Don't use console.log like this

[graduate]

I've spent some time solving this problem till I understood that `%` symbol has a special meaning in `console.log`, the same like in function `printf` in C programming language.

var str = '%apple', n = 25;console.log( str, n ); // You expect "apple 25" but you get "25pple"

[thescientist]

I don't see the %.

[graduate]

Sorry, I changed my first post.

[thescientist]

actually, I wouldn't be sure what to expect since you are using two variables delimited by a comma. perhaps the comma is open to interpretation. doing this achieves your expected result however.

var str = '%apple', n = 25;console.log(str + n); // outputs apple25

[graduate]

Yes, when you dont use the %, the function just join its arguments into a string, with % it uses the first argument as a pattern and includes into it the others.