How does < and > operators works

Question

I recently got into a golf kata that asked to check if a letter is uppercase or not in no more than 9 characters.

The users answers were kind of scary (those example also works with > in reverse order):

f=s=>s<{}

f=c=>c<{}

f=_=>_<f

f=Z=>Z<f

Some variable names don't works with f:

f=a=>a<f

f=z=>z<f

Why and how does < and > operators works when comparing functions to basic strings? Maybe pointers comparison?

Try by yourself:

// working
f1=s=>s<{}
f2=c=>c<{}
f3=_=>_<f3
f4=Z=>Z<f4

// not working
f5=s=>s<f5

console.log('f1', f1('A'))
console.log('f1', f1('a'))

console.log('f2', f2('A'))
console.log('f2', f2('a'))

console.log('f3', f3('A'))
console.log('f3', f3('a'))

console.log('f4', f4('A'))
console.log('f4', f4('a'))

console.log('f5', f5('A'))
console.log('f5', f5('a'))

Answer 1

Comparing a string to {} is effectively comparing the string to the results of ({}).toString(), which is always "[object Object]". The important part is the [ character, which is the first code point after the last upper-case letter (Z). Thus, if you know that the input string is some letter, comparing to a string beginning with [ will tell you that it's an upper-case letter.

To be clear,

s => s<{}

is effectively

s => s<"[object Object]"

It has nothing to do with variable names chosen when you use an object ({}), but it's important for function-reference versions. In those, the "_" is important because it's a character after "Z" but before all the lower-case letters.