Comment by eesmith

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.