文法存儲(chǔ)及輸出

1、精選優(yōu)質(zhì)文檔-傾情為你奉上實(shí)驗(yàn)報(bào)告學(xué)號(hào)專業(yè)計(jì)算機(jī)班級(jí)姓名課程名稱編譯原理課程類型必修課實(shí)驗(yàn)名稱實(shí)驗(yàn)一、文法存儲(chǔ)及輸出實(shí)驗(yàn)?zāi)康模?1.掌握文法在計(jì)算機(jī)內(nèi)的存儲(chǔ)表示； 2.輸入任意的上下文無(wú)關(guān)文法；3.以四元組的形式輸出Chomsky 文實(shí)驗(yàn)內(nèi)容：敘述本次實(shí)驗(yàn)的實(shí)現(xiàn)方法、分析、實(shí)現(xiàn)了哪些功能、給出一個(gè)或多個(gè)測(cè)試用例及其預(yù)期運(yùn)行結(jié)果，要求能夠反映本次實(shí)驗(yàn)的成果。1.本次實(shí)驗(yàn)的實(shí)現(xiàn)方法為：采用數(shù)組的形式實(shí)現(xiàn)文法在機(jī)內(nèi)的存儲(chǔ)表示。2.本次試驗(yàn)的問(wèn)題分析為：一是文法的輸入形式；二是文法在計(jì)算機(jī)內(nèi)的存儲(chǔ)表示形式。3.本次試驗(yàn)應(yīng)實(shí)現(xiàn)以下功能：輸入一個(gè)文法的識(shí)別符號(hào)與重寫(xiě)規(guī)則，輸出文法的四要素VN，VT，P，Z

2、。4.測(cè)試實(shí)例：實(shí)例：輸入文法GE:E:=E+T|T T:=T*F|FF:=(E)|i實(shí)驗(yàn)代碼如下：#include <stdio.h> char GName10; typedef char SymbolT; SymbolT StartSymbol; typedef struct SymbolT LeftSym; SymbolT Rightpart20; int RightLength; RuleTC; RuleTC GrammarC20; SymbolT VT20; SymbolT VN20; int NVN=0,NVT=0; typedef struct int LeftSym

3、Num; int RightpartNum20; int RightLength; RuleT; RuleT Grammar20; int RuleN; char StartSym; char Input 20= "E:=E+T|T", "T:=T*F|F", "F:=(E)|i", "0" ; int LineNumofG=3; void GetGrammar() char LeftS,Sym; int k,j,m; StartSym=GrammarC1.LeftSym; for(k=1; k<=RuleN

4、; k+) LeftS=GrammarCk.LeftSym; for(j=1; j<=NVN; j+) if(LeftS=VNj) Grammark.LeftSymNum=j+100; break; for(j=1; j<=GrammarCk.RightLength; j+) Sym=GrammarCk.Rightpartj; for(m=1; m<=NVN; m+)if(Sym=VNm) Grammark.RightpartNumj=m+100;break; if(m<=NVN) continue; for(m=1; m<=NVT; m+) if(Sym=VTm

5、) Grammark.RightpartNumj=m; break; Grammark.RightLength=GrammarCk.RightLength; void storeG( ) int row,colum, k,j,m; char sym,U; RuleN=0; row=0; NVN=NVT=0; while(row<LineNumofG) colum=0; sym=Inputrowcolum; for(k=1; k<=NVN; k+) if(sym=VNk) break; if(k>NVN) VN+NVN=sym; GrammarC+RuleN.LeftSym=s

6、ym; colum=colum+4; j=0; while(Inputrowcolum!='0') sym=Inputrowcolum+; if(sym='|') GrammarCRuleN.RightLength=j;U=GrammarCRuleN.LeftSym; RuleN=RuleN+1; GrammarCRuleN.LeftSym=U; j=0; continue; j=j+1; GrammarCRuleN.Rightpartj=sym; for(k=1;k<=NVT; k+)if( sym=VTk) break;if(k>NVT)VT+N

7、VT=sym; GrammarCRuleN.RightLength=j; row=row+1; for( m=1; m<=NVT; m+) for(k=1; k<=NVN; k+) if(VTm=VNk) VTm=' ' break; for(m=1;m<=NVT; m+) if(VTm=' ') for(j=m+1; j<=NVT; j+) VTj-1=VTj; NVT=NVT-1; if(VTm=' ') m=m-1; continue; GetGrammar( ); DispG( ) int k,j,m; print

8、f("n 文法的識(shí)別符號(hào) Z 為:%cn", StartSym); printf("重寫(xiě)規(guī)則集 P 為:n"); for(k=1; k<=RuleN; k+) printf("%8c%c:=",' ', VNGrammark.LeftSymNum-100); for(j=1; j<=Grammark.RightLength; j+) m=Grammark.RightpartNumj; if(m<100) printf("%c",VTm); else printf("%c&

9、quot;,VNm-100); printf("n"); printf("文法的非終結(jié)符號(hào)集為：VN= "); for(k=1; k<=NVN-1; k+) printf("%c", VNk); printf(","); printf("%c n",VNNVN); printf("文法的終結(jié)符號(hào)集為：VT= "); for(k=1; k<=NVT-1; k+) printf("%c", VTk); printf(",");

10、printf("%c n",VTNVT); void main( ) char c='Y' int j; printf("n 文法重寫(xiě)規(guī)則:n"); for(j=0; j<LineNumofG; j+) printf("%sn",Inputj); printf("重新輸入文法的重寫(xiě)規(guī)則?(Y/N): "); scanf("%c",&c); if(c='Y' | c='y') printf("輸入行數(shù)(一個(gè)規(guī)則一行):"); scanf("%d",&LineNumofG); for(j=0; j<LineNumofG; j+) scanf("%s",Inputj); storeG( ); DispG( ); 實(shí)驗(yàn)結(jié)果：實(shí)例一：輸入文法GE:E:=E+T|T T:=T*F|FF:=(E)|i實(shí)例二：輸入文法G2Z:Z:=aZb|ab Z:=CSd S:=CSd|Cd實(shí)例三：輸入文法G3E:E:=T|E+T|E-TT:=F|T*F|T/F F:=(E)|i實(shí)例四：輸入文法：G4Z:Z:=Ab|c A:=Aa