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Programming-Idioms

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

Be concise.

Be useful.

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Please try to avoid dependencies to third-party libraries and frameworks.

Other implementations
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";
}
implicit none

print *,x
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}
use 5.012;
print $x if $x; # won't compile, see explanation
if 'x' in locals():
	print x
try:
    x
except NameError:
    print("does not exist")
puts x if defined?(x)