模型检测ndash例子

1、模型检测 例子计算机科学国家重点实验室1程序正确的重要性应用广泛航空航天金融设备的控制日常生活软件错误的可能后果火箭 Ariane 5 Explosion (1997)火星气候轨道器 NASA Mars Climate Orbiter (1999)2程序正确性方法测试黑盒测试白盒测试形式验证推理验证模型检测可执行代码程序代码程序代码或其抽象模型程序运行环境模型规则算法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; 对ISR的要求:输入为0和20

10、之间的整数时，输出为其平方根的整数部分。输入为0和20之间的整数时，输出为其平方根的整数部分。5测试黑盒测试白盒测试Program testing can be used to show the presence of bugs, but never to show their absence! - Edsger W. Dijkstra6形式验证推理验证模型检测正确性正确性+不正确性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; 对ISR的要求:(at line 18):(m*m)n)8程序的模型检测(例子)9模型检测./verds c isr.c sp isr.sp验证结果反例10验证过程C ProgramModelAutomaticTranslatorVERDSMo

15、del CheckerNegative ConclusionSpecificationError Trace11反例分析以下输入产生不正确结果1 3 5 7 9 11 13 15 17 19 0 2 4 6 8 10 12 14 16 18 4不正确运行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模型检测./verds c isr2.c sp isr.sp验证结果14验证过程C Progr

20、amModelAutomaticTranslatorVERDSModel CheckerSpecificationPositive Conclusion15模型检测的缺点与优点缺点可直接验证的程序规模小可直接验证的程序结构简单优点: 自动验证对不正确的程序可生成诊断信息16形式验证推理验证模型检测推理验证与模型检测相结合17推理验证+模型检测#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推理验证+模型检测#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模型检测./verds c isr.c sp isr.sp fsp isr.fsp 时间优势对比:验证结果20验证过程C ProgramModelAutomaticTranslatorVERDSModel

27、CheckerSpecificationPositive ConclusionFunctionSpecification21并发模型的模型检测(例子)22并发模型的模型检测(例子)a=s0b=t0 x=0y=0t=023并发模型的模型检测验证问题Model建模24并发模型(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并发模型(进程模块说明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并发模型(进程模块说明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模型检测./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模型检测结论Property 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进程公平性说明30并发模型(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并发模型(进程模块说明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并发模型(进程模块说明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模型检测结论Property 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进程公平性说明235并发模型(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)