I have specific questions for my project
input = "3d6"
I want to convert this string some parts to integer. For instance I want to use input[0] like integer.
How can I do this?
There's two problems here:
How to convert a string to an integer
The most straightforward method is the Atoi (ASCII to integer) function in the strconv package., which will take a string of numeric characters and coerce them into an integer for you.
How to extract meaningful components of a known string pattern
In order to use strconv.Atoi, we need the numeric characters of the input by themselves. There's lots of ways to slice and dice a string.
You can just grab the first and last characters directly - input[:1] and input[2:] are the ticket.
You could split the string into two strings on the character "d". Look at the split method, a member of the strings package.
For more complex problems in this space, regular expressions are used. They're a way to define a pattern the computer can look for. For example, the regular expression ^x(\d+)$ will match on any string that starts with the character x and is followed by one or more numeric characters. It will provide direct access to the numeric characters it found by themselves.
Go has first class support for regular expressions via its regexp package.
For example,
package main
import (
"fmt"
)
func main() {
input := "3d6"
i := int(input[0] - '0')
fmt.Println(i)
}
Playground: https://play.golang.org/p/061miKcXdIF
Output:
3
Related
This question already has answers here:
How to convert an int value to string in Go?
(10 answers)
Closed last year.
this is my day 1 using goLang, I am currently trying to consume a data but I encounter an error, and that's converting integer to string
func createLink(title string, page int) string {
url := url.URL{
Scheme: "https",
Host: "jsonmock.hackerrank.com",
Path: "/api/movies/search/",
}
query := url.Query()
query.Set("page", string(page))
query.Set("title", title)
url.RawQuery = query.Encode()
return url.String()
}
you can try that code, and the result is
actual result :
https://jsonmock.hackerrank.com/api/movies/search/?page=%01&title=spiderman
expected result :
https://jsonmock.hackerrank.com/api/movies/search/?page=1&title=spiderman
There's %01 , an that's something that I do not want. i believe that I made a mistake in converting an integer to string
You should use strconv.Itoa() method to format your integers as strings. This is better explained in the linked answer. For the sake of completeness, here's how you end up with %01 in your result:
first, int 1 gets "plain-converted" to string by following this conversion rule:
Converting a signed or unsigned integer value to a string type yields
a string containing the UTF-8 representation of the integer. Values
outside the range of valid Unicode code points are converted to
"\uFFFD".
then the resulting string (with character of unicode code point equal to 1) gets URL-encoded, ending up with %01 as its representation.
As a sidenote, you're warned about this if you run go vet over your code:
hello.go:19:20: conversion from int to string yields a string of one
rune, not a string of digits (did you mean fmt.Sprint(x)?)
While this doesn't always give you absolutely the best advice on how to fix your error, it at least pushed you into the right direction. And it's strongly recommended to get used to the idea of running this (or similar) kind of checks from day 1 of learning language.
I am trying to convert small strings into their respective ascii decimal values. Like converting the string "Ag" to "065103".
I tried using
integer_variable : Integer := Integer'Value(Ag);
but that gives me constraint error: bad input for 'Value: "Ag".
Is there something else I can use to make this work? Could I just use enumeration?
Strings in Ada are arrays of Characters, thus if you want to convert String to Integer values you have to do this for each Character separately, by taking its position in enumeration (as suggested in the comment to your question). In your example it could be:
integer_variable1 : Natural := Character'Pos('A');
integer_variable2 : Natural := Character'Pos('g');
I have a long string which contains only decimal numbers with two signs after comma
str = "123,457568,22321,5484123,77"
The numbers in string only decimals with two signs after comma. How I can separate them in different numbers like that
arr = ["123,45" , "7568,22" , "321,54" , "84123,77"]
You could try a regex split here:
str = "123,457568,22321,5484123,77"
nums = str.split(/(?<=,\d{2})/)
print nums
This prints:
123,45
7568,22
321,54
84123,77
The logic above says to split at every point where a comma followed by two digits precedes.
Scan String for Commas Followed by Two Digits
This is a case where you really need to know your data. If you always have floats with two decimal places, and commas are decimals in your locale, then you can use String#scan as follows:
str.scan /\d+,\d{2}/
#=> ["123,45", "7568,22", "321,54", "84123,77"]
Since your input data isn't consistent (which can be assumed by the lack of a reliable separator between items), you may not be able to guarantee that each item has a fractional component at all, or that the component has exactly two digits. If that's the case, you'll need to find a common pattern that is reliable for your given inputs or make changes to the way you assign data from your data source into str.
I am wondering how to make something where if X=5 and Y=2, then have it output something like
Hello 2 World 5.
In Java I would do
String a = "Hello " + Y + " World " + X;
System.out.println(a);
So how would I do that in TI-BASIC?
You have two issues to work out, concatenating strings and converting integers to a string representation.
String concatenation is very straightforward and utilizes the + operator. In your example:
"Hello " + "World"
Will yield the string "Hello World'.
Converting numbers to strings is not as easy in TI-BASIC, but a method for doing so compatible with the TI-83+/84+ series is available here. The following code and explanation are quoted from the linked page:
:"?
:For(X,1,1+log(N
:sub("0123456789",ipart(10fpart(N10^(-X)))+1,1)+Ans
:End
:sub(Ans,1,length(Ans)-1?Str1
With our number stored in N, we loop through each digit of N and store
the numeric character to our string that is at the matching position
in our substring. You access the individual digit in the number by
using iPart(10fPart(A/10^(X, and then locate where it is in the string
"0123456789". The reason you need to add 1 is so that it works with
the 0 digit.
In order to construct a string with all of the digits of the number, we first create a dummy string. This is what the "? is used
for. Each time through the For( loop, we concatenate the string from
before (which is still stored in the Ans variable) to the next numeric
character that is found in N. Using Ans allows us to not have to use
another string variable, since Ans can act like a string and it gets
updated accordingly, and Ans is also faster than a string variable.
By the time we are done with the For( loop, all of our numeric characters are put together in Ans. However, because we stored a dummy
character to the string initially, we now need to remove it, which we
do by getting the substring from the first character to the second to
last character of the string. Finally, we store the string to a more
permanent variable (in this case, Str1) for future use.
Once converted to a string, you can simply use the + operator to concatenate your string literals with the converted number strings.
You should also take a look at a similar Stack Overflow question which addresses a similar issue.
For this issue you can use the toString( function which was introduced in version 5.2.0. This function translates a number to a string which you can use to display numbers and strings together easily. It would end up like this:
Disp "Hello "+toString(Y)+" World "+toString(X)
If you know the length of "Hello" and "World," then you can simply use Output() because Disp creates a new line after every statement.
I need to actually print a Dollar sign in Dart, ahead of a variable. For example:
void main()
{
int dollars=42;
print("I have $dollars."); // I have 42.
}
I want the output to be: I have $42. How can I do this? Thanks.
Dart strings can be either raw or ... not raw (normal? cooked? interpreted? There isn't a formal name). I'll go with "interpreted" here, because it describes the problem you have.
In a raw string, "$" and "\" mean nothing special, they are just characters like any other.
In an interpreted string, "$" starts an interpolation and "\" starts an escape.
Since you want the interpolation for "$dollars", you can't use "$" literally, so you need to escape it:
int dollars = 42;
print("I have \$$dollars.");
If you don't want to use an escape, you can combine the string from raw and interpreted parts:
int dollars = 42;
print(r"I have $" "$dollars.");
Two adjacent string literals are combined into one string, even if they are different types of string.
You can use a backslash to escape:
int dollars=42;
print("I have \$$dollars."); // I have $42.
When you are using literals instead of variables you can also use raw strings:
print(r"I have $42."); // I have $42.