Are Int, String etc. considered to be 'primitives' in Swift? - swift

Question

Types representing numbers, characters and strings implemented using structures in swift.
An excerpt from the official documentation:
Data types that are normally considered basic or primitive in other
languages—such as types that represent numbers, characters, and
strings—are actually named types, defined and implemented in the Swift
standard library using structures.
Does that mean the following:
Int
Float
String
// etc
... are not considered as primitives?

Answer 1

Yes and no...
As other answers have noted, in Swift there's no difference at the language level between the things one thinks of as "primitives" in other languages and the other struct types in the standard library or the value types you can create yourself. For example, it's not like Java, where there's a big difference between int and Integer and it's not possible to create your own types that behave semantically like the former. In Swift, all types are "non-primitive" or "user-level": the language features that define the syntax and semantics of, say, Int are no different from those defining CGRect or UIScrollView or your own types.
However, there is still a distinction. A CPU has native instructions for tasks like adding integers, multiplying floats, and even taking vector cross products, but not those like insetting rects or searching lists. One of the things people talk about when they name some of a language's types "primitively" is that those are the types for which the compiler provides hooks into the underlying CPU architecture, so that the things you do with those types map directly to basic CPU instructions. (That is, so operations like "add two integers" don't get bogged down in object lookups and function calls.)
Swift still has that distinction — certain standard library types like Int and Float are special in that they map to basic CPU operations. (And in Swift, the compiler doesn't offer any other means to directly access those operations.)
The difference with many other languages is that for Swift, the distinction between "primitive" types and otherwise is an implementation detail of the standard library, not a "feature" of the language.
Just to ramble on this subject some more...
When people talk about strings being a "primitive" type in many languages, that's a different meaning of the word — strings are a level of abstraction further away from the CPU than integers and floats.
Strings being "primitive" in other languages usually means something like it does in C or Java: The compiler has a special case where putting something in quotes results in some data getting built into the program binary, and the place in the code where you wrote that getting a pointer to that data, possibly wrapped in some sort of object interface so you can do useful text processing with it. (That is, a string literal.) Maybe the compiler also has special cases so that you can have handy shortcuts for some of those text processing procedures, like + for concatenation.
In Swift, String is "primitive" in that it's the standard string type used by all text-related functions in the standard library. But there's no compiler magic keeping you from making your own string types that can be created with literals or handled with operators. So again, there's much less difference between "primitives" and user types in Swift.

Answer 2

Swift does not have primitive types.
In all programming languages, we have basic types that are available part of the language. In swift, we have these types availed through the Swift standard library, created using structures. These include the types for numbers, characters and strings.

Answer 3

No, they're not primitives in the sense other languages define primitives. But they behave just like primitives, in the sense that they're passed by value. This is consequence from the fact that they're internally implemented by structs. Consider:
import Foundation
struct ByValueType {
var x: Int = 0;
}
class ByReferenceType {
var x: Int = 0;
}
var str: String = "no value";
var byRef: ByReferenceType = ByReferenceType();
var byVal: ByValueType = ByValueType();
func foo(var type: ByValueType) {
type.x = 10;
}
func foo(var type: ByReferenceType) {
type.x = 10;
}
func foo(var type: String) {
type = "foo was here";
}
foo(byRef);
foo(byVal);
foo(str);
print(byRef.x);
print(byVal.x);
print(str);
The output is
10
0
no value

Answer 4

Just like Ruby, Swift does not have primitive types.
Int per example is implemented as structand conforms with the protocols:
BitwiseOperationsType
CVarArgType
Comparable
CustomStringConvertible
Equatable
Hashable
MirrorPathType
RandomAccessIndexType
SignedIntegerType
SignedNumberType
You can check the source code of Bool.swift where Bool is implemented.

Answer 5

There are no primitives in Swift.
However, there is a distinction between "value types" and "reference types". Which doesn't quite fit with either C++ or Java use.

Answer 6

They are partial primitive.
Swift not exposing the primitive like other language(Java). But Int, Int16, Float, Double are defined using structure which behaves like derived primitives data type(Structure) not an object pointer.
Swift Documentation favouring to primitive feature :
1.They have Hashable extension say
Axiom: x == y implies x.hashValue == y.hashValue.
and hashValue is current assigned value.
2.Most of the init method are used for type casting and overflow check (remember swift is type safe).
Int.init(_ other: Float)
3. See the below code Int have default value 0 with optional type. While NSNumber print nil with optional type variable.
var myInteger:Int? = Int.init()
print(myInteger) //Optional(0)
var myNumber:NSNumber? = NSNumber.init()
print(myNumber) //nil
Swift Documentation not favouring to primitive feature :
As per doc, There are two types: named types and compound types. No concept of primitives types.
Int support extension
extension Int : Hashable {}
All these types available through the Swift standard library not like standard keyword.