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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.)

use 5.012;
print $x if $x; # won't compile, see explanation
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";
}
import std.stdio;
static if (is(typeof(x = x.init)))
    writeln(x);
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
$x = 'foo';

if (isset($x)) {
    echo $x;
}
{$if DECLARED(x)}
writeln(x);
{$endif}
if 'x' in locals():
	print x
puts x if defined?(x)

New implementation...
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