The Alpha AXP, part 16: What are the dire consequences of having 32-bit values in non-canonical form?

Raymond Chen

On the Alpha AXP, 32-bit values are typically represented in so-called canonical form. But what happens if you use a non-canonical representation?

Well, it depends on what instruction consumes the non-canonical representation.

If the consuming instruction is an explicit 32-bit instruction, such as ADDL or STL, then the upper 32 bits are ignored, and the operation proceeds with the lower 32 bits. In that case, the non-canonical representation causes no harm. For example, consider this calculation:

    ; Calculate Rc = Ra + Rb + 0x1234 (32-bit result)
    LDA     Rc, 0x1234(zero)    ; Rc = 0x00000000`00001234
    ADDL    Rc, Rb, Rc          ; Rc = Rb + 0x1234
    ADDL    Rc, Ra, Rc          ; Rc = Ra + Rb + 0x1234

If we are willing to use a non-canonical form temporarily, we could simplify this to

    ; Calculate Rc = Ra + Rb + 0x1234 (32-bit result)
    LDA     Rc, 0x1234(Rb)      ; Rc = Rb + 0x1234 (64-bit intermediate)
    ADDL    Rc, Ra, Rc          ; Rc = Ra + Rb + 0x1234 (32-bit result)

The LDA will put Rc into non-canonical 32-bit form if Rb is in the range 0x7FFFEDCC to 0x7FFFFFFF because the LDA instruction is 64-bit only, and the result would be in the range 0x00000000`80000000 through 0x00000000`80001233, which are non-canonical. But all is forgiven at the ADDL instruction, since it considers only the 32-bit portion of the addends (ignoring the non-canonical part) and generates a 32-bit result in canonical form.

On the other hand, if the instruction that consumes the non-canonical 32-bit value is a 64-bit instruction, then the non-canonical value will cause trouble.

Consider this simple function:

void f(int x)
    if (x == 0) DoSomething();

The Windows ABI for Alpha AXP requires that all 32-bit values be passed and returned in canonical form. You are welcome to use non-canonical values inside your function, but all communication with the outside world must use canonical form for 32-bit values.

This function might assemble to something like this:

    BEQ     a0, DoSomething ; tail call optimization
    RET     zero, (ra), 1   ; return without doing anything

The first instruction checks whether x is zero. If so, then it jumps directly to the Do­Something function, leaving the return address unchanged, so that when Do­Something returns, it returns to the caller of f. (This is a tail call optimization.)

If the value is not zero, then it returns to the caller.

There is no 32-bit version of the BEQ instruction; it always tests the full 64 bits.

If the value of x were not canonical, then the branch instruction could suffer false negatives: Even though the lower 32 bits are zero, there may be nonzero bits set in the upper half. That cause the BEQ to report “sorry, not zero” even though the 32-bit part of a0 was zero.

There are a number of instructions which do not have a 32-bit version and which always operate on the full 64-bit register value. Another example:

void f(int x, int y)
    if (x < y) DoSomething();

This function might assemble to something like this:

    CMPLT   a0, a1, t0      ; t0 = 1 if a0 < a1
    BNE     t0, DoSomething ; tail call optimization
    RET     zero, (ra), 1   ; return without doing anything

In this version, the compiler performs a signed less-than operation and branches based on the result. The CMPLT instruction always operates on the full 64-bit register value; there is no 32-bit version. Consequently, passing a non-canonical value can result in the debugger reporting strange things like “Well, even though you passed x = 1 and y = 2, the less-than comparison returned false because x was passed in the non-canonical form of 0xFFFFFFFF`00000001.

Using sign-extended values for canonical form for 32-bit values has the nice property that signed and unsigned comparisons of 32-bit values have the same results as signed and unsigned comparisons of their corresponding canonical forms.

If zero-extension had been used for canonical form, then unsigned comparisons would be preserved, but signed comparisons would not agree: The 32-bit signed comparison of 0x00000000 with 0xFFFFFFFF would report that the first value is larger (0 > −1) but the 64-bit signed comparison 0x00000000`00000000 with 0x00000000`FFFFFFFF of the corresponding zero-extended values would report that the second value is larger (0 < 4,294,967,295).

I’m pretty sure this was not a coincidence.

Bonus chatter: Non-canonical values introduce another case where uninitialized variables can result in strange behavior. Consider:

int f()
    int v;
    ... a bunch of code that somehow forgot to set v ...
    ... but in a complicated way that eluded code flow analysis ...
    return (v < 0) ? -1 : 0;

This might get compiled to the following:

    ; compiler chooses t0 to represent v
    SRA     t0, #32, v0     ; v0 = 0xFFFFFFFF`FFFFFFFF if t0 was negative
                            ; v0 = 0x00000000`00000000 if t0 was nonnegative
    RET     zero, (ra), 1   ; return the result

If the code forgets to assign a value to v, then it will have the value left over from whatever code ran earlier. Suppose that leftover value happened to be the non-canonical value 0x12345678`12345678. In that case, the result of the SRA would be 0x00000000`12345678, and the function f ends up returning some value that seems to be impossible from reading the code: According to the code, the function always returns either -1 or 0, yet sometimes we crash because it returned the crazy value 0x12345678!


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