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運算式求值與轉換 -stack 之應用

運算式求值與轉換 -stack 之應用. Lai Ah Fur. 前序式 (prefix) 求值. 處理過程 : 由右而左掃描運算式 : 二種處理方式 ( 1) 由右而左讀取運算元 / 子 (2) 將運算元 / 子由左而右 push 到 stack, 再由 stack 中 pop ( 順序即為由右而左 ) 掃描到運算元 (0..9),push 到 stack 掃描到運算子 (+-*/), 由 stack 中 pop 出二元素 . 並進行計算 , 最後將計算結果 push 到 stack

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運算式求值與轉換 -stack 之應用

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  1. 運算式求值與轉換-stack之應用 Lai Ah Fur

  2. 前序式(prefix)求值 • 處理過程: • 由右而左掃描運算式:二種處理方式 (1)由右而左讀取運算元/子 (2)將運算元/子由左而右push到stack,再由stack中pop (順序即為由右而左) • 掃描到運算元(0..9),push到stack • 掃描到運算子(+-*/),由stack中pop出二元素. 並進行計算,最後將計算結果push到stack • 當掃描運算式結束,將由stack中pop出最後運算式結果.若stack中仍有元素表示運算式寫法有誤

  3. 前序式求值example • *+47-82 • 處理過程: • Scan 2, push into stack • Scan 8 , push into stack • Scan -, pop 2 operands from stack, then evaluate, finally push the result into stack • Scan 7, push into stack • Scan 4, push into stack • Scan +, pop 2 operands from stack, then evaluate, finally push the result into stack • Scan *, pop 2 operands from stack, then evaluate, finally push the result into stack • Finally, pop the result from the stack The first operand: Subtrahend The second :Minuend(被)

  4. 後序式(postfix)求值 • 處理過程: • 由左而右掃描運算式 • 掃描到運算元(0..9),push到stack • 掃描到運算子(+-*/),由stack中pop出二元素. 並進行計算,最後將計算結果push到stack • 當掃描運算式結束,將由stack中pop出最後運算式結果.若stack中仍有元素表示運算式寫法有誤

  5. 後序式求值example • 47+82-* • 處理過程: • Scan 4, push into stack • Scan 7, push into stack • Scan +, pop 2 operands from stack, then evaluate, finally push the result into stack • Scan 8, push into stack • Scan 2, push into stack • Scan -, pop 2 operands from stack, then evaluate, finally push the result into stack • Scan *, pop 2 operands from stack, then evaluate, finally push the result into stack • Finally, pop the result from the stack The first operand: :Minuend(被) The second:Subtrahend 2 8 11 8-2

  6. 8 7 4 * + 中序式求值example • 4+7*8-2 • 處理過程: • Scan 4, push into operand stack • Scan +, push into operator stack • Scan 7, push into operand stack • Scan *,priority:*(目前)>+(stack top), push into operator stack • Scan 8, push into stack • Scan -, priority:-(目前)<=*(stack top), pop 2 operands from stack and pop * from operator stack, then evaluate, push the result into operand stack; priority:-(目前)<=+(stack top) ,pop 2 operands from stack and pop + from operator stack, then evaluate, push the result into operand stack; finally push – into operator stack. • Scan 2, push into stack • Scan over, pop 2 operands from stack and pop - from operator stack • 自行練習: 3+4*5*6-7-8處理過程 • Problem: 未處理“()”問題 4+7*8-2

  7. 中序式(infix)求值(1) • 處理過程: • 由左而右掃描運算式,且須二個stack • 掃描到運算元(0..9),push到operand stack • 掃描到運算子(+-*/), push到operator Stack先檢查 : (1)if operator stack is empty,將目前之運算子push到operator stack;(2)比較目前之運算子優先權是否大於operator stack top之運算子,若是則直接將目前之運算子push到operator;否則(<=),由operand stack中pop出二運算元, operator stack中pop出運算子,並進行計算, 計算結果push到operand stack.最後再將目前之運算子push到operator stack • 當掃描運算式結束,將operand stack中pop出二元素, operator stack中pop出運算子,並進行計算, 計算結果push到stack.

  8. 中序式求值(2) • (4+7)*(8-2) • 處理過程: • 掃描到運算子“(“, 無條件push到operator stack. • 掃描到運算子“)“, pop出operator stack之 operator,進行計算,直到遇到”(” • 其他,同前

  9. 中序轉後序 A Example: A+B*C/D-E 轉換程序:由左而右掃描運算式,且須一個operator stack • Scan A:輸出至後序運算式之後端 • Scan +:push至operator stack (目前堆疊is empty) • Scan B:輸出至後序運算式之後端 • Scan *:push至operator stack (目前堆疊top之operator’s priority <“*” priority) • Scan C:輸出至後序運算式之後端 • Scan /:pop堆疊top之operator且輸出至後序運算式之後端, 最後再將“/”push至operator stack (目前堆疊top之operator’s priority =“/” priority) • Scan D:輸出至後序運算式之後端 • Scan -:pop堆疊top之operator ”/”及“+”且輸出至後序運算式之後端, 最後再將“-”push至operator stack (目前堆疊top之operator and next top operators’ priority >=“/” priority) • Scan E:輸出至後序運算式之後端 • pop堆疊top之operator ”-”,輸出至後序運算式之後端 AB ABC ABC*D ABC*D/+E-

  10. 中序轉後序 Example: A+(B*C/D-E-K)*F-G*H SOLUTION: ABC*D/E-K-F*+GH*-

  11. 中序轉前序 E Example: A+B*C/D-E 轉換程序:由右而左掃描運算式,且須一個operator stack • Scan E:輸出至後序運算式之前端 • Scan -:push至operator stack (目前堆疊is empty) • Scan D:輸出至後序運算式之前端 • Scan /:push至operator stack (目前堆疊top之operator’s priority <=“/” priority) • Scan C:輸出至後序運算式之前端 • Scan *:將“*”push至operator stack (目前堆疊top之operator’s priority =“*” priority) • Scan B:輸出至後序運算式之前端 • Scan +:pop堆疊top之operator ”*”及“/”且輸出至後序運算式之前端, 最後再將“+”push至operator stack (目前堆疊top之operator and next top operators’ priority >“+” priority) • Scan A:輸出至後序運算式之前端 • pop堆疊top之operator ”+”及”-” 且輸出至後序運算式之前端 DE CDE /*BCDE A/*BCDE -+A/*BCDE

  12. 中序轉前序 Example: A+(B*C/D-E-K)*F-G*H SOLUTION: -+A*--/*BCDEKF*GH

  13. 作業四 • 功能 • 中序轉前序(數字/變數) • 中序轉後序(數字/變數) • 前序求值 • 後序求值 • 中序求值 • 介面: applet (java) / GUI (C++) • 須顯示轉序/求值詳細過程中:顯示stack及輸出字串之內容

  14. Homework 5:simple BASIC simulator (interpreter) Syntax: • 接受之variable:A~Z共52個(分大小寫) • Basic Command:print,input(如input a), 指定敘述(含++,--) • 接受Math expression,如a+b*3, 56-4*9/2 • Operator:+,-,*,/,^ • Statement前加行號,可不按順序 Interpreter 操作環境 • Load ‘…bas’命令:載入.bas程式檔,顯示於input box • Save命令:將 input box存檔 • run命令:依行號順序,解譯程式,顯示於output .遇syntax error ,顯示訊息於output • cls命令:清除memory程式,及所有variable內容 • quit:離開interpreter

  15. Testing example for homework • 10 a=10*5-(10-4) • 20 c=30 • 31 d=(a*c-10)/2-7*c • 25 print a,c,a*c,45/(5+9-2)*2 • 30 input c • 40 print d,a,c • 45 print “a=“,a,”c=“,c

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