編譯原理語法分析實驗報告(編輯修改稿)
2024-11-14 21:57 本頁面
【文章內(nèi)容簡介】 j, k = 0。printf(“請輸入文法規(guī)則數(shù)：”)。scanf(“%d”, amp。r)。printf(“請輸入文法規(guī)則：n”)。for(i = 0。i} for(i = 0。i} for(i = 0。ifor(j = 0。st[i][j]!= 39。39。j++)if((st[i][j]39。Z39。)amp。amp。 st[i][j]!= 39。39。amp。amp。st[i][j]!= lable[k++] = st[i][j]。for(j = 0。st[i][j]!= 39。39。j++){} if(st[i][0]39。Z39。){} if(st[i][j] = 39。A39。amp。amp。st[i][j] } if(st[i][j + 1] = 39。A39。amp。amp。st[i][j + 1] } printf(“文法error!n”)。exit(1)。printf(“文法error!n”)。exit(1)。scanf(“%s”, st[i])。//存儲文法規(guī)則，初始化FIRSTVT集和LASTVT集*/first[i][0] = 0。/*first[i][0]和last[i][0]分別表示st[i][0]非終極符的last[i][0] = 0。FIRSTVT集和LASTVT集中元素的個數(shù)*/39。39。amp。amp。st[i][j]!= 39。|39。)lable[k] = 39。39。lable[k + 1] = 39。39。} table()。//printf(“FIRST集為：n”)。//輸出每個非終結(jié)符的FIRST集 for(i = 0。i} printf(“LAST集為：n”)。//輸出每個非終結(jié)符的LAST集 for(i = 0。i} printf(“算符優(yōu)先分析表如下:n”)。for(i = 0。lable[i]!= 39。39。i++)printf(“t%c”, lable[i])。printf(“n”)。for(i = 0。i} printf(“請輸入文法輸入符號串以結(jié)束:”)。scanf(“%s”, input)。deal()。cout char text[20][10]。//存儲改寫后的文法int i, j, k, t, l, x = 0, y = 0。int m, n。x = 0。for(i = 0。ifirstvt(st[i][0])。lastvt(st[i][0])。} for(i = 0。i{text[x][y] = st[i][0]。y++。for(j = 1。st[i][j]!= 39。39。j++){if(st[i][j] == 39。|39。)//{text[x][y] = 39。39。x++。y = 0。text[x][y] = st[i][0]。y++。text[x][y++] = 39。39。text[x][y++] = 39。39。}else{text[x][y] = st[i][j]。y++。}}text[x][y] = 39。39。x++。y = 0。} r1 = x。//printf(“轉(zhuǎn)化后的文法為:n”)。for(i = 0。iprintf(“%sn”, text[i])。} for(i = 0。i{//輸出轉(zhuǎn)化后的文法規(guī)則串 /*求每個終結(jié)符的推導(dǎo)結(jié)果(去掉“”后的} str[i][0] = text[i][0]。for(j = 3, l = 1。text[i][j]!= 39。39。j++, l++)str[i][l] = text[i][j]。str[i][l] = 39。39。for(i = 0。ifor(j = 1。text[i][j + 1]!= 39。39。j++){if(terminal_symbol(text[i][j])amp。amp。 terminal_symbol(text[i][j + 1])){} if(text[i][j + 2]!= 39。39。amp。amp。terminal_symbol(text[i][j])amp。amp。 {} if(terminal_symbol(text[i][j])amp。amp。!terminal_symbol(text[i][j + 1]))//終結(jié){} if(!terminal_symbol(text[i][j])amp。amp。 terminal_symbol(text[i][j + 1]))//非終{ for(k = 0。k} m = location(text[i][j])。for(t = 0。t} n = location(first[k][t + 1])。data[m][n] = 39。break。m = location(text[i][j]
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