C语言表达式求值(带详细注释).doc

/*表达式求值，输入一个表达式，如1+2*3#，程序可计算出结果为7支持以下符号：+ - * / ( ) .可以计算整数、小数其中表示次方，25表示2的5次方*/*头文件*/#include #include #include #include #include /*宏定义*/#define INIT_STACK_SIZE 100#define SET_NUM 8#define N 100/*字符优先级表*/unsigned char priorSET_NUMSET_NUM = /* + - * / ( ) # */ /*+*/, , , , , , , , , , , , , , , , , , , , , , , , , /*(*/, , , , , =, , , , , , , , , , /*#*/, , , , , , =, , , , , , , ; unsigned char priorSetSET_NUM = +, -, *, /, (, ), #, ;/*结构体定义，这是用来存放字符的栈*/typedef structchar *base;char *top;int stacksize; SqStackC;/*结构体定义，这是用来存放数字的栈*/typedef structdouble *base;double *top;int stacksize; SqStackN;void initStackN(SqStackN &);void initStackC(SqStackC &);void pushN(SqStackN &, double);void pushC(SqStackN &, double);void popN(SqStackN &, double &);void popC(SqStackN &, double &);double calculate(double, char, double);int findInSet(char);char compare(char, char);void getSolution();/*主函数*/void main()getSolution();/*初始化数字栈*/void initStackN(SqStackN &S)S.base = (double*) malloc(INIT_STACK_SIZE * sizeof(double);S.top = S.base;S.stacksize = INIT_STACK_SIZE;/*初始化字符栈*/void initStackC(SqStackC &S)S.base = (char*) malloc(INIT_STACK_SIZE * sizeof(char);S.top = S.base;S.stacksize = INIT_STACK_SIZE;/*向数字栈中存放数字*/void pushN(SqStackN &S, double x)if (S.top-S.base = S.stacksize) return;*(S.top+) = x;/*向字符栈中存放字符*/void pushC(SqStackC &S, char x)if (S.top - S.base = S.stacksize) return;*(S.top+) = x;/*从数字栈中取出数字*/void popN(SqStackN &S, double &x)if (S.top=S.base) return;x = *(-S.top);/*从字符栈中取出字符*/void popC(SqStackC &S, char &x)if (S.top = S.base) return;x = *(-S.top);/*这个函数返回a operation b的值。假如operation为+，则返回a+b的值*/double calculate(double a, char operation, double b)/*判断operation，返回对应的计算结果*/switch (operation) case +:return a + b;case -:return b - a;case *:return a * b;case /:return b / a;case :return pow(b, a);default:return 0;/*查找字符c在priorSet中的什么位置*/*priorSet是所支持的所有字符的集合*/int findInSet(char c)int i;for (i = 0; i = 0 & tempi = 0 & tempi :/*取出字符栈顶运算符，和数字栈顶的两个数字*/popC(OPTR, operation);popN(OPND, a);popN(OPND, b);/*计算结果*/sum = calculate(a, operation, b);/*结果入栈*/pushN(OPND, sum);break;/*如果优先级小*/case :/*入栈，继续循环*/pushC(OPTR, tempi);i+;break;/*如果优先级相等*/case =:/*优先级相等只