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Programming-Idioms

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  • Perl

Idiom #258 Convert list of strings to list of integers

Convert the string values from list a into a list of integers b.

@b = map { $_ += 0 } @a

Technically, Perl doesn't have types, however, deep in its guts it does know if a variable has been used in a string context. This is for interacting with other languages and protocols. The '+=0' 'numifies' the entry.
System.Linq;
var b = a.Select(i => int.Parse(i)).ToList();
New implementation...
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programming-idioms.org 
@b = map { $_ += 0 } @a
var b = a.Select(i => int.Parse(i)).ToList();
var b = a.map(int.parse).toList();
  allocate (b(size(a)))
  read (unit=a,fmt=*) b

b := make([]int, len(a))
var err error
for i, s := range a {
	b[i], err = strconv.Atoi(s)
	if err != nil {
		return err
	}
}
let b = a.map(Number)
b = a.stream()
    .map(Integer::parseInt)
    .toList();
for (String s : a) b.add(parseInt(s));
b = stream(s)
    .mapToInt(Integer::parseInt)
    .toArray();
var
  a: TStringList;
  b: specialize TList<Integer>;
...
for i := 0 to a.count-1 do b.add(IntToStr(a[i]));
b = [int(elem) for elem in a]
b = [*map(int, a)]
b = a.map(&:to_i)
let b: Vec<i64> = a.iter().map(|x| x.parse::<i64>().unwrap()).collect();
let b: Vec<i32> = a.iter().flat_map(|s| s.parse().ok()).collect();

For i = LBound(a) To UBound(a)
    b(i) = CInt(a(i))
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