Functions argument default value

Question

function foo( a, b ) {

  a = a || '123';
  b = b || 55;  
  document.write( a + ',' + b );
}

foo(); // prints: 123,55
foo('bar'); // prints: bar,55
foo('x', 'y'); // prints x,y

but:

foo(0,''); // prints: 123,55

why dont it print 0 ,55?

0 is falsy. I'd use var a = typeof a !== 'undefined' ? a : '123'; — Blender
– Blender, Commented Apr 12, 2013 at 6:21

deceze · Accepted Answer · 2013-04-12 06:16:52Z

4

Because || tests for truthiness, and 0 is among the values that are considered to be false.

answered Apr 12, 2013 at 6:16

deceze♦

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Comments

Joseph · Accepted Answer · 2013-04-12 06:17:03Z

3

Because the value 0 is a "falsy" value and is considered a false

answered Apr 12, 2013 at 6:17

Joseph

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Comments

Parthik Gosar · Accepted Answer · 2013-04-12 06:17:52Z

2

0 and "" also compute to false when checked. Hence you need to change your condition to

a = a != null ? a : '123';
b = b != null ? a : 55;

answered Apr 12, 2013 at 6:17

Parthik Gosar

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