kc3-lang/kc3/doc/1_KC3/1.6_Variable.en.md
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Thomas de Grivel
Date : 2024-10-14T17:14:30
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1.6 Variables
Variables in KC3 can be defined using the litteral value for a variable
which is always ?. You can cast this litteral value and it will not
really be casted but it will give you a typed variable litteral value.
E.g. (List) ?.
The typed variable litteral value will only accept to be set once to
one value of the variable’s type (in this example the type of a linked
list).
It’s actually a syntax so you cannot rename ? by mistake and
so is an easy task to do static analysis of variable creation.
The default type for a variable which you can also specify explicitly
is Tag which is an enum-tagged union type of any other KC3 types
currently defined in the environment. So ? is exactly equivalent to
(Tag) ? and they will both accept to be set once to one value of any
type.
A variable is settable once and cannot be changed afterwards (there is an exception if you write C code and link to it but it is not easy nor silent).
This way you do not need to lock or trust any behaviour, once your variable is set to a value the value of the variable will never change, it really is read-only.
You can also use the assignment operator which is <- which in turn calls
tag_init_copy. It works like the C assignment operator (=).
Examples :
# Declare a unsigned byte 8 bits variable "x".
x = (U8) ?
# Set the variable "x" to zero.
x <- 0
# Allocate again for the same binding name "x"
x = (U8) ?
# Also set the new variable "x" to zero with just one Unicode symbol
# that is AltGr+Y on my keyboard.
x ← 0So how do I change anything if it is read-only ?
You can always reset an existing binding at will to another variable litteral and another variable will be created for the same name and it will be in a different memory location, settable once and then read-only again so you can use it without locking.
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# 1.6 Variables Variables in KC3 can be defined using the litteral value for a variable which is always `?`. You can cast this litteral value and it will not really be casted but it will give you a typed variable litteral value. E.g. `(List) ?`. The typed variable litteral value will only accept to be set once to one value of the variable's type (in this example the type of a linked list). It's actually a syntax so you cannot rename `?` by mistake and so is an easy task to do static analysis of variable creation. The default type for a variable which you can also specify explicitly is `Tag` which is an enum-tagged union type of any other KC3 types currently defined in the environment. So `?` is exactly equivalent to `(Tag) ?` and they will both accept to be set once to one value of any type. A variable is settable once and cannot be changed afterwards (there is an exception if you write C code and link to it but it is not easy nor silent). This way you do not need to lock or trust any behaviour, once your variable is set to a value the value of the variable will never change, it really is read-only. You can also use the assignment operator which is `<-` which in turn calls `tag_init_copy`. It works like the C assignment operator (`=`). Examples : ```elixir # Declare a unsigned byte 8 bits variable "x". x = (U8) ? # Set the variable "x" to zero. x <- 0 # Allocate again for the same binding name "x" x = (U8) ? # Also set the new variable "x" to zero with just one Unicode symbol # that is AltGr+Y on my keyboard. x ← 0 ``` ## So how do I change anything if it is read-only ? You can always reset an existing binding at will to another variable litteral and another variable will be created for the same name and it will be in a different memory location, settable once and then read-only again so you can use it without locking. --- Top : [KC3 documentation](/doc/) Previous : [1.5 List](1.5_List) Next : [2 HTTPd](/doc/2_HTTPd)