compute for's end only once

examples/puzzles/aoc2024_04.zr

@@ -39,11 +39,11 @@ func part1[lines: Array] : void
 
 func part2[lines: Array] : void
     out := 0
+    s := _stackalloc(5) as str
 
     for x in 1..lines->size-1
         for y in 1..str.len(array.nth(lines, x))-1
             if array.nth(lines, x)[y] == 'A'
-                s := _stackalloc(5) as str
                 s[0] = array.nth(lines, x - 1)[y - 1]
                 s[1] = array.nth(lines, x + 1)[y - 1]
                 s[2] = array.nth(lines, x + 1)[y + 1]

examples/puzzles/aoc2025_03.zr

@@ -1,5 +1,6 @@
 func part1[lines: Array] : void
     sum := 0
+    s := _stackalloc(3) as str
 
     for i in 0..lines->size
         line : str = array.nth(lines, i)
@@ -7,7 +8,6 @@ func part1[lines: Array] : void
         largest := 0
         for j in 0..str.len(line)
             for k in (j+1)..str.len(line)
-                s := _stackalloc(3) as str
                 s[0] = line[j]
                 s[1] = line[k]
                 s[2] = 0