Question

根据NEG和NEGU的定义，

NEG $X,Y,$Z (negate signed): s($X) := Y - s($Z).
NEGU $X,Y,$Z (negate unsigned): u($X) := (Y - u($Z)) mod 2^64.

我们假设$Z = s(-1)或u(2^64 - 1)。然后第一个操作码会在$X时将值1放入Y = 0寄存器，而最近的操作码会因u(-(2^64 - 1)) mod 2^64 = 1而得到相同的结果。我对么？ NEG时，$Z = -2^63指令是否会引发溢出异常？

Answer 1

简短回答“当$ Z = -2 ^ 63时，NEG指令是否应该引发溢出异常？”

yes, but you probably already suspected that.

逻辑上，NEG $ X，0，-2 ^ 63应该给出2 ^ 63，这超出了有符号正整数的范围，因而溢出。但是，如果你像我一样，你需要证明实际上发生了整数溢出。这是：

t       IS      $255

    LOC     #20          //handle the integer overflow event
    PUSHJ   255,Err
    PUT     rJ,$255
    GET     $255,rB
    RESUME

        LOC     #100
Main    SET     t,#4000
        PUT     rA,t        //set the integer overflow event bit
        SETH    $0,#8000
        NEG     $1,0,$0
        GETA    t,End
        TRAP    0,Fputs,StdOut
        TRAP    0,Halt,0
End     BYTE    "End of program",#a,0

Err     SET     $0,$255           //overflow subroutine, prints out message
        GETA    t,Emes
        TRAP    0,Fputs,StdOut
        GET     t,rW
        INCL    t,4
        PUT     rW,t
        SET     $255,$0
        POP     0,0
Emes    BYTE    "Error: integer overflow",#a,0

Mmix NEG和NEGU操作码

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