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It’s because parseInt is expecting a string, so the decimal gets converted to a string, and
0.0000005.toString()returns5e-7.Common Dynamic Typing L
More like javascript L, even python would throw a TypeError for this example.
And to further expand on that, if you do pass in a
stingstring, it handles it correctly.> parseInt('0.0000005') 0What if I pass in a Stewart Copeland?
😆 I’ll be watching you…
or a Honda civic
