In this situation, arguments.callee.name get nothing.
foo = function(){
console.log(arguments.callee.name);
}
Is there any solution?
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2The solution is correct. Your function has no name. – Bergi Feb 05 '15 at 02:21
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Why are you trying to do this? Argument callee is black magic and won't work in strict mode. – hugomg Feb 05 '15 at 02:24
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Why do you need the name of the function? – Felix Kling Feb 05 '15 at 03:14
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see also http://stackoverflow.com/q/28260389/1048572, which details why this isn't even *possible*. – Bergi Feb 05 '15 at 15:48
1 Answers
0
You didn't name it, thus it has no name.
foo = function(){
return (arguments.callee.name);
};
bar = function foobar() {
return (arguments.callee.name);
};
function foobar () {
return (arguments.callee.name);
}
console.log(foo()); //""
console.log(bar()); //"foobar"
console.log(foobar()) //"foobar"
The difference between foo() and foobar() is that foo() is a function expression, whereas foobar() is a declared function. The difference between foo() and bar() and is that one has a name, the other doesn't. Both are function expressions. Declared functions need a name, function expressions do not. See this canonical question for more info:
var functionName = function() {} vs function functionName() {}
