2f"
res0: String = 5,01
My question is, how do I set the format so that I get the result 5.01? I would like to be able to set the locale only for that String, i.e. so that I don't change the locale for the whole environment.
Cheers,
Johan
Using the same Java library number formatting support accessible
from StringOps enriched String class, you could specify another locale just for that output:
"%.2f".formatLocal(java.util.Locale.US, a)
(as described in "How to convert an Int to a String of a given length with leading zeros to align?")
The Scala way would be to use the string f interpolator (Scala 2.10+), as in the OP's question, but it is using the "current locale", without offering an easy way to set that locale to a different one just for one call.
Locale.setDefault(Locale.US)
println(f"$a%.2f")
I'd like to take a String e.g. "1234" and convert it to an Integer which represents the sum of all the characters.
I thought perhaps treating the String as a List of characters and doing a reduce / inject, would be the simplest mechanism. However, In all my attempts I have not managed to succeed in getting the syntax correct.
I attempted something along these lines without success.
int sum = myString.inject (0, { Integer accu, Character value ->
return accu + Character.getNumericValue(value)
})
Can you help me determine a simple syntax to resolve this problem (I can easily solve it in an java like verbose way with loops etc)
Try:
"1234".collect { it.toInteger() }.sum()
Solution by #dmahapatro
"1234".toList()*.toInteger().sum()
This question already has answers here:
Is there an easy way to return a string repeated X number of times?
(21 answers)
Closed 9 years ago.
Just a curiosity I was investigating.
The matter: simply repeating (multiplying, someone would say) a string/character n times.
I know there is Enumerable.Repeat for this aim, but I was trying to do this without it.
LINQ in this case seems pretty useless, because in a query like
from X in "s" select X
the string "s" is being explored and so X is a char. The same is with extension methods, because for example "s".Aggregate(blablabla) would again work on just the character 's', not the string itself. For repeating the string something "external" would be needed, so I thought lambdas and delegates, but it can't be done without declaring a variable to assign the delegate/lambda expression to.
So something like defining a function and calling it inline:
( (a)=>{return " "+a;} )("a");
or
delegate(string a){return " "+a}(" ");
would give a "without name" error (and so no recursion, AFAIK, even by passing a possible lambda/delegate as a parameter), and in the end couldn't even be created by C# because of its limitations.
It could be that I'm watching this thing from the wrong perspective. Any ideas?
This is just an experiment, I don't care about performances, about memory use... Just that it is one line and sort of autonomous. Maybe one could do something with Copy/CopyTo, or casting it to some other collection, I don't know. Reflection is accepted too.
To repeat a character n-times you would not use Enumerable.Repeat but just this string constructor:
string str = new string('X', 10);
To repeat a string i don't know anything better than using string.Join and Enumerable.Repeat
string foo = "Foo";
string str = string.Join("", Enumerable.Repeat(foo, 10));
edit: you could use string.Concat instead if you need no separator:
string str = string.Concat( Enumerable.Repeat(foo, 10) );
If you're trying to repeat a string, rather than a character, a simple way would be to use the StringBuilder.Insert method, which takes an insertion index and a count for the number of repetitions to use:
var sb = new StringBuilder();
sb.Insert(0, "hi!", 5);
Console.WriteLine(sb.ToString());
Otherwise, to repeat a single character, use the string constructor as I've mentioned in the comments for the similar question here. For example:
string result = new String('-', 5); // -----
For the sake of completeness, it's worth noting that StringBuilder provides an overloaded Append method that can repeat a character, but has no such overload for strings (which is where the Insert method comes in). I would prefer the string constructor to the StringBuilder if that's all I was interested in doing. However, if I was already working with a StringBuilder, it might make sense to use the Append method to benefit from some chaining. Here's a contrived example to demonstrate:
var sb = new StringBuilder("This item is ");
sb.Insert(sb.Length, "very ", 2) // insert at the end to append
.Append('*', 3)
.Append("special")
.Append('*', 3);
Console.WriteLine(sb.ToString()); // This item is very very ***special***