模型檢測(cè)例子

1、模型檢測(cè) 例子計(jì)算機(jī)科學(xué)國(guó)家重點(diǎn)實(shí)驗(yàn)室1程序正確的重要性應(yīng)用廣泛航空航天金融設(shè)備的控制日常生活軟件錯(cuò)誤的可能后果火箭 Ariane 5 Explosion (1997)火星氣候軌道器 NASA Mars Climate Orbiter (1999)2程序正確性方法測(cè)試黑盒測(cè)試白盒測(cè)試形式驗(yàn)證推理驗(yàn)證模型檢測(cè)可執(zhí)行代碼程序代碼程序代碼或其抽象模型程序運(yùn)行環(huán)境模型規(guī)則算法3例子: ISR#include /*/int in();int isr(int x,int k);int isk(int n,int k);/*/int main()int n=0;int m=0;int k=1; printf

2、(INFO: system is now activen); while (1) n=in();m=isr(n,k);k=isk(n,k); printf(RESULT: %inn,m);/*/int isr(int x,int k) int y1=0; int y2=0; int y3=0; y1=0; y2=1; y3=1; if (x=2|(x2&k=20) x=x-1; while (y3=x) y1=y1+1; y2=y2+2; y3=y3+y2; return y1;/*/int isk(int n,int k) if (k!=20) if (k!=n) k=21; else if

3、 (k=19) k=0; else k=k+2; else k=21; return k;/*/int in() char c=0; int k=0; while (1) k=0; putc(N,stdout); putc(:,stdout); putc(9,stdout); c=getc(stdin); if (c=n) printf(INFO: the input must be 1 or 2 digitsnn); continue; if (c9) while (1) c=getc(stdin); if (c=n) break; printf(INFO: the input must b

4、e 1 or 2 digitsnn); continue; k=c-0; c=getc(stdin); if (c=n) return k; if (c9) while (1) c=getc(stdin); if (c=n) break; printf(INFO: the input must be 1 or 2 digitsnn); continue; if (k2) k=k*10+(c-0); else if (k=2&c=0) k=20; else while (1) c=getc(stdin); if (c=n) break; printf(INFO: the input number

5、 must be in 0,.,20nn ); continue; c=getc(stdin); if (c!=n) while (1) c=getc(stdin); if (c=n) break; printf(INFO: the input must be 1 or 2 digitsnn); continue; return k; ./isrINFO: system is now activeN: 2RESULT: 1N: 10RESULT: 3N: 20RESULT: 4N: 30INFO: the input number must be in 0,.,20N: 5RESULT: 2N

6、:4例子: ISR#include /*/int in();int isr(int x,int k);int isk(int n,int k);/*/int main()int n=0;int m=0;int k=1; printf(INFO: system is now activen); while (1) n=in();m=isr(n,k);k=isk(n,k); printf(RESULT: %inn,m);/*/int isr(int x,int k) int y1=0; int y2=0; int y3=0; y1=0; y2=1; y3=1; if (x=2|(x2&k=20)

7、x=x-1; while (y3=x) y1=y1+1; y2=y2+2; y3=y3+y2; return y1;/*/int isk(int n,int k) if (k!=20) if (k!=n) k=21; else if (k=19) k=0; else k=k+2; else k=21; return k;/*/int in() char c=0; int k=0; while (1) k=0; putc(N,stdout); putc(:,stdout); putc(9,stdout); c=getc(stdin); if (c=n) printf(INFO: the inpu

8、t must be 1 or 2 digitsnn); continue; if (c9) while (1) c=getc(stdin); if (c=n) break; printf(INFO: the input must be 1 or 2 digitsnn); continue; k=c-0; c=getc(stdin); if (c=n) return k; if (c9) while (1) c=getc(stdin); if (c=n) break; printf(INFO: the input must be 1 or 2 digitsnn); continue; if (k

9、2) k=k*10+(c-0); else if (k=2&c=0) k=20; else while (1) c=getc(stdin); if (c=n) break; printf(INFO: the input number must be in 0,.,20nn ); continue; c=getc(stdin); if (c!=n) while (1) c=getc(stdin); if (c=n) break; printf(INFO: the input must be 1 or 2 digitsnn); continue; return k; 對(duì)ISR的要求:輸入為0和20

10、之間的整數(shù)時(shí)，輸出為其平方根的整數(shù)部分。輸入為0和20之間的整數(shù)時(shí)，輸出為其平方根的整數(shù)部分。5測(cè)試黑盒測(cè)試白盒測(cè)試Program testing can be used to show the presence of bugs, but never to show their absence! - Edsger W. Dijkstra6形式驗(yàn)證推理驗(yàn)證模型檢測(cè)正確性正確性+不正確性7例子: ISR#include /*/int in();int isr(int x,int k);int isk(int n,int k);/*/int main()int n=0; int m=0;int k=

11、1; printf(INFO: system is now activen); while (1) n=in();m=isr(n,k);k=isk(n,k); printf(RESULT: %inn,m);/*/int isr(int x,int k) int y1=0; int y2=0; int y3=0; y1=0; y2=1; y3=1; if (x=2|(x2&k=20) x=x-1; while (y3=x) y1=y1+1; y2=y2+2; y3=y3+y2; return y1;/*/int isk(int n,int k) if (k!=20) if (k!=n) k=21

12、; else if (k=19) k=0; else k=k+2; else k=21; return k;/*/int in() char c=0; int k=0; while (1) k=0; putc(N,stdout); putc(:,stdout); putc(9,stdout); c=getc(stdin); if (c=n) printf(INFO: the input must be 1 or 2 digitsnn); continue; if (c9) while (1) c=getc(stdin); if (c=n) break; printf(INFO: the inp

13、ut must be 1 or 2 digitsnn); continue; k=c-0; c=getc(stdin); if (c=n) return k; if (c9) while (1) c=getc(stdin); if (c=n) break; printf(INFO: the input must be 1 or 2 digitsnn); continue; if (k2) k=k*10+(c-0); else if (k=2&c=0) k=20; else while (1) c=getc(stdin); if (c=n) break; printf(INFO: the inp

14、ut number must be in 0,.,20nn ); continue; c=getc(stdin); if (c!=n) while (1) c=getc(stdin); if (c=n) break; printf(INFO: the input must be 1 or 2 digitsnn); continue; return k; 對(duì)ISR的要求:(at line 18):(m*m)n)8程序的模型檢測(cè)(例子)9模型檢測(cè)./verds c isr.c sp isr.sp驗(yàn)證結(jié)果反例10驗(yàn)證過(guò)程C ProgramModelAutomaticTranslatorVERDSMo

15、del CheckerNegative ConclusionSpecificationError Trace11反例分析以下輸入產(chǎn)生不正確結(jié)果1 3 5 7 9 11 13 15 17 19 0 2 4 6 8 10 12 14 16 18 4不正確運(yùn)行12修正后的例子: ISR2#include /*/int in();int isr(int x,int k);int isk(int n,int k);/*/int main()int n=0; int m=0;int k=1; printf(INFO: system is now activen); while (1) n=in();k=i

16、sk(n,k);m=isr(n,k); printf(RESULT: %inn,m);/*/int isr(int x,int k) int y1=0; int y2=0; int y3=0; y1=0; y2=1; y3=1; if (x=2|(x2&k=20) x=x-1; while (y3=x) y1=y1+1; y2=y2+2; y3=y3+y2; return y1;/*/int isk(int n,int k) if (k!=20) if (k!=n) k=21; else if (k=19) k=0; else k=k+2; else k=21; return k;/*/int

17、 in() char c=0; int k=0; while (1) k=0; putc(N,stdout); putc(:,stdout); putc(9,stdout); c=getc(stdin); if (c=n) printf(INFO: the input must be 1 or 2 digitsnn); continue; if (c9) while (1) c=getc(stdin); if (c=n) break; printf(INFO: the input must be 1 or 2 digitsnn); continue; k=c-0; c=getc(stdin);

18、 if (c=n) return k; if (c9) while (1) c=getc(stdin); if (c=n) break; printf(INFO: the input must be 1 or 2 digitsnn); continue; if (k2) k=k*10+(c-0); else if (k=2&c=0) k=20; else while (1) c=getc(stdin); if (c=n) break; printf(INFO: the input number must be in 0,.,20nn ); continue; c=getc(stdin); if

19、 (c!=n) while (1) c=getc(stdin); if (c=n) break; printf(INFO: the input must be 1 or 2 digitsnn); continue; return k; ./isr2INFO: system is now activeN: 2RESULT: 1N: 10RESULT: 3N: 20RESULT: 4N: 30INFO: the input number must be in 0,.,20N: 5RESULT: 2N:13模型檢測(cè)./verds c isr2.c sp isr.sp驗(yàn)證結(jié)果14驗(yàn)證過(guò)程C Progr

20、amModelAutomaticTranslatorVERDSModel CheckerSpecificationPositive Conclusion15模型檢測(cè)的缺點(diǎn)與優(yōu)點(diǎn)缺點(diǎn)可直接驗(yàn)證的程序規(guī)模小可直接驗(yàn)證的程序結(jié)構(gòu)簡(jiǎn)單優(yōu)點(diǎn): 自動(dòng)驗(yàn)證對(duì)不正確的程序可生成診斷信息16形式驗(yàn)證推理驗(yàn)證模型檢測(cè)推理驗(yàn)證與模型檢測(cè)相結(jié)合17推理驗(yàn)證+模型檢測(cè)#include /*/int in();int isr(int x,int k);int isk(int n,int k);/*/int main()int n=0; int m=0;int k=1; printf(INFO: system is now

21、 activen); while (1) n=in();k=isk(n,k);m=isr(n,k); printf(RESULT: %inn,m);/*/int isr(int x,int k) int y1=0; int y2=0; int y3=0; y1=0; y2=1; y3=1; if (x=2|(x2&k=20) x=x-1; while (y3=x) y1=y1+1; y2=y2+2; y3=y3+y2; return y1;/*/int isk(int n,int k) if (k!=20) if (k!=n) k=21; else if (k=19) k=0; else k=

22、k+2; else k=21; return k;/*/int in() char c=0; int k=0; while (1) k=0; putc(N,stdout); putc(:,stdout); putc(9,stdout); c=getc(stdin); if (c=n) printf(INFO: the input must be 1 or 2 digitsnn); continue; if (c9) while (1) c=getc(stdin); if (c=n) break; printf(INFO: the input must be 1 or 2 digitsnn);

23、continue; k=c-0; c=getc(stdin); if (c=n) return k; if (c9) while (1) c=getc(stdin); if (c=n) break; printf(INFO: the input must be 1 or 2 digitsnn); continue; if (k=0;k=0;k=20;18推理驗(yàn)證+模型檢測(cè)#include /*/int in();int isr(int x,int k);int isk(int n,int k);/*/int main()int n=0; int m=0;int k=1; printf(INFO

24、: system is now activen); while (1) n=in();k=isk(n,k);m=isr(n,k); printf(RESULT: %inn,m);/*/int isr(int x,int k) int y1=0; int y2=0; int y3=0; y1=0; y2=1; y3=1; if (x=2|(x2&k=20) x=x-1; while (y3=x) y1=y1+1; y2=y2+2; y3=y3+y2; return y1;/*/int isk(int n,int k) if (k!=20) if (k!=n) k=21; else if (k=1

25、9) k=0; else k=k+2; else k=21; return k;/*/int in() char c=0; int k=0; while (1) k=0; putc(N,stdout); putc(:,stdout); putc(9,stdout); c=getc(stdin); if (c=n) printf(INFO: the input must be 1 or 2 digitsnn); continue; if (c9) while (1) c=getc(stdin); if (c=n) break; printf(INFO: the input must be 1 o

26、r 2 digitsnn); continue; k=c-0; c=getc(stdin); if (c=n) return k; if (c9) while (1) c=getc(stdin); if (c=n) break; printf(INFO: the input must be 1 or 2 digitsnn); continue; if (k=0;k=0;k=0&r=20;19模型檢測(cè)./verds c isr.c sp isr.sp fsp isr.fsp 時(shí)間優(yōu)勢(shì)對(duì)比:驗(yàn)證結(jié)果20驗(yàn)證過(guò)程C ProgramModelAutomaticTranslatorVERDSModel

27、CheckerSpecificationPositive ConclusionFunctionSpecification21并發(fā)模型的模型檢測(cè)(例子)22并發(fā)模型的模型檢測(cè)(例子)a=s0b=t0 x=0y=0t=023并發(fā)模型的模型檢測(cè)驗(yàn)證問(wèn)題Model建模24并發(fā)模型(主程序)VVMVAR x: 0.1; y: 0.1; t: 0.1;INIT x=0; y=0; t=0;PROC p0: p0m(); p1: p1m();SPEC AG(!(p0.a=s2&p1.b=t2); AG(!p0.a=s1|AF(p0.a=s2|p1.b=t2)&(!p1.b=t1|AF(p0.a=s2|p1.

28、b=t2); AG(!p0.a=s1|AF(p0.a=s2)&(!p1.b=t1|AF(p1.b=t2); AG(!p0.a=s1|EF(p0.a=s2)&(!p1.b=t1|EF(p1.b=t2);25并發(fā)模型(進(jìn)程模塊說(shuō)明1)MODULE p0m()VAR a: s0,s1,s2,s3;INIT a=s0;TRANS a=s0: (y,t,a):=(1,1,s1); a=s1&(x=0|t=0): (a):=(s2); a=s1&!(x=0|t=0):(a):=(s1); a=s2: (y,a):=(0,s3); a=s2: (a):=(s2); a=s3: (y,t,a):=(1,1,s

29、1); 26并發(fā)模型(進(jìn)程模塊說(shuō)明2)MODULE p1m()VAR b: t0,t1,t2,t3;INIT b=t0; TRANS b=t0: (x,t,b):=(1,0,t1); b=t1&(y=0|t=1): (b):=(t2); b=t1&!(y=0|t=1):(b):=(t1); b=t2: (x,b):=(0,t3); b=t2: (b):=(t2); b=t3: (x,t,b):=(1,0,t1);27模型檢測(cè)./verds -ck 1 me002.vvmVERSION: verds 1.42 - DEC 2012FILE: me001.vvmPROPERTY: A G ! (a

30、 = 2 )& (b = 2 )bound = 1 time = 0- time = 0bound = 2 time = 0- time = 0bound = 3 time = 0- time = 0bound = 4 time = 0- time = 0bound = 5 time = 0- time = 0bound = 6 time = 0- time = 0CONCLUSION: TRUE (time=0)28模型檢測(cè)結(jié)論P(yáng)roperty ConclusionAG(!(p0.a=2&p1.a=2)trueAG(!p0.a=1|AF(p0.a=2|p1.a=2)&(!p1.a=1|AF(

31、p0.a=2|p1.a=2)falseAG(!p0.a=1|AF(p0.a=2)&(!p1.a=1|AF(p1.a=2)falseAG(!p0.a=1|EF(p0.a=2)&(!p1.a=1|EF(p1.a=2)true29進(jìn)程公平性說(shuō)明30并發(fā)模型(主程序)VVMVAR x: 0.1; y: 0.1; t: 0.1;INIT x=0; y=0; t=0;PROC p0: p0m(); p1: p1m();SPEC AG(!(p0.a=s2&p1.b=t2); AG(!p0.a=s1|AF(p0.a=s2|p1.b=t2)&(!p1.b=t1|AF(p0.a=s2|p1.b=t2); AG(!

32、p0.a=s1|AF(p0.a=s2)&(!p1.b=t1|AF(p1.b=t2); AG(!p0.a=s1|EF(p0.a=s2)&(!p1.b=t1|EF(p1.b=t2);31并發(fā)模型(進(jìn)程模塊說(shuō)明1a)MODULE p0m()VAR a: s0,s1,s2,s3;INIT a=s0;TRANS a=s0: (y,t,a):=(1,1,s1); a=s1&(x=0|t=0): (a):=(s2); a=s1&!(x=0|t=0):(a):=(s1); a=s2: (y,a):=(0,s3); a=s2: (a):=(s2); a=s3: (y,t,a):=(1,1,s1);FAIRNES

33、S running; 32并發(fā)模型(進(jìn)程模塊說(shuō)明2a)MODULE p1m()VAR b: t0,t1,t2,t3;INIT b=t0; TRANS b=t0: (x,t,b):=(1,0,t1); b=t1&(y=0|t=1): (b):=(t2); b=t1&!(y=0|t=1):(b):=(t1); b=t2: (x,b):=(0,t3); b=t2: (b):=(t2); b=t3: (x,t,b):=(1,0,t1);FAIRNESS running;33模型檢測(cè)結(jié)論P(yáng)roperty ConclusionAG(!(p0.a=2&p1.a=2)trueAG(!p0.a=1|AF(p0.a=2|p1.a=2)&(!p1.a=1|AF(p0.a=2|p1.a=2)trueAG(!p0.a=1|AF(p0.a=2)&(!p1.a=1|AF(p1.a=2)falseAG(!p0.a=1|EF(p0.a=2)&(!p1.a=1|EF(p1.a=2)true34進(jìn)程公平性說(shuō)明235并發(fā)模型(主程序)VVMVAR x: 0.1; y: 0.1; t: 0.1;INIT x=0; y=0; t=0;PROC p0: p0m(); p1: p1m();SPEC AG(!(p0.a=s2&p1.b=t2)