[c]
//
#include
#include
//运算符栈的长度
#define OPSTACK_LENGTH 5
//操作数栈的长度
#define NUMSTACK_LENGTH 100
//输入串的最大长度
#define MAX_STRING_LENGTH 100

//运算符结构体
struct operatorStruct
{
//运算符名称
char name;
//优先级
int priority;
//目数，即操作数个数，例如单目运算符为1，双目运算符2
int opnum;
};

typedef struct operatorStruct OPERATOR;

//运算符栈
OPERATOR opStack[OPSTACK_LENGTH];
//运算符栈顶指针
int opStackTop = -1;
//操作数栈
double numStack[NUMSTACK_LENGTH];
//操作数栈顶指针
int numStackTop = -1;

//获取一个字符所代表的运算符的优先级
int getPriority(char name)
{
if (name == '(' || name == ')')
{
return 0;
}
if (name == '!')
{
return 3;
}
if (name == '*' || name == '/')
{
return 2;
}
if (name == '+' || name == '-')
{
return 1;
}
exit(1);
}
//获取一个字符所代表的运算符的目数
int getOpNum(char name)
{
if (name == '*' || name == '/' || name == '+' || name == '-')
{
return 2;
}
if (name == '!')
{
return 1;
}
if (name == '(' || name == ')')
{
return 0;
}
exit(1);
}

//运算符压栈
void pushOperator(OPERATOR op)
{
if (opStackTop < OPSTACK_LENGTH - 1)
{
opStack[++opStackTop] = op;
}
else
{
exit(1);
}
}
//运算符出栈
OPERATOR popOperator()
{
if (opStackTop >= 0)
{
return opStack[opStackTop--];
}
else
{
exit(1);
}
}
//操作数压栈
void pushNumber(double num)
{
if (numStackTop < NUMSTACK_LENGTH - 1)
{
numStack[++numStackTop] = num;
}
else
{
exit(1);
}
}
//操作数出栈
double popNumber()
{
if (numStackTop >= 0)
{
return numStack[numStackTop--];
}
else
{
exit(1);
}
}
//将输入字符串中的以0-9开头、到下一个运算符结束的一段转化为浮点型
//i加上浮点型对应的字符串的长度
double getNumFromString(char *s, int *i)
{
int j = 0;
static char numstr[MAX_STRING_LENGTH];
while ((*s) >= '0' && *s <= '9')
{
numstr[j++] = (*s);
s++;
}
if ((*s) == '.')
{
numstr[j++] = (*s);
s++;
while ((*s) >= '0' && *s <= '9')
{
numstr[j++] = (*s);
s++;
}
}
(*i) = (*i) + j;
numstr[j] = '\0';
return atof(numstr);
}

//从操作数栈中弹出两个操作数，完成一次双目运算
double opertate2Num(OPERATOR op)
{
double num2 = popNumber();
double num1 = popNumber();
if (op.name == '+')
{
return num1 + num2;
}
if (op.name == '-')
{
return num1 - num2;
}
if (op.name == '*')
{
return num1 * num2;
}
if (op.name == '/')
{
return num1 + num2;
}
exit(1);
}
//从操作数栈中弹出一个操作数，完成一次单目运算
double opertate1Num(OPERATOR op)
{
double num = popNumber();
if (op.name == '!')
{
double result = 1;
while (num > 1)
{
result *= num;
num--;
}
return result;
}
exit(1);
}
//完成一次运算 www.2cto.com
double operate(OPERATOR op)
{
if (op.opnum == 1)
{
return opertate1Num(op);
}
else if (op.opnum == 2)
{
return opertate2Num(op);
}
exit(1);
}

int main()
{
char string[MAX_STRING_LENGTH];//表达式的输入串
int i;
OPERATOR op, topOp;//op为从当前输入串中提取的一个运算符，topOp为运算符栈栈顶的运算符

topOp.name = '#';
topOp.priority = 0;
topOp.opnum = 0;
pushOperator(topOp);//压入#作为初始运算符

scanf("%s", string);