Array Bubble Sort in Assembly Language - arrays

Question

Problem is that why is only sorting the last element and then putting 5 on every index?
I am posting for the first time so I apologize if my question is unclear.
arr BYTE 5,4,3,2,1
arr1 BYTE 5 DUP(?)
temp BYTE ?
.code
main PROC
mov ecx,LENGTHOF arr
mov esi,Offset arr
mov edi,offset arr+1
L1:
mov al,[esi]
mov edx,ecx
mov ecx,4
L2:
mov bl,[edi]
cmp al,bl
JG L3
jmp L5
L3:
mov temp,al
mov [esi],bl
;mov dl,temp
mov [edi],al
inc edi
L5:
loop L2
mov ecx,edx
inc esi
loop L1
;Printing my array
mov ecx,LENGTHOF arr
mov al,arr[0]
CALL dumpRegs
mov al,arr[1]
CALL dumpRegs
mov al,arr[2]
CALL dumpRegs
mov al,arr[3]
CALL dumpRegs
CALL dumpRegs
exit
main ENDP
END main

Answer 1

Problem is that why is only sorting the last element and then putting 5 on every index?
Your L2 InnerLoop works exclusively with the 1st array element since, once it is loaded in AL, you don't advance the pointer in ESI. And because this particular value happens to be the greatest value in the array, it gets stored everywhere.
Your L2 InnerLoop also works with a fixed count of 4 where it should be working with a count of CurrentOuterLoopCount - 1.
The correct solution does not need the 2 different pointer registers. The array elements that are to be compared are always adjacent to each other and so a simple offset of +1 will do the trick.
This is a working BubbleSort. Find out how it works, don't just copy it!
arr BYTE 5,4,3,2,1
.code
main PROC
mov ecx, LENGTHOF arr
sub ecx, 1
jbe Done ; Array is empty or has but 1 element
OuterLoop:
mov edx, ecx ; 5 elements means 4 comparisons
mov esi, Offset arr
InnerLoop:
mov al, [esi]
mov bl, [esi+1]
cmp al, bl
jng NoSwap
mov [esi], bl
mov [esi+1], al
NoSwap:
inc esi
dec edx
jnz InnerLoop
dec ecx ; Next time 3 comparisons, then 2, and then 1
jnz OuterLoop
Done: