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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.)

$x = 'foo';

if (isset($x)) {
    echo $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";
}
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
{$if DECLARED(x)}
writeln(x);
{$endif}
use 5.012;
print $x if $x; # won't compile, see explanation
if 'x' in locals():
	print x
puts x if defined?(x)

New implementation...
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