一个简单的四则运算程序C语言实现--实现处理括号

 

说明：经过今天的测试解决了括号处理的一些问题。比如由于增加了paren 结构元素之后，复制函数没有更新。

及非直接括号。等类型。如(2+4)等非必须性括号。

没有使用高级的数据结构如堆栈啦。

但程序本还是挺好的。注释也挺 不错。

写得也不错。 哈哈。两天的成果啊！完全自己写的啊！

/*** * @author banxi1988 * @mail banxi1988 at gmail.com * @qq 787928000 * 一个简单的表达式四则运算器。C语言实现。实现处理括号。 * @bug-report : if have any suggestion and find any bug please send mail to my mail or send massage to my qq! thanks */#include<stdio.h>#include<stdlib.h>#include<string.h>#include<time.h>//#define BANXI_DEBUG 1 // 是否为调试模式#define MAX_LEN 1024 //最长的运算表达式。#define D_BIT_LEN 32 //最大数值位包括小数点。/*** * 一些有用的宏，关于字符操作 */#define is_numerical(ch) (strchr("0123456789.",(ch)) != NULL) // 决断是否是组成数值的元素#define is_operator(ch) (strchr("+-*/",(ch)) != NULL) // 判断是否是运算符#define is_muldiv(ch) (strchr("*/",(ch)) != NULL) // 判断是否是乘法或除法#define is_paren(ch) (strchr("()",(ch)) != NULL) //判断是否是括号#define is_end_input(ch) (ch == EOF || ch == '/n') //判断是否是输入的结尾。一次只允许输入一行。所以..typedef enum OP_TYPE_TAG{OPERAND,OPERATOR,LPAREN,RPAREN}OP_TYPE; // 表达式元素类型，lparen 意为左括号。rparen意为右括号typedef struct OP_ITEM_TAG{ // 表达式元素项类型double operand; // 用来存储操作数char operator; // 存储操作符char paren; // 存储括号OP_TYPE op_type; // 存储元素类型}OP_ITEM;OP_ITEM op_items[MAX_LEN] = {{0,0,0,0}};// 表达式元素项数组。int items_count = 0; // 元素项数int operand_count = 0; // 操作数数int operator_count = 0;// 操作符数int lparen_count = 0; //左括号数int rparen_count = 0; //右括号数/*** * @function: 移除表达式元素之间空白符。 * 因为'/n'也是空白符。一般输入当中也容易把作为输入结束标志。 * 所以。如果处理'/n'当作结束标志。 */ int remove_space(){char ch;while(isspace(ch=getchar())&&(!is_end_input(ch)));ungetc(ch,stdin);}//end remove_space/*** * @function: 如果输入的表达式操作数位数过大，进行提示。 * 虽然double型数据可以达到正负37位。但是此程序只处理32位。 * 所以提示。因为小程序，不想处理那么大的数值 */ int bits_length_tips(int index){if(index > 37){printf("WARNING:YOU HAVE INPUT A TOO LARGER NUMBER,IF HAVE NO POINT!/n");}else if(index > D_BIT_LEN){printf("WARNING:YOU HAVA INPUT A LARGER NUMBER!/n");}//}//end bits_length_tips/*** * @function: 扫描标准输入。组建表达式。并判断是否是有效的运算表达式。 * 此函数比较关键。逻辑相当复杂点。经过重构之后，显得稍微清楚了点。 * 合理利用库函数。是一大亮点。 */ int scan_input(){char buffer[D_BIT_LEN] = {'/0'};char operator = 'a';double operand = 0.0;char ch = 'a';OP_TYPE op_type;int index = 0; //for digit bit count int flag = 0; //for float . token countint fail = 0;while(1){remove_space();ch = getchar();if(is_end_input(ch)){break;}//end end ifif(is_numerical(ch)){/// 接受到非数字组成字符跳出，或者空格也跳出index = 0; do{if(isdigit(ch)){buffer[index++] = ch;}else{flag++;if(flag > 1){printf("ERROR:too many /'./' char,need a number/n");exit(EXIT_FAILURE);}else{buffer[index++] = ch;}}// }while(is_numerical(ch=getchar()));//end numerical while ungetc(ch,stdin);// 将空格或者其它字符放回，以便下面处理bits_length_tips(index); // 超多位数可能越界提示index = (index > 36)?36:index;buffer[index] = '/0';operand = atof(buffer);op_items[items_count].operand = operand;op_items[items_count].op_type = OPERAND;operand_count++;items_count++;}else if(is_operator(ch)){op_items[items_count].operator = ch;op_items[items_count].op_type = OPERATOR;operator_count++;items_count++;}else if(is_paren(ch)){op_items[items_count].paren = ch;//op_type = (ch == '(')?LPAREN:RPAREN; if(ch == '('){op_type = LPAREN;lparen_count++;}else{op_type = RPAREN;rparen_count++;}op_items[items_count].op_type = op_type;items_count++;}else{printf("ERROR:NON VALID CHAR %c/n",ch);exit(EXIT_FAILURE);}//end ch judge}//end scan whileif(items_count % 2 == 0){printf("ERROR:non valid expression/n");fail = 1;}//if(lparen_count != rparen_count){printf("ERROR:PARENTHESIS NOT MATCH!/n");fail = 1;}//if(operand_count != (operator_count + 1)){printf("ERROR:OPERAND OR OPERATOR NOT ENOUGH/n");fail = 1;}//if(fail){exit(EXIT_FAILURE);}//return 0;}// end scan_input/*** * @function:用于调试时打印op_items中的数据项。 */voidprint_data(){int i = 0;for(i = 0; i < items_count; i++){if(op_items[i].op_type == OPERAND){printf("%d/t%lf/t",i,op_items[i].operand);}//else if(op_items[i].op_type == OPERATOR){ printf("%d/t%c/t",i,op_items[i].operator);}else{ printf("%d/t%c/t",i,op_items[i].paren);}//printf("/n");}//end for}//end print_data/*** * @function :根据传入的两个数及运算符即操作符，返回运算结果 * @return : double * 说明：均为从左到右运算 */ double cal(double left,char op,double right){switch(op){case '+':return left+right;break;case '-':return left - right ;break;case '*':return left*right;break;case '/': if(right == 0){ printf("divide not be 0/n");exit(EXIT_FAILURE); }// else{ return left/right; }break;default:printf("non valid operator %c /n",op);exit(EXIT_FAILURE); break;}//end op switch}//end cal/** * @function :将下标from处的op_items内容复制到dest下标处。dest处的内容将被覆盖。 */void copy_items(int dest,int from){op_items[dest].operand = op_items[from].operand;op_items[dest].operator = op_items[from].operator;op_items[dest].paren = op_items[from].paren;op_items[dest].op_type = op_items[from].op_type;}//end copy_items/** * @function: 从start下标处开始。将steps后的op_items中的内容往前移steps个单位。 * 即start后面的steps个数组内容没有了将后面的向前移动以便对齐。 * */int move_items(int start,int steps){int index = 0;for(index = start ; (index+steps) < items_count; index++){copy_items(index,index+steps);}//return 0;}//end move_items/*** * @fuction: 按顺序查找操作符的在op_items中的下标。 * 依赖items_count */int find_operator(){int i = -1;int end = items_count -1;for(i = 1; i < end; i++){if(op_items[i].op_type == OPERATOR){return i;}//END IF }//end forreturn -1;}///** * @function : 扫描op_items数组，查找乘法或者除法。如果返回找到第一个的下标。 *如果没有找到返回-1 */int find_muldiv(){int i = 0;int end = items_count -1;for(i = 1; i < end; i++){if(op_items[i].op_type == OPERATOR){ if(is_muldiv(op_items[i].operator)){return i; }//end }//end op_type if}// return -1;}///*** * @function :查找最里面的左括号。如果找到返回相应op_items下标。否则返回-1 * */ int find_inner_lparen(){int innerest = -1;int i = 0;int end = items_count - 1;for(i = 0; i < end; i++){if(op_items[i].op_type == LPAREN){innerest = i;}//}//end i forreturn innerest;}///*** * @function :查找最里面的左括号。如果找到返回相应op_items下标。否则返回-1 * */ int find_inner_rparen(){int i = 0;int end = items_count - 1;for(i = 0; i < end; i++){if(op_items[i].op_type == RPAREN){return i;}//}//end i forreturn -1;}///*** * @function : 带优先级的表达式计算。括号，最优，乘除次之。接着加减。 * 一种思路就是，先扫描整个op_items 数组。找到优先级最高的计算。 * 然后将后面前移。 */double cal_exp_with_pri(){int index = 0;int rparen_index = 0;int i = 0;int op_index = 0; // 操作符下标 int result_index = 0;// 运算结果存储位置下标。int steps = 0; // 前移项数int end = 0;double result = 0;op_index = 0;end = operator_count + 1; //因为把计算部分统一移到了for的开头。所以要多一次循环。for(i = 0; i < end; i++){if(op_index > 0){result = cal(op_items[op_index - 1].operand,op_items[op_index].operator,op_items[op_index+1].operand); op_items[result_index].operand = result;op_items[result_index].op_type = OPERAND;move_items(result_index+1,steps);items_count -= steps;#ifdef BANXI_DEBUGprintf("/n=====CURRENT OP_ITEMS=============================/n");print_data();#endif}//end index = find_inner_lparen();rparen_index = find_inner_rparen(); // 处理外围大括号情况。if((index > -1) && (rparen_index - index)== 4){op_index = index + 2;result_index = index;steps = 4;continue; //}//end if index = find_muldiv();if(index > 0){op_index = index;result_index = index -1;steps = 2;continue; //}//index = find_operator(); // 执行到这里说明已经没有括号，乘除了。返回的是加减法操作符。if(index > 0){op_index = index;result_index = index -1;steps = 2;continue;}//end if }//end operator_count for if(items_count > 2)return op_items[1].operand; // 出现外围的括号无法依靠上面的处理完全消除(12) 这样的情况return op_items[0].operand;}//endint main(void){double result = 0.0;scan_input();#ifdef BANXI_DEBUGprintf("/n=======CURRENT OP_ITEMS================================/n");print_data();#endif // BANXI_DEBUGresult = cal_exp_with_pri();printf("/t %.2lf/n",result);printf("calculate takes time %ld seconds /n",clock()/CLOCKS_PER_SEC);return EXIT_SUCCESS;}//end /***** 运行及测试结果： 调试模式输出如下： banxi1988@banxi:~/cpp/c/exp_cal$ gcc -g -o simple simple.c banxi1988@banxi:~/cpp/c/exp_cal$ lsexp_calSession.vim simple01.c simple.c test_simple.txtsimple simple01_with_pri.c simple_with_pri.cbanxi1988@banxi:~/cpp/c/exp_cal$ vi test_simple.txt banxi1988@banxi:~/cpp/c/exp_cal$ cat test_simple.txt | ./simple 0(1(22.0000003+41.0000005)6*73.0000008+93.00000010)0(13.0000002*33.0000004+53.0000006)==============================================================0(19.0000002+33.0000004)==============================================================0(112.0000002)============================================================== 0.00calculate takes time 0 seconds banxi1988@banxi:~/cpp/c/exp_cal$ gcc -g -o simple simple.c banxi1988@banxi:~/cpp/c/exp_cal$ cat test_simple.txt | ./simple =====CURRENT OP_ITEMS ====================================0(1(22.0000003+41.0000005)6*73.0000008+93.00000010)=====CURRENT OP_ITEMS ====================================0(13.0000002*33.0000004+53.0000006)=====CURRENT OP_ITEMS ====================================0(19.0000002+33.0000004)=====CURRENT OP_ITEMS ====================================0(112.0000002) 12.00calculate takes time 0 seconds banxi1988@banxi:~/cpp/c/exp_cal$ gcc -g -o simple simple.c banxi1988@banxi:~/cpp/c/exp_cal$ cat test_simple.txt ((2+1)*3+3)banxi1988@banxi:~/cpp/c/exp_cal$ cat test_simple.txt | ./simple =======CURRENT OP_ITEMS================================0(1(22.0000003+41.0000005)6*73.0000008+93.00000010)=====CURRENT OP_ITEMS=============================0(13.0000002*33.0000004+53.0000006)=====CURRENT OP_ITEMS=============================0(19.0000002+33.0000004)=====CURRENT OP_ITEMS=============================0(112.0000002) 12.00calculate takes time 0 seconds banxi1988@banxi:~/cpp/c/exp_cal$ gcc -g -o simple simple.c banxi1988@banxi:~/cpp/c/exp_cal$ cat test_simple.txt | ./simple 12.00calculate takes time 0 seconds banxi1988@banxi:~/cpp/c/exp_cal$ * */