Why ['1', '7', '11'].map(parseInt) returns [1, NaN, 3] in JavaScript
1–10 of 163 posts
Re: Why ['1', '7', '11'].map(parseInt) returns [1, NaN, 3] in JavaScript
#2Stuff like this is what drives me crazy when I work with dynamic languages like JS or PHP. I have seen a lot of code that looked perfectly fine but suffered from unexpected casts or defaults.
I much prefer languages like C#, C++ or TypeScript where the compiler warns me of such problems.
Re: Why ['1', '7', '11'].map(parseInt) returns [1, NaN, 3] in JavaScript
#3To quote the tl;dr: ['1', '7', '11'].map(parseInt) doesn’t work as intended because map passes three arguments into parseInt() on each iteration. The second argument index is passed into parseInt as a radix parameter. So, each string in the array is parsed using a different radix.
That's hilarious. Everybody loves the syntactic sugar that makes things easy, until the unexpected point where it makes things very hard.
Re: Why ['1', '7', '11'].map(parseInt) returns [1, NaN, 3] in JavaScript
#4This isn't strange or surprising. parseInt takes two arguments, the second one is the radix and map will call with three arguments, the value, the index and the whole array. You just have to know this and it might be different in other languages. [ ... ].map(x => ...) is the right way to do this.
Re: Why ['1', '7', '11'].map(parseInt) returns [1, NaN, 3] in JavaScript
#5Stuff like this is what drives me crazy when I work with dynamic languages like JS or PHP. I have seen a lot of code that looked perfectly fine but suffered from unexpected casts or defaults. I much prefer languages like C#, C++ or TypeScript where the compiler warns me of such problems.
This has nothing to do with casts or defaults. Map passes multiple parameters and parseInt accepts multiple parameters. Being unaware of the functions you're using is the problem, not some sort of weird gotchas of the language. The code in the title makes multiple assumptions, and those assumptions all proved wrong.
Re: Why ['1', '7', '11'].map(parseInt) returns [1, NaN, 3] in JavaScript
#6This isn't strange or surprising. parseInt takes two arguments, the second one is the radix and map will call with three arguments, the value, the index and the whole array. You just have to know this and it might be different in other languages. [ ... ].map(x => ...) is the right way to do this.
Yeah.. Didn't even need to load the article. You're passing the index as the radix. Makes sense
Re: Why ['1', '7', '11'].map(parseInt) returns [1, NaN, 3] in JavaScript
#7TL;DR, parseInt() takes the number base as a second argument, and map() passes three arguments (value, index, whole array).
Using directly like this a function with map() is just incorrect, the correct way to do it is:
['1', '7', '11'].map(x => parseInt(x))
Re: Why ['1', '7', '11'].map(parseInt) returns [1, NaN, 3] in JavaScript
#8This has less to do with "parseInt" than it has to do with "map" in JavaScript. It doesn't work exactly like .map in Java, Ruby, Elixir, etc., because ".map" in JS also passes in the index, and the full array.
Re: Why ['1', '7', '11'].map(parseInt) returns [1, NaN, 3] in JavaScript
#9This isn't strange or surprising. parseInt takes two arguments, the second one is the radix and map will call with three arguments, the value, the index and the whole array. You just have to know this and it might be different in other languages. [ ... ].map(x => ...) is the right way to do this.
Yeah, I’m not surprised at all by this. I get that there are a lot of weird things in JS but I don’t think this is a JS oddity.
Surprise! You need to know the basics of how your language works.
Re: Why ['1', '7', '11'].map(parseInt) returns [1, NaN, 3] in JavaScript
#10Easy fix ['1', '7', '11'].map(Number)