Creating implementation for idiom 108 : Determine if variable name is defined

Idiom

# 108 Determine if variable name is defined

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Lead paragraph

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.)

New implementation

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Other implementations

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";
}

Demo

import std.stdio;

static if (is(typeof(x = x.init)))
    writeln(x);

implicit none

print *,x

if (typeof x !== 'undefined') {
    console.log(x);
}

Demo
Origin

try {
	console.log(x);
} catch (e) {
	if (!e instanceof ReferenceError) {
		throw e;
	}
}

import static java.lang.System.out;
import java.lang.reflect.Field;

try {
    Field f = getClass().getDeclaredField("x");
    out.println(f.get(this));
} catch (NoSuchFieldException e) {

} catch (IllegalAccessException e) {

}

if x then print(x) end

$x = 'foo';

if (isset($x)) {
    echo $x;
}

if(@$foo) print($foo);

{$if DECLARED(x)}
writeln(x);
{$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

try:
    x
except NameError:
    print("does not exist")

if 'x' in locals():
	print(x)

puts x if defined?(x)

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