Why does this return 'foo', not 'foobar'? I need function g to modify (non-global) var v, yet function g is a global function. Thanks.
f();
function f() {
var v = 'foo';
g(v);
alert(v);
}
function g(v) {
v = v+'bar';
return v;
}
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3 Answers
Because you return v from g(v) call but you do not reassign v
f();
function f() {
var v = 'foo';
v = g(v); //you need to assign what is returned
alert(v);
}
function g(v) {
v = v+'bar';
return v;
}
Comments
Because javascript only works by value, not by reference. See John Hartsock's answer.
Comments
Primitive ALL arguments in javascript (the string argument to g, in this case) are pass-by-value instead of pass-by-reference, which means the v you're working with in function g(v) is a copy of the v in function f.
Edit: all arguments are passed by value, not just primitives.
3 Comments
Pointy
All arguments in JavaScript are passed by value.
Kevin Qi
My mistake, you're right. Non-primitives are passed by value as well, but you can modify their properties.
Pointy
Yup. That's a really frequent argument that pops up, but the key is that the term "call-by-reference" (like "call-by-value" of course) has a pretty specific meaning, and the fact that values can be references to objects is a frequent source of confusion. Too bad they didn't call it "call-by-back-pointer" or something :-)