E.toString/E.toUint8Array - are these working correctly?

[espruino-discuss]

Posted at 2016-03-17 by DrAzzy

>scene
=new Uint16Array(32)
>scene[0]=0;
=0
>scene[1]=5000;
=5000
>scene[4]=44444;
=44444
>scene
=new Uint16Array([0, 5000, 0, 0, 44444,  ... 0, 0, 0, 0, 0])
>buff=new Uint8Array(64);
=new Uint8Array(64)
>E.toUint8Array(scene);
=new Uint8Array([0, 136, 0, 0, 156,  ... 0, 0, 0, 0, 0]) //truncated
>temp=new Uint8Array(64);
=new Uint8Array(64)
>temp.set(scene);
=undefined
>temp
=new Uint8Array([0, 136, 0, 0, 156,  ... 0, 0, 0, 0, 0]) //truncated
>E.toString(scene)
="\x00\x88\x00\x00\x9C\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00" //truncated
>temp = E.toUint8Array(E.toString(scene));
=new Uint8Array([0, 136, 0, 0, 156,  ... 0, 0, 0, 0, 0])
>temp=new Uint8Array(scene.buffer)
=new Uint8Array([0, 0, 136, 19, 0,  ... 0, 0, 0, 0, 0]) //not truncated (what I expected the other things to do)

Are those ones that truncate the values instead of splitting them up supposed to behave that way?

[gfwilliams]

Posted at 2016-03-17 by @gfwilliams

By truncating you mean producing an 8 bit number - turning 5000 into 136, rather than 19 and 136?

I think so - E.toX takes whatever you give it, and should produce something with the same number of elements. I think if you did:

a = new Uint16Array(4);
a[1] = 4;
b = E.toUint8Array(a);
console.log(b[1]);

You'd expect to get 4 returned.

As you found, if you want to get at the untruncated byte data of a UintXArray, you can just access array.buffer