The Intel instruction "v1 = ushr v2, v2" will implicitly fix the output register for v2 to %rcx because the output is tied to the first input operand and the second input operand is fixed to %rcx. Make sure we handle this transitive constraint when checking for interference with the globally live registers. Fixes #218
53 lines
1.6 KiB
Plaintext
53 lines
1.6 KiB
Plaintext
test regalloc
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set is_64bit
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isa intel haswell
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; Test combinations of constraints.
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;
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; The Intel ushr instruction requires its second operand to be passed in %rcx and its output is
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; tied to the first input operand.
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;
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; If we pass the same value to both operands, both constraints must be satisfied.
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; Found by the Binaryen fuzzer in PR221.
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;
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; Conditions triggering the problem:
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;
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; - The same value used for a tied operand and a fixed operand.
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; - The common value is already in %rcx.
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; - The tied output value is live outside the EBB.
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;
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; Under these conditions, Solver::add_tied_input() would create a variable for the tied input
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; without considering the fixed constraint.
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function %pr221(i64 [%rdi], i64 [%rsi], i64 [%rdx], i64 [%rcx]) -> i64 [%rax] {
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ebb0(v0: i64, v1: i64, v2: i64, v3: i64):
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v4 = ushr v3, v3
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jump ebb1
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ebb1:
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return v4
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}
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; Found by the Binaryen fuzzer in PR218.
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;
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; This is a similar situation involving combined constraints on the ushr instruction:
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;
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; - The %rcx register is already in use by a globally live value.
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; - The ushr x, x result is also a globally live value.
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;
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; Since the ushr x, x result is forced to be placed in %rcx, we must set the replace_global_defines
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; flag so it can be reassigned to a different global register.
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function %pr218(i64 [%rdi], i64 [%rsi], i64 [%rdx], i64 [%rcx]) -> i64 [%rax] {
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ebb0(v0: i64, v1: i64, v2: i64, v3: i64):
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; check: regmove $v3, %rcx ->
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v4 = ushr v0, v0
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; check: $v4 = copy
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jump ebb1
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ebb1:
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; v3 is globally live in %rcx.
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; v4 is also globally live. Needs to be assigned something else for the trip across the CFG edge.
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v5 = iadd v3, v4
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return v5
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}
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