Need carification on the JavaScript code behaviour in strict mode

Question

In this MDN document about the Strict Mode of JavaScript, under "Semantic differences -> this in function calls", it's mentioned that:

When a function was called with call or apply, if the value was a primitive value, this one was boxed into an object (or the global object for undefined and null). In strict mode, the value is passed directly without conversion or replacement.

I need clarification for this statement. When I test this, I don't see any difference based on the mode (strict or sloppy) of the code.

Please let me know if I have misunderstood the statement.

This is how I've tested:

(function() { function a() { console.log(this); } function b() { "use strict"; a.call(2); } function c() { a.call(3); } b(); c(); })();

Result :

It works as expected if you put "use strict"; at top-level, the beginning of the IIFE, or in a(). But not if you put it in b() or c(). Not sure why. — Barmar
– Barmar, Commented Jul 7, 2021 at 14:46

Barmar · Accepted Answer · 2021-07-07 14:54:27Z

The behavior being described depends on whether the function being called is in strict mode, not the caller.

(function() { function strict() { "use strict"; console.log(this); } function sloppy() { console.log(this); } function b() { strict.call(2); } function c() { sloppy.call(3); } b(); c(); })();

Normally you put an entire script into strict mode, so the distinction between caller and callee doesn't matter.

Thank you for the answer. I think I've misunderstood the statement. Now it makes sense.

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