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Programming-Idioms

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Idiom #108 Determine if variable name is defined

Print the value of variable x, but only if x has been declared in this program.
This makes sense in some languages, not all of them. (Null values are not the point, rather the very existence of the variable.)

implicit none

print *,x
import std.stdio;
static if (is(typeof(x = x.init)))
    writeln(x);
int x = 42;

void printIfDefined(alias name)()
{
    import std.stdio: writeln;
    static if( __traits(compiles, writeln(mixin(name))))
        writeln(mixin(name));
}

void main(string[] args)
{
    printIfDefined!"x";
    printIfDefined!"Foo.bar";
}
if (typeof x !== 'undefined') {
    console.log(x);
}
try {
	console.log(x);
} catch (e) {
	if (!e instanceof ReferenceError) {
		throw e;
	}
}
if x then print(x) end
if(@$foo) print($foo);
$x = 'foo';

if (isset($x)) {
    echo $x;
}
{$if DECLARED(x)}
writeln(x);
{$endif}
no strict 'vars';  # the default
print $x;  # compile error

print $x if $main::{x}; # prints $x if it was declared


use strict 'vars';
# or
use v5.11; # or higher

print $x if $main::{x};  # unavoidable compile error

if 'x' in locals():
	print x
try:
    x
except NameError:
    print("does not exist")
puts x if defined?(x)

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