FWIW, here's the modern Python equivalent to "1..10.odd.sqr.lt(50)"
>>> [s for i in range(50) if i&1 and (s:=i*i) < 50]
[1, 9, 25, 49]
And here's my interpretation of the C equivalent:
#include <stdio.h>
int main() {
int i, s;
for(i = 1; i <= 10; i++) {
if (i & 1 && (s = i*i) < 50) {
printf("%d ", s);
}
}
return 0;
}
Not quite a-la 1972, but then again the example 1972 code from that page wouldn't compile then either - variables had to be declared at the start of the function.
I wonder what the APL looks like. I hacked this solution:
(x<50)/x←(1=2|x)/x←((⍳10)*2)
but I'm guessing the real solution would be 1/3 the size.
FWIW, here's the modern Python equivalent to "1..10.odd.sqr.lt(50)"
And here's my interpretation of the C equivalent: Not quite a-la 1972, but then again the example 1972 code from that page wouldn't compile then either - variables had to be declared at the start of the function.I wonder what the APL looks like. I hacked this solution:
but I'm guessing the real solution would be 1/3 the size.