189 lines
6.2 KiB
C
189 lines
6.2 KiB
C
/*
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Program: evalExpression-v2.c (Report comments/bugs to chikh@yuntech.edu.tw)
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Function: 整合書範例infixTopostfix.c程式與第三章末習題第四題的要求，使得程式可接受使用者
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從鍵盤輸入中序表示式字串，隨即顯示對應的後序表示式，再依此進行運算並顯示結果值。
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與evalExpression-v1.c功能相同，唯一差異在evaluate()函數的實作方式；此版以strtok()
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作字串剖析，以空白作分隔字元，分離出所需的運算元與運算子。
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Notes:
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10+20*(50-30)/2^4-6*8
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10+20*(-50+30)/-2^4-6*8
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10.5+20.8*(50.1-30.6)/2.5^4-6.6*8.2
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10.5+20.8*(-50.1+30.6)/2.5^-4-6.6*8.2
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10-30*(+50-20)/-2^-4+7.5*-6
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10+30*(+50.2+3.8)/+2^2.5+15*-6.5
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*/
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#include <stdio.h>
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#include <string.h> /* for strtok() */
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#include <math.h> /* for pow() */
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#include <ctype.h> /* for isdigit() */
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#define MAX 120
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void infix_to_postfix(char [], int, char []); /* 由中序轉後序函數 */
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int compare(char, char); /* 比較兩個運算子函數 */
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float evaluate(char []); /* 新訂的函數，為後序式實作所需的數學運算並回傳結果值 */
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/* 在中序表示法佇列及暫存堆疊中，運算子的優先順序表，其優先值為INDEX/2 */
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char infix_priority[9] = {'#', ')', '+', '-', '*', '/', '^', '('};
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char stack_priority[8] = {'#', '(', '+', '-', '*', '/', '^'};
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int main()
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{
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int i, index = -1;
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char infix_q[MAX], postExpr[2*MAX] = {0}; /* 儲存使用者輸入中序式字元串，轉換為後序式字串存於postExpr */
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printf("\n請輸入中序表示式(支援+-*/^()運算) => ");
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while (infix_q[index] != '\n')
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infix_q[++index] = getchar();
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infix_q[index] = '#'; /* 原輸入的最後一個字元取代為'#'，作為運算式字串結尾 */
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printf("\n對應的後序表示式：");
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infix_to_postfix(infix_q,index,postExpr);
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//printf("\n --> %s\n",postExpr);
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printf("\n\n運算結果=%g",evaluate(postExpr));
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//printf("\t(正確值=%g)\n\n",10+20*(50-30)/pow(2,4)-6*8);
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//printf("\t(正確值=%g)\n\n",10+20*(-50+30)/pow(-2,4)-6*8);
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printf("\t(正確值=%g)\n\n",10.5+20.8*(50.1-30.6)/pow(2.5,4)-6.6*8.2);
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//printf("\t(正確值=%g)\n\n",10.5+20.8*(-50.1+30.6)/pow(2.5,-4)-6.6*8.2);
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//printf("\t(正確值=%g)\n\n",10-30*(+50-20)/pow(-2,-4)+7.5*-6);
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//printf("\t(正確值=%g)\n\n",10+30*(+50.2+3.8)/pow(2,2.5)+15*-6.5);
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return 0;
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}
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void infix_to_postfix(char infix_q[], int index, char postExpr[])
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{
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int top = 0, ctr, tag = 1, i = 0;
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char c, stack_t[MAX]; /* 儲存還不必輸出的運算子 */
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stack_t[top] = '#'; /* 於堆疊最底下加入#為結束符號 */
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for (ctr = 0; ctr <= index; ctr++) { /* 運算式視為字元串，從第0字元開始，逐字掃描，直至最後一個字元為止 */
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switch (infix_q[ctr]) { /* 檢視ctr索引值對應的字元內容，查看是否吻合底下任一情況 */
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case ')': /* 輸入為')'，將輸出堆疊內運算子，直到彈出'('為止 */
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printf(" "); /* 螢幕顯示加入空白間隔 */
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sprintf(postExpr,"%s ",postExpr);
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while (stack_t[top] != '(') {
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printf("%c ",c=stack_t[top--]);
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sprintf(postExpr,"%s%c ",postExpr,c);
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}
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top--; /* 把'('一併移除 */
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break;
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case '#': /* 輸入為#，代表已掃描到最末字元，於是把堆疊內尚未輸出的運算子通通彈出 */
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printf(" ");
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sprintf(postExpr,"%s ",postExpr); /* 導入空白於後序式字串 */
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while (stack_t[top] != '#') {
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printf("%c ",c=stack_t[top--]);
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sprintf(postExpr,"%s%c ",postExpr,c);
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}
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break;
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case '(':
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case '^':
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case '*':
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case '/':
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printf(" ");
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sprintf(postExpr,"%s ",postExpr);
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while (compare(stack_t[top],infix_q[ctr])) {
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printf("%c ",c=stack_t[top--]);
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sprintf(postExpr,"%s%c ",postExpr,c);
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}
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stack_t[++top] = infix_q[ctr];
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tag = 1;
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break;
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case '+':
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case '-':
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if (tag == 1) { /* 判斷 '(' '^' '*' '/' 後的 +- 號是否表示正負的一元運算子 */
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stack_t[++top] = infix_q[ctr];
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tag = 2;
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}
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else {
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printf(" ");
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sprintf(postExpr,"%s ",postExpr);
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while (compare(stack_t[top],infix_q[ctr])) {
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printf("%c ",c=stack_t[top--]);
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sprintf(postExpr,"%s%c ",postExpr,c);
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}
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stack_t[++top] = infix_q[ctr];
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tag = 1;
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}
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break;
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case ' ':
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break; /* do nothing，直接略過 */
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/* 輸入為運算元，則直接輸出 */
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default:
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for (i = 0; isdigit(infix_q[ctr+i]) || infix_q[ctr+i]=='.'; i++) {
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printf("%c",c=infix_q[ctr+i]);
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sprintf(postExpr,"%s%c",postExpr,c);
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}
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ctr += (i-1);
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if (tag == 2) { /* 將存放在堆疊的負號輸出 */
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printf(" %c ",c=stack_t[top--]); /* 視同彈出堆疊頂端的元素 */
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sprintf(postExpr,"%s %c ",postExpr,c=='-'?'_':'@');
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}
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tag = 0;
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break;
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}
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}
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}
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/* 比較兩運算子優先權，若輸入運算子小於等於堆疊中運算子，則傳回值為 1，否則為 0 */
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int compare(char stack_o, char infix_o)
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{
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int index_s = 0, index_i = 0;
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while (stack_priority[index_s] != stack_o)
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index_s++;
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while (infix_priority[index_i] != infix_o)
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index_i++;
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return index_s/2 >= index_i/2 ? 1 : 0;
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}
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float evaluate(char expr[])
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{
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char *token;
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float stack[MAX], a, b, c;
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int top = -1;
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token = strtok(expr," "); /* 取得第一個符記 */
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while (token != NULL) { /* 若不為空，則嘗試繼續剖析字串 */
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if (atof(token)!=0.0)
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stack[++top] = atof(token);
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else {
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b = stack[top--];
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a = stack[top--];
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switch(token[0]) {
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case '+':
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c = a+b;
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break;
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case '-':
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c = a-b;
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break;
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case '@':
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stack[++top] = a; /* 多取出的資料推回堆疊，因為一元運算 */
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c = b;
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break;
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case '_':
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stack[++top] = a; /* 多取出的資料推回堆疊，因為一元運算 */
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c = -b;
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break;
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case '*':
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c = a*b;
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break;
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case '/':
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c = (float)a/b;
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break;
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case '^':
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c = (float)pow(a,b);
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break;
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}
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stack[++top] = c;
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}
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token = strtok(NULL," "); /* 很特別的寫法，請同學多加留意 */
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}
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return stack[top];
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}
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