如何仅使用IF和GOTO语句实现一维数组的冒泡排序？

Hey there! Let's work through your bubble sort implementation—nested loops using just IF and GOTO are totally doable, you just need to align the label flow with bubble sort's core logic and fix a few key issues in your current code.

First, let's break down what's off in your existing code:

• Wrong sorting logic: You're comparing arr(i) with every arr(j), which is closer to selection sort, not bubble sort. Bubble sort relies on swapping adjacent elements to "bubble" the largest unsorted value to the end each pass.
• Broken swap: Lines 10-12 don't actually swap elements. You overwrite arr(j) with arr(i) before assigning arr(i) to the modified arr(j)—so no change happens.
• Inefficient loop bounds: The inner loop should only run through the unsorted portion of the array (up to arr_size - i), since each outer pass leaves the last i elements sorted.
• Misplaced jumps: After modifying j, you're jumping back to the outer loop check instead of the inner loop, which breaks the nested flow.

Here's a corrected, working version that sticks strictly to your constraints, plus a breakdown of how the nested loops work:

integer i,j,arr_size
      character*26 arr(1000), tmp
c Initialize outer loop counter
      i = 1
c Outer loop guard: check if we need another pass
299   if(i .lt. arr_size) go to 300
      go to 305  ! Exit sorting, all elements are ordered

c Start a new outer pass: reset inner loop counter to start of unsorted section
300   j = 1
c Inner loop guard: check if j is within the unsorted range
301   if(j .lt. arr_size - i) go to 302
      go to 304  ! End inner pass, move to next outer iteration

c Compare adjacent elements: swap if out of order
302   if(arr(j) .gt. arr(j+1)) go to 303
      go to 306  ! No swap needed, move to next pair

303   tmp = arr(j)
      arr(j) = arr(j+1)
      arr(j+1) = tmp  ! Correct swap logic using temporary variable

c Increment inner loop counter and jump back to inner guard
306   j = j + 1
      go to 301

c Increment outer loop counter and jump back to outer guard
304   i = i + 1
      go to 299

c Exit point for the subroutine
305   return
      end

How the Nested Loops Work with IF/GOTO: #

• Outer loop: Label 299 acts as the loop condition check. When i reaches arr_size, we exit to 305 (sorting complete). After each inner loop finishes, we increment i at 304 and jump back to 299 to start the next pass.
• Inner loop: For each outer pass, we reset j to 1 at 300. Label 301 checks if we're still in the unsorted part of the array. After each comparison (swap or not), we increment j and jump back to 301 to check the next adjacent pair.
• Swap correction: We now use tmp to store the original value of arr(j) before overwriting it with arr(j+1), then assign tmp to arr(j+1)—this properly swaps the two elements.

This structure exactly mimics the nested DO loops you'd use in modern Fortran, but uses only IF and GOTO for flow control, matching your requirements.

内容的提问来源于stack exchange，提问作者Huzo