Data Flow Anomaly Analysis

Transcription

1 Pof. D. Liggesmeye, 1 Contents Data flows an ata flow anomalies State machine fo ata flow anomaly analysis Example withot loops Example with loops Data Flow Anomaly Analysis Softwae Qality Assance Softwae Qality Assance Pof. D. Liggesmeye, 2 Data Flows an Data Flow Anomalies Data Flows an Data Flow Anomalies The ata flow anomaly analysis gaantees the ientification of cetain falts (so-calle ata flow anomalies) The ata flow w..t. to a cetain vaiable on a paticla exection path can be escibe by its seqence of efinitions, efeences (p-ses an c-ses) an nefinitions (see ata flow testing) Rles fo ata flows A vale mst not be assigne twice to a vaiable (-anomaly) An nefine vaiable mst not be efeence (-anomaly) The vale of a vaiable mst not be elete iectly afte the vale has been assigne (-anomaly) These ata flow anomalies can be etecte by static analysis x is efine: (efine) The vaiable x is assigne a vale (e.g. x = 5;) x is efeence: (efeence) The vale of the vaiable x is ea in a comptation o in a ecision, i.e., the vale of x oes not change (e.g. y = x + 1; o if (x > 0)...) x is nefine: (nefine) The vale of the vaiable x is elete (e.g., eletion of local vaiables within a fnction o pocee at its temination). At pogam stat all vaiables ae nefine x is not se: e (empty) The instction of the noe ne consieation oes not inflence the vaiable x. x is not efine, efeence o nefine Softwae Qality Assance Pof. D. Liggesmeye, 3 Softwae Qality Assance Pof. D. Liggesmeye, 4

2 Pof. D. Liggesmeye, 5 Data Flows an Data Flow Anomalies State Machine fo Data Flow Anomaly Analysis Let s consie the two following ata flows w..t. a vaiable (: nefinition, : efinition, : efeence) 1: 2: Seqence 1 begins with the patten. The vaiable has a anom vale at the time of the efeence, as it was not efine befoe. Thee is a ata flow anomaly of the type ; the efeence of a vaiable with nefine, anom vale Seqence 2 contains two sccessive vaiable efinitions. The fist efinition has no effect, as the vale is always ovewitten by the secon efinition. The ata flow anomaly is of the type Seqence 2 ens with a efinition followe by an nefinition. The vale assigne by the efinition is not se, as it is immeiately elete aftewas. This ata flow anomaly is of the type ata flow anomaly stat (S) (-anomaly) (-anomaly) efine (-anomaly) (D) nefine (U) If the state ata flow anomaly is eache o at the en of a ata flow anomaly analysis the state nefine is not eache, a ata flow anomaly is etecte. The state machine efines a egla gamma. Sch gammas ae a stana case fo compile constction. In compiles they seve as a basis fo the lexical analysis. Ths, ata flow analysis can be integate into compiles (which is the case in some compiles yo shol check whethe yo compile can o it) efeence (R) Softwae Qality Assance Softwae Qality Assance Pof. D. Liggesmeye, 6 Example withot Loops Example withot Loops The opeation MinMax gets two nmbes via ateface which ae to be etne oee accoing to size voi MinMax (int& Min, int& Max) int Help; Max = Min; Help = Min; en MinMax; Assignment of the ata flow attibtes w..t. the vaiables to the noes of the contol flow gaph eletion of a vale (nefine) vale assignment (efine) eaing a vale (efeence) Analysis of the ata flows fo the vaiables on the paths of the contol flow gaph Softwae Qality Assance Pof. D. Liggesmeye, 7 Softwae Qality Assance Pof. D. Liggesmeye, 8

3 Pof. D. Liggesmeye, 9 Example withot Loops Example withot Loops Contol flow gaph of MinMax n stat (Min), (Max), (Help) Data flows of MinMax Impot of Min an Max (Min), (Max) Path (Min), (Max) Vaiable n stat n ot n stat n ot (Help), (Max) Min Max = Min; (Min), (Max) Max Help= Min; (Min), (Help) Help Expot of Min an Max n ot (Min), (Max) (Min), (Max), (Help) nefine efine - efeence Softwae Qality Assance Softwae Qality Assance Pof. D. Liggesmeye, 10 Example withot Loops Example withot Loops The coecte vesion of the opeation eas as follows voi MinMax (int& Min, int& Max) int Help; Help = Min; Min = Max; END MinMax; Impot of Min an Max Help = Min; Min = Max; Expot of Min an Max n stat n ot (Min), (Max), (Help) (Min), (Max) (Min), (Max) (Min), (Help) (Max), (Min) (Help), (Max) (Min), (Max) (Min), (Max), (Help) Softwae Qality Assance Pof. D. Liggesmeye, 11 Softwae Qality Assance Pof. D. Liggesmeye, 12

4 Pof. D. Liggesmeye, 13 Example withot Loops Path n stat Vaiable Min Max Help n ot n stat n ot Assmption: Data flow anomaly analysis mst be one fo all paths if the nmbe of paths is to lage, ata flow anomaly analysis may not be feasible (eason: loops, see path testing) Fotnately this assmption is not coect Concening the ata flow anomaly analysis it is sfficient to analyze the paths p to the fist iteation of loops (the secon exection of the loop boy) If no ata flow anomalies occe ntil then, it is ense that also on the paths with a highe nmbe of loop iteations no anomalies will occ nefine efine - efeence Softwae Qality Assance Softwae Qality Assance Pof. D. Liggesmeye, 14 An opeation ses Newtonian iteation as an appoximation pocee in oe to etemine the sqae oot The opeation shol etemine the sqae oot fo the non-negative inpts Fo negative inpts the vale 0.0 is to be etne De to the appoximation pocee it is ifficlt to give an estimation fo the maximm nmbe of iteations. This may case afinite nmbe of paths (emak: If the loop is well-esigne it shol teminate fo evey inpt, so the nmbe of iteation will be finite; bt in this special case, this is ha to pove.) oble Sqt(oble X) oble etnvale; if (X > 0.0) oble W; while (ABS(W*W-X) > 0.01) W = W - ((W*W-X) / (2.0 * W)); etnvale = W; else etnvale = 0.0; etn (etnvale); Softwae Qality Assance Pof. D. Liggesmeye, 15 Softwae Qality Assance Pof. D. Liggesmeye, 16

5 Pof. D. Liggesmeye, 17 n stat if (X > 0.0)... while (ABS (W * W) X) > W = W ((W * W X) / (2.0 * W)) etnvale = W else etnvale = 0.0 etn (etnvale) n ot (X), (W), (etnvale) (X) (X) (X), (W) (X), (W), (W) (W), (etnvale) (etnvale) (etnvale) The analysis of the path which is passe thogh fo non-positive inpt vales is elatively simple X: W: etnvale: None of these ata flows contains anomalies Fo positive inpts the loop is execte. The ata flow fo the vaiable X begins with the sb-seqence p to the loop ecision. If the loop is not entee, the sb-seqence follows iectly. If the loop is entee, the sb-seqence eges itself in. This sb-seqence epeats with evey fthe loop exection. Ths, the ata flows on these paths can be give complete fom. The ata flow fo the vaiable X is: () n, with n >= 0. Vale n epesents the nmbe of loop exections. Fo the vaiables W an etnvale also complete expessions fo the ata flows ae eceive similaly X: () n, n>=0 W: () n, n >=0 etnvale: (X), (W), (etnvale) Softwae Qality Assance Softwae Qality Assance Pof. D. Liggesmeye, 18 Qestion: Which vales n have to be consiee w..t. ata flow anomaly analysis Cetainly the case n=0 has to consiee, as a new seqence eslts e to the isappeaance of the backete sb-seqence The case n=1 also has to be consiee, as two new sb-seqence emege at the beginning of the backete expession an at its en Fthemoe the case n=2 is to be consiee. Fo claification the seqence... () n... shol be looke at which fo n=0 an n=1 has no ata flow anomalies, bt fo n=2 (......) shows a -anomaly Fo geate vales n no potential new ata flow anomalies eslt. If no ata flow anomaly on the paths p to the secon loop exection has occe yet, none will occ actally. The infinite nmbe of paths has no inflence on this The opeation sqt fo the vaiable W shows a ata flow anomaly. The ata flow () n begins with a -anomaly. The vale of the vaiable W is not initialise yet at the time of the fist eaing access. Howeve, the opeation woks coectly fo anom positive initial vales of W, so that ynamic testing oes not etect the falt eliably. Fo negative initial vales of W the negative oot is etemine. If W by accient is initially eqal zeo, the pogam cashes, as a ivie by zeo occs. While by ynamic testing this falt can be etecte only neliably, it is ientifiable by ata flow anomaly analysis eliably an at vey low costs Softwae Qality Assance Pof. D. Liggesmeye, 19 Softwae Qality Assance Pof. D. Liggesmeye, 20

6 Pof. D. Liggesmeye, 21 n stat (X), (W), (etnvale) (X) if (X > 0.0)... (X) W = 1.0 (W) while (ABS (W * W) X) > (X), (W) W = W ((W * W X) / (2.0 * W)) (X), (W), (W) etnvale = W (W), (etnvale) else etnvale = 0.0 (etnvale) etn (etnvale) (etnvale) n ot (X), (W), (etnvale) Softwae Qality Assance