OK. The upper left corner box character is 16{a. and displays (at
least on my screen) as follows:
16{a.
┌
a.i.8 u:4 u:16
16
a.i. '┌' NB. Why isn`t the expression above the same as this?
226 148 140
4 u: should give a wchar which is passed to 8 u: which should give U8.
But it doesn't. It's still 16{a. which is not a valid U8 character.
But if I take the character displayed it is valid U8. According to the
Vocabulary it seems that the result of the first expression should be
the same as the second.
For 6 u: it says that pairs of characters are converted to wchar. On
Windows/Intel where the low-order byte of a 16 bit integer comes
first:
a.i.6 u:'A',0{a.
65
Would the above expression still be the same on processors where the
high-order byte comes first? It seems to me that it should. I
shouldn't have to test the type of processor I'm using to determine
how 6 u: works.
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