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  3. MASM 9.0 bug.

MASM 9.0 bug.

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  • M Member 4194593

    Where should I post a MASM 9.0 bug. It caused a very hard to find error (did not crash). Don't tell me MS, and don't ask why not. Dave.

    N Offline
    N Offline
    newton saber
    wrote on last edited by
    #8

    I think you could've explained it by now. I built a component with MASM which calls op code

    mov eax, [ebx]

    When it executes the processor melts. Maybe something like that. Keep it interesting and we'll read it. :) Good luck.

    M 1 Reply Last reply
    0
    • N newton saber

      I think you could've explained it by now. I built a component with MASM which calls op code

      mov eax, [ebx]

      When it executes the processor melts. Maybe something like that. Keep it interesting and we'll read it. :) Good luck.

      M Offline
      M Offline
      Member 4194593
      wrote on last edited by
      #9

      Thank you for the invitation. Just a warning to all. I had a function which had always been working. I modified it to align a short loop (added ALIGN OWORD). The function fails to function correctly (did not cause an error - just bad results which caused other errors - quite difficult to determine just what was happening). The data was 8 DWORDS which were all 0 except most significant which was 40000000h, and I was subtracting a single DWORD with a value of 1. I expected to get 7 DWORDS of FFFFFFFFh and the most significant DWORD with 3FFFFFFFh:

      This is the source code:

      ;*******************************************************************************
      ;
      ; SUB a LONG_NUMB value from another. This is a routine called to subtract a
      ; long number from another after all scaling and validations have been done.
      ;
      ; esi has the source OFFSET of the data.
      ; edi has the already scaled OFFSET of the destination data.
      ; ecx has the DWORD count to subtract.
      ;
      ; Returns nothing with edi pointing to the last word modified. Note that this
      ; may not be the highest DWORD in the destination if a short source is
      ; subtracted from a long destination and there are no borrows past the last
      ; DWORD in the destination or source. This is primarily a check for a size
      ; extension due to a final borrow.
      ;
      ;*******************************************************************************

      ALIGN OWORD

      SUBIt PROC PRIVATE USES eax ebx esi

      pushfd
      clc
      

      ;
      ; subtract all the value DWORDS.
      ;
      ALIGN OWORD

      DoAll:
      mov ebx,[esi]
      lea esi,[esi + (SIZEOF DWORD)]
      sbb [edi],ebx
      lea edi,[edi + (SIZEOF DWORD)]
      dec ecx
      jnz DoAll
      jnc Exit
      mov eax,0
      ;
      ; Propogate the borrow.
      ;
      ALIGN OWORD

      Again:
      sbb [edi],eax
      lea edi,[edi + (SIZEOF DWORD)]
      jc Again
      ;
      ; Exit SUBIt, adjust the destination pointer to the last modified DWORD.
      ;
      Exit:
      sub edi,(SIZEOF DWORD)
      popfd
      test dTestZero,0
      ret
      SUBIt ENDP

      This is the generated code from MASM 9.0 (Visual Studio "Disassembly" tab):

      ALIGN OWORD

      SUBIt PROC PRIVATE USES eax ebx esi
      00404FC0 push eax
      00404FC1 push ebx
      00404FC2 push esi

      pushfd
      

      00404FC3 pushfd
      clc
      00404FC4 clc
      00404FC5 lea esp,[esp]
      00404FCC lea esp,[esp]
      ;
      ; subtract all the value DWORDS.
      ;
      ALIGN OWORD

      DoAl

      L 1 Reply Last reply
      0
      • M Member 4194593

        Thank you for the invitation. Just a warning to all. I had a function which had always been working. I modified it to align a short loop (added ALIGN OWORD). The function fails to function correctly (did not cause an error - just bad results which caused other errors - quite difficult to determine just what was happening). The data was 8 DWORDS which were all 0 except most significant which was 40000000h, and I was subtracting a single DWORD with a value of 1. I expected to get 7 DWORDS of FFFFFFFFh and the most significant DWORD with 3FFFFFFFh:

        This is the source code:

        ;*******************************************************************************
        ;
        ; SUB a LONG_NUMB value from another. This is a routine called to subtract a
        ; long number from another after all scaling and validations have been done.
        ;
        ; esi has the source OFFSET of the data.
        ; edi has the already scaled OFFSET of the destination data.
        ; ecx has the DWORD count to subtract.
        ;
        ; Returns nothing with edi pointing to the last word modified. Note that this
        ; may not be the highest DWORD in the destination if a short source is
        ; subtracted from a long destination and there are no borrows past the last
        ; DWORD in the destination or source. This is primarily a check for a size
        ; extension due to a final borrow.
        ;
        ;*******************************************************************************

        ALIGN OWORD

        SUBIt PROC PRIVATE USES eax ebx esi

        pushfd
        clc
        

        ;
        ; subtract all the value DWORDS.
        ;
        ALIGN OWORD

        DoAll:
        mov ebx,[esi]
        lea esi,[esi + (SIZEOF DWORD)]
        sbb [edi],ebx
        lea edi,[edi + (SIZEOF DWORD)]
        dec ecx
        jnz DoAll
        jnc Exit
        mov eax,0
        ;
        ; Propogate the borrow.
        ;
        ALIGN OWORD

        Again:
        sbb [edi],eax
        lea edi,[edi + (SIZEOF DWORD)]
        jc Again
        ;
        ; Exit SUBIt, adjust the destination pointer to the last modified DWORD.
        ;
        Exit:
        sub edi,(SIZEOF DWORD)
        popfd
        test dTestZero,0
        ret
        SUBIt ENDP

        This is the generated code from MASM 9.0 (Visual Studio "Disassembly" tab):

        ALIGN OWORD

        SUBIt PROC PRIVATE USES eax ebx esi
        00404FC0 push eax
        00404FC1 push ebx
        00404FC2 push esi

        pushfd
        

        00404FC3 pushfd
        clc
        00404FC4 clc
        00404FC5 lea esp,[esp]
        00404FCC lea esp,[esp]
        ;
        ; subtract all the value DWORDS.
        ;
        ALIGN OWORD

        DoAl

        L Offline
        L Offline
        Lost User
        wrote on last edited by
        #10

        Yes, add eax, 0 would reset the carry flag. Weird issue, but fortunately the fix is easy: don't align that loop. I know, it's common knowledge that loops should be aligned, and that used to help. But these days we have loop buffers and µop caches so it doesn't matter anymore, unless the loop is huge but then the effect is comparatively small.

        J M 2 Replies Last reply
        0
        • OriginalGriffO OriginalGriff

          Member 4194593 wrote:

          Don't tell me MS

          OK I won't - you clearly know they are the only ones who can do anything about it. Not that they will...

          Member 4194593 wrote:

          don't ask why not

          Can I ask why we shouldn't ask why not? And if not, can I ask why we shouldn't ask why we shouldn't ask why not?

          Bad command or file name. Bad, bad command! Sit! Stay! Staaaay...

          P Offline
          P Offline
          PhilLenoir
          wrote on last edited by
          #11

          Have you been at the Kool Aid again Griff?

          Life is like a s**t sandwich; the more bread you have, the less s**t you eat.

          OriginalGriffO 1 Reply Last reply
          0
          • L Lost User

            Yes, add eax, 0 would reset the carry flag. Weird issue, but fortunately the fix is easy: don't align that loop. I know, it's common knowledge that loops should be aligned, and that used to help. But these days we have loop buffers and µop caches so it doesn't matter anymore, unless the loop is huge but then the effect is comparatively small.

            J Offline
            J Offline
            Jeremy Falcon
            wrote on last edited by
            #12

            Well I'm glad you know what hell he's talking about because none of the rest of us do. I'm just gonna go crawl back into my JavaScript and PHP corner now...

            Jeremy Falcon

            1 Reply Last reply
            0
            • Z ZurdoDev

              What is MASM? You sure it isn't SPASM?

              There are only 10 types of people in the world, those who understand binary and those who don't.

              N Offline
              N Offline
              Nelek
              wrote on last edited by
              #13

              As far as it is not SPAM...

              M.D.V. ;) If something has a solution... Why do we have to worry about?. If it has no solution... For what reason do we have to worry about? Help me to understand what I'm saying, and I'll explain it better to you Rating helpful answers is nice, but saying thanks can be even nicer.

              1 Reply Last reply
              0
              • M Member 4194593

                Where should I post a MASM 9.0 bug. It caused a very hard to find error (did not crash). Don't tell me MS, and don't ask why not. Dave.

                S Offline
                S Offline
                S Douglas
                wrote on last edited by
                #14

                Isn't the best place to post a Microsoft bug to, Microsoft? You know it will never be fixed thus never closed. Remain there forever :)


                Common sense is admitting there is cause and effect and that you can exert some control over what you understand.

                D 1 Reply Last reply
                0
                • P PhilLenoir

                  Have you been at the Kool Aid again Griff?

                  Life is like a s**t sandwich; the more bread you have, the less s**t you eat.

                  OriginalGriffO Offline
                  OriginalGriffO Offline
                  OriginalGriff
                  wrote on last edited by
                  #15

                  It's a PITA to chop and line up for snorting... :laugh:

                  Bad command or file name. Bad, bad command! Sit! Stay! Staaaay...

                  "I have no idea what I did, but I'm taking full credit for it." - ThisOldTony
                  "Common sense is so rare these days, it should be classified as a super power" - Random T-shirt

                  1 Reply Last reply
                  0
                  • S S Douglas

                    Isn't the best place to post a Microsoft bug to, Microsoft? You know it will never be fixed thus never closed. Remain there forever :)


                    Common sense is admitting there is cause and effect and that you can exert some control over what you understand.

                    D Offline
                    D Offline
                    DJ van Wyk
                    wrote on last edited by
                    #16

                    Just think of the warm-and-fuzzy feeling you will get for submitting it to Microsoft. I know they are still going to look at all those reports I submitted with Windows 95 and thank me later.

                    My plan is to live forever ... so far so good

                    M 1 Reply Last reply
                    0
                    • D DJ van Wyk

                      Just think of the warm-and-fuzzy feeling you will get for submitting it to Microsoft. I know they are still going to look at all those reports I submitted with Windows 95 and thank me later.

                      My plan is to live forever ... so far so good

                      M Offline
                      M Offline
                      Member 4194593
                      wrote on last edited by
                      #17

                      My feeling exactly, I was an ORCAS beta tester. Dave.

                      1 Reply Last reply
                      0
                      • L Lost User

                        Yes, add eax, 0 would reset the carry flag. Weird issue, but fortunately the fix is easy: don't align that loop. I know, it's common knowledge that loops should be aligned, and that used to help. But these days we have loop buffers and µop caches so it doesn't matter anymore, unless the loop is huge but then the effect is comparatively small.

                        M Offline
                        M Offline
                        Member 4194593
                        wrote on last edited by
                        #18

                        Harold,

                        harold aptroot wrote:

                        so it doesn't matter anymore

                        I made some timing tests and it took 22% longer to execute an unaligned loop vs an aligned loop, and that was for a short count of loops needed to complete the job, and I had even longer counts I wanted to run. It does make a difference, align your loops! Dave.

                        L 1 Reply Last reply
                        0
                        • M Member 4194593

                          Harold,

                          harold aptroot wrote:

                          so it doesn't matter anymore

                          I made some timing tests and it took 22% longer to execute an unaligned loop vs an aligned loop, and that was for a short count of loops needed to complete the job, and I had even longer counts I wanted to run. It does make a difference, align your loops! Dave.

                          L Offline
                          L Offline
                          Lost User
                          wrote on last edited by
                          #19

                          Ok, show your tests then, and I'll show you mine, which showed no difference. I've tested all 16 different alignments using these

                          global cmp0
                          proc_frame cmp0
                          [endprolog]
                          mov ecx, -10000000
                          xor eax, eax
                          jmp _cmp0_loop
                          align 16

                          _cmp0_loop:
                          add ecx, 1
                          jnz _cmp0_loop
                          ret
                          endproc_frame

                          ; etc

                          global cmp15
                          proc_frame cmp15
                          [endprolog]
                          mov ecx, -10000000
                          xor eax, eax
                          jmp _cmp15_loop
                          align 16
                          nop
                          nop
                          nop
                          nop
                          nop
                          nop
                          nop
                          nop
                          nop
                          nop
                          nop
                          nop
                          nop
                          nop
                          nop
                          _cmp15_loop:
                          add ecx, 1
                          jnz _cmp15_loop
                          ret
                          endproc_frame

                          Measured over 100 runs, taking the minimum, they all take 10000000 cyles + change (function call overhead and so on). No consistent or significant differences.

                          M 1 Reply Last reply
                          0
                          • L Lost User

                            Ok, show your tests then, and I'll show you mine, which showed no difference. I've tested all 16 different alignments using these

                            global cmp0
                            proc_frame cmp0
                            [endprolog]
                            mov ecx, -10000000
                            xor eax, eax
                            jmp _cmp0_loop
                            align 16

                            _cmp0_loop:
                            add ecx, 1
                            jnz _cmp0_loop
                            ret
                            endproc_frame

                            ; etc

                            global cmp15
                            proc_frame cmp15
                            [endprolog]
                            mov ecx, -10000000
                            xor eax, eax
                            jmp _cmp15_loop
                            align 16
                            nop
                            nop
                            nop
                            nop
                            nop
                            nop
                            nop
                            nop
                            nop
                            nop
                            nop
                            nop
                            nop
                            nop
                            nop
                            _cmp15_loop:
                            add ecx, 1
                            jnz _cmp15_loop
                            ret
                            endproc_frame

                            Measured over 100 runs, taking the minimum, they all take 10000000 cyles + change (function call overhead and so on). No consistent or significant differences.

                            M Offline
                            M Offline
                            Member 4194593
                            wrote on last edited by
                            #20

                            Harold, Your code is not doing the same thing, it is decrementing a register. My code was propagateing a borrow through multiple DWORDS. My timing indicated that the aligned code took 3.57 seconds to loop 100,000,000 (one hundred million) times, and 4.36 seconds to loop the same number of times for the unaligned test. That indicates that it took 1.22% longer. In addition, The loop had to re-initialize all of the DWORDS with the correct data for each loop. The loop included this initialization for both the base time and also for the test time, thus the time increment was due to only the time increment for the unaligned loop, meaning that the % difference is really in the neighborhood of 44%. I will give you the timing differences and the code changes between the two tests. If you want to see the the execution time differences (visual studio "disassembly" tab, then just ask and I will include that also.

                            RSA-Test TimeSbbLoop - 100M*TEST32: 3.57

                            ;
                            ; Code from Masm 9.0 .lst file for aligned test.
                            ;
                            C ;*******************************************************************************
                            C ;
                            C ; Timing test for alignment.
                            C ;
                            C ; esi has the source OFFSET of the data.
                            C ; edi has the already scaled OFFSET of the destination data.
                            C ; ecx has the DWORD count to subtract.
                            C ;
                            C ;*******************************************************************************
                            C ALIGN OWORD
                            00001AEE C .data
                            C
                            C ALIGN OWORD
                            00001AF0 00000020 [ C TestData DWORD 32 DUP (0) ; 128 BYTES, 8 xmm regs
                            00000000
                            ]
                            C
                            00001B70 00000001 C TestDWORD DWORD 1
                            C
                            C ALIGN WORD
                            00001B74 0023 C WORD (LENGTHOF szTestCase - 1)
                            00001B76 52 53 41 2D 54 C szTestCase BYTE "RSA-Test TimeSbbLoop - 100M*TEST32:",0
                            65 73 74 20 54
                            69 6D 65 53 62
                            62 4C 6F 6F 70
                            20 2D 20 31 30
                            30 4D 2A 54 45
                            53 54 33 32 3A
                            00
                            C
                            000006A0 C .code
                            C
                            C ;
                            C ; Get start time.
                            C ;
                            000006A0 C Start:
                            000006A0 E8 000054BB C CALL GetStartTime
                            C ; jnz Exit
                            C ;
                            C ; Clear the clear regs.
                            C ;
                            000006A5 66| 0F EF C0 C pxor xmm0,xmm0
                            000006A9 66| 0F 6F C8 C movdqa xmm1,xmm0
                            000006AD 66| 0F 6F D0 C

                            L 1 Reply Last reply
                            0
                            • M Member 4194593

                              Harold, Your code is not doing the same thing, it is decrementing a register. My code was propagateing a borrow through multiple DWORDS. My timing indicated that the aligned code took 3.57 seconds to loop 100,000,000 (one hundred million) times, and 4.36 seconds to loop the same number of times for the unaligned test. That indicates that it took 1.22% longer. In addition, The loop had to re-initialize all of the DWORDS with the correct data for each loop. The loop included this initialization for both the base time and also for the test time, thus the time increment was due to only the time increment for the unaligned loop, meaning that the % difference is really in the neighborhood of 44%. I will give you the timing differences and the code changes between the two tests. If you want to see the the execution time differences (visual studio "disassembly" tab, then just ask and I will include that also.

                              RSA-Test TimeSbbLoop - 100M*TEST32: 3.57

                              ;
                              ; Code from Masm 9.0 .lst file for aligned test.
                              ;
                              C ;*******************************************************************************
                              C ;
                              C ; Timing test for alignment.
                              C ;
                              C ; esi has the source OFFSET of the data.
                              C ; edi has the already scaled OFFSET of the destination data.
                              C ; ecx has the DWORD count to subtract.
                              C ;
                              C ;*******************************************************************************
                              C ALIGN OWORD
                              00001AEE C .data
                              C
                              C ALIGN OWORD
                              00001AF0 00000020 [ C TestData DWORD 32 DUP (0) ; 128 BYTES, 8 xmm regs
                              00000000
                              ]
                              C
                              00001B70 00000001 C TestDWORD DWORD 1
                              C
                              C ALIGN WORD
                              00001B74 0023 C WORD (LENGTHOF szTestCase - 1)
                              00001B76 52 53 41 2D 54 C szTestCase BYTE "RSA-Test TimeSbbLoop - 100M*TEST32:",0
                              65 73 74 20 54
                              69 6D 65 53 62
                              62 4C 6F 6F 70
                              20 2D 20 31 30
                              30 4D 2A 54 45
                              53 54 33 32 3A
                              00
                              C
                              000006A0 C .code
                              C
                              C ;
                              C ; Get start time.
                              C ;
                              000006A0 C Start:
                              000006A0 E8 000054BB C CALL GetStartTime
                              C ; jnz Exit
                              C ;
                              C ; Clear the clear regs.
                              C ;
                              000006A5 66| 0F EF C0 C pxor xmm0,xmm0
                              000006A9 66| 0F 6F C8 C movdqa xmm1,xmm0
                              000006AD 66| 0F 6F D0 C

                              L Offline
                              L Offline
                              Lost User
                              wrote on last edited by
                              #21

                              Of course it's not doing the same thing, it's just testing the effect of alignment by itself. There is way too much shit here to be sure of anything.

                              M 1 Reply Last reply
                              0
                              • L Lost User

                                Of course it's not doing the same thing, it's just testing the effect of alignment by itself. There is way too much shit here to be sure of anything.

                                M Offline
                                M Offline
                                Member 4194593
                                wrote on last edited by
                                #22

                                Harold, My feelings exactly. I will use what I have and beware of any ALIGN in the code section except for the entry at PROC. I will insure that there is no entry to an aligned Label except by a jump or conditional jump. Dave.

                                1 Reply Last reply
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