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1 1 Regression Verification for Multi-Threaded Programs Sagar Chaki, SEI-Pittsburgh Arie Gurfinkel, SEI-Pittsburgh Ofer Strichman, Technion-Haifa Originally presented at VMCAI 2012 VSSE 2012

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2 2 Regression Verification Formal equivalence checking of two similar programs Three selling points: Specification: – not needed Complexity: – dominated by the change, not by the size of the programs. Invariants: – Easy to derive automatically high-quality loop/recursion invariants Tool support: RVT (Technion) SymDiff (MSR)

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3 3 Possible applications of regression verification Understand how changes propagate through the API Validate refactoring Semantic Impact Analysis Generic Translation Validation Proving Determinism P is equivalent to itself, P is deterministic many more…

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4 4 Definition: Partial Equivalence Let f 1 and f 2 be functions with equivalent I/O signature f 1 and f 2 are partially equivalent, executions of f 1 and f 2 on equal inputs …which terminate, result in equal outputs. Undecidable

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5 5 Multi-Threaded Programs (MTPs) Finite set of root functions : P = f 1 || … || f n Each executed in a separate thread Communicate via shared variables is P partially equivalent to itself? …no! int base = 1; void f 1 (int n, int *r) { if (n < 1) *r = base; else { int t; f 1 (n-1, &t); *r = n * t; } void f 2 () { base = 2; } int base = 1; void f 1 (int n, int *r) { if (n < 1) *r = base; else { int t; f 1 (n-1, &t); *r = n * t; } void f 2 () { base = 2; } P P shared variable Computes n! or 2*n! thread

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6 6 Towards partial equivalence of MTPs P is a multi-threaded program (P): the set of terminating computations of P R (P) = { ( in, out ) | ∃ ∈ (P ). begins in in and ends in out } Example: R (P) = { ( n, n !), ( n, 2n !) | n ∈ Z } int base = 1; void f 1 (int n, int *r) { if (n < 1) *r = base; else { int t; f 1 (n-1, &t); *r = n * t; } void f 2 () { base = 2; } int base = 1; void f 1 (int n, int *r) { if (n < 1) *r = base; else { int t; f 1 (n-1, &t); *r = n * t; } void f 2 () { base = 2; } P P

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7 7 Partial equivalence of MTPs MTPs P 1, P 2 are partially equivalent if R (P 1 ) = R (P 2 ). Denoted p.e.n. (P 1, P 2 ) Partial Equivalence of Nondeterministic programs Claim: 8 P. ² p.e.n.(P,P)

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8 8 Towards a sound proof rule for partial equivalence of MTPs - The sequential case - Challenges in extending it to p.e.n. - A proof rule

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9 9 First: The sequential case [GS’08] Loops/recursion are not a big problem in practice: Recursive calls are replaced with calls to the same uninterpreted function: Abstracts the callees By construction assumes the callees are partially equivalent Sound by induction f1f1 f2f2 = f1f1 f2f2 UFUF UFUF = isolation

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10 First: The sequential case [GS’08] Now we have two isolated functions f1, f2 Checking partial equivalence is decidable: in1 = in2 = *; o1 = f1(in); o2 = f2(in); assert (o1==o2) The algorithm traverses up the call graphs while abstracting equal callees. How can this method be extended to proving p.e.n(f1,f2) ?

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11 f2(int in) { x1 = in; x2 = in; } What affects partial equivalence of MTPs? Before: in = 1 o1 = 1, o2 = 0; (1, h 1, 0 i ) 2 R(p) f1() { o1 = x1; o2 = x2; } || x1 = x2 = 0 After: in = 1 o1 = 1 ) o2 = 1 (1, h 1, 0 i ) R(p) Swap write order x2 = in; x1 = in;

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12 f1(int in1) { x1 = in1; t1 = x2; o1 = t1; } f2(int in2) { t2 = x1; x2 = in2; o2 = t2; } After: in1 = 1, in2 = 2 o2 = 1 ) x1 = 1 < t2 = x1 ) t1 = 0 ) o1 = 0 ( h 1,2 i, h 2,1 i ) R(p) || x1 = x2 = 0 Before: in1 = 1, in2 = 2 x1 = 1; t2 = x1 = 1; x2 = in2 = 2; t1 = x2 = 2; o1 = t1 = 2; o2 = t2 = 1; ( h 1,2 i, h 2,1 i ) 2 R(p) Swap R/W order t1 = x2; x1 = in1; What affects partial equivalence of MTPs?

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13 f2(int in) { x1 = in; x2 = in; } Before: in = 1 o1 = 0, o2 = 1; (1, h 0, 1 i ) 2 R(p) f1() { o1 = x1; o2 = x2; } || x1 = x2 = 0 After: in = 1 o1 = 0 ) o2 = 0 (1, h 0, 1 i ) R(p) Swap read order o2 = x2; o1 = x1; What affects partial equivalence of MTPs?

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14 Preprocessing 1.Loops ) recursive functions 2.Mutual recursion ) simple recursion 3.Non-recursive functions ) inlined 4.Function’s return value ) parameter 5.Each shared variable x only appears in t = x or x = exp (add auxiliaries) 6.Each auxiliary variable is read once int base = 1; void f 1 (int n, int *r) { if (n < 1) *r = base; else { int t; f 1 (n-1, &t); *r = n * t; } void f 2 () { base = 2; } int base = 1; void f 1 (int n, int *r) { if (n < 1) *r = base; else { int t; f 1 (n-1, &t); *r = n * t; } void f 2 () { base = 2; } P P

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15 Mapping Assume each function is used in a single thread. Otherwise, duplicate it Find a mapping between the non-basic types Find a bijective map between: threads shared variables functions (same proto-type), in mapped functions: read globals, written-to globals Without such a mapping: goto end-of-talk.

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16 The Observable Stream Consider a function f and input in The observable stream of f ( in )’s run is its sequence of function calls read/write of shared variables Example: let x be a shared variable: x x x x

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17 Observable Equivalence of Functions Denote by observe-equiv( f, f ’) Assume: outputs are defined via shared variables Hence: observable equivalence ) partial equivalence f, f ’ are observably equivalent, 8 in. f ( in ), f ’( in ) have equal sets of finite observable streams

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18 We prove observe-equiv( f, f ’) by isolating f, f ’. proving observ. equivalence of all (isolated) pairs Given f, we build an isolated version [ f ], in which: Function calls are replaced with calls to uninterpreted functions Shared variables are read from an “input” stream The observable stream is recorded Observable Equivalence of Functions

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19 Reading from an input stream… For each shared variable x: R(x) Ã R(*) // read R(x) W(x) g(…) W(x) R(x) R(*)

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20 R(x) local W(x) g(…) W(x) R(x) W(x) local g(…) R(x) local R(x) R(* 1 ) R(* 2 ) R(* 3 ) Enforce: equal locations ) same ‘*’ value … if their streams up to that point were equal. Reading from an input stream…

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21 Summary: from f to [ f ] Transform function f and f ’ to [ f ] and [ f ’] by: Function calls are replaced with calls to UFs – t = g() ) t = UF g () – For (g,g’) 2 map, UF g = UF g’ Shared variables are read from an “input” stream: – t = x ) t = UF x ( loc ) // loc is the location in the stream – For (x,x’) 2 map, Uf x = Uf x’ The Observable Stream is recorded.

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22 Example: from f to [ f ] int base = 1; void f (int n, int *r) { int t; if (n < 1) { t = base; *r = t;} else { f(n-1,&t); *r = n * t; } int base = 1; void f (int n, int *r) { int t; if (n < 1) { t = base; *r = t;} else { f(n-1,&t); *r = n * t; } read write function call list out; void [f] (int n, int *r) { int t, loc = 0; if (n < 1) { t = UF base (loc); out += (R,”base”); loc++; out += (W,”r”,t); loc++;} else { t = UF f,t (n-1); out += (C,f,n-1); out += (W,”r”, n * t); loc++; } list out; void [f] (int n, int *r) { int t, loc = 0; if (n < 1) { t = UF base (loc); out += (R,”base”); loc++; out += (W,”r”,t); loc++;} else { t = UF f,t (n-1); out += (C,f,n-1); out += (W,”r”, n * t); loc++; } Output treated as shared Observable stream f f [f][f] [f][f]

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23 Checking observable equivalence of [ f ], [ f ’] Generate sequential program S: in1 = in2 = *; [f(in1)];[f’(in2)]; rename ([f’].out); // according to map assert([f].out == [f’].out); Validity of S is decidable f f’f’ [f][f] [ f ’] SCBMC

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24 A Proof Rule Compositional: check one function pair at a time Sequential: no thread composition General: supports loops/recursion + arbitrary # of threads 8 f, f ’ 2 map. observe-equiv([ f ], [ f ’]) p.e.n. (P 1, P 2 ) 8 f, f ’ 2 map. observe-equiv([ f ], [ f ’]) p.e.n. (P 1, P 2 )

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25 Summary and Future Thoughts We suggested foundations of regression verification for MTPs Notion of partial equivalence of multi-threaded programs A proof rule Challenges ahead: Synchronization primitives: Locks, semaphores, atomic blocks Dynamic thread creation Make more complete

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26 Proof Rule 2 Premises are weaker than Rule 1 but harder to discharge More “complete” than Rule 1 Details in the paper 8 f,f’ 2 map. ¢ (f,f’) p.e. (P 1, P 2 ) 8 f,f’ 2 map. ¢ (f,f’) p.e. (P 1, P 2 ) Observable equivalence of f and f’ under environments compatible with the other threads in the program Key insights: No loops ) compute k x : max #times f reads x Let be the threads writing to x. Compute the k x -recursion-bounded abstraction Assume Uf x returns values produced by

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