a = function(x) {
return new Date().getTime();
}
b = function(x) {
return new Date().getTime();
}
c = function(x,y) {
return x+' '+y
}
console.log(c(a(),b()));
In the above code, functions a and b are called simultaneously by function c. Does function a execute first and then function b, or do both functions execute simultaneously? It's hard to tell since the time stamps returned are the same.
c(a(),b()) - this is a function invocation expression.
Invocation expressions consist of following parts:
When function invocation expression is evaluated, function expression is evaluated first and then the argument expressions, if there are any, are evaluated (from left to right) to produce values that are to be passed to the function that is being called.
c(a(),b()) is evaluated as:
First of all, identifier c is evaluated. If c wasn't defined, javascript would have thrown a ReferenceError.
Since in this case, c is a function, so argument expressions are evaluated from left to right.
First a() is evaluated which itself is an invocation expression. So a is evaluated first and since there are no arguments, function a is invoked and the body of the function a executes. Since a returns new Date().getTime(), it becomes the value of a() invocation expression.
After that b() is evaluated in the same way as a() is evaluated.
Finally, c() is called with the return values of a() and b() as its arguments. If c wasn't a function, javascript would have thrown a TypeError.
c is not defined, with a ReferenceError. If c is defined, the arguments are evaluated before trying to invoke it with those values, so if it's not actually callable you only find out with a TypeError around step 5."foo"((function () { throw new Error("whoops!"); })()) you'll see the uncaught whoops! error, not a TypeError that a string isn't callable.
Javascript: The Definitive Guide, says about Invocation Expressions in his book. If you read the fourth line, it says that function expression is evaluated before argument expressions.TypeError is thrown only after evaluating the argument expressions but function expression itself is evaluated before argument expressions, right?Javascript is always running in one Thread, so they are called after one another.
The reason why the timestamp is the same is because the clock of your computer (or the client computer) has only a specific resolution. I have something of 16ms in mind which is definitively bigger than the call to the two functions.
Anyway I'm not exactly sure what function is called first I would assume a and than b, but this can be dependent to the Javascript interpreter.
c()does not call eithera()orb(); they're called in preparation for the call toc(), and they're called one at a time, left to right.c, the arguments tocare resolved left-to-right by callingathenb.c(a(),b()). If it's not clear what's happening, don't do it that way. Good programmers write code that humans can understand.