Question

用堆栈来实现四则运算.（C语言）

数据结构的问题

Answer 1

here it is，1个半小时才搞定的，好辛苦，呵呵：

#include <stdio.h>
#include <ctype.h>
#include <stdlib.h>

double var[1024];
char opt[1024];
double* sp1 = var;
char* sp2 = opt;

#define PUSH(s, v) (*s++ = v)
#define POP(s) (*--s)
#define EMPTY1() (sp1 == var)
#define EMPTY2() (sp2 == opt)
#define TOP(s) (s[-1])

#define isopt1(c) (c == '+' || c == '-')
#define isopt2(c) (c == '*' || c == '/')
#define isopt(c) (isopt1(c) || isopt2(c))

#define isbkt(c) (c == '(')
double docal(double lhs, double rhs, char op)
{
switch(op) {
case '+':return lhs + rhs;
case '-':return lhs - rhs;
case '*':return lhs * rhs;
}
if(op == '/' && rhs == 0) {
printf("divided by zero encountered.\n");
exit(0);
}
return lhs / rhs;
}

void eval1(char* expr)
{
double t1, t2;
while(sp1 - var > 1 && !EMPTY2() && TOP(sp2) != '(') {
t2 = POP(sp1);
t1 = POP(sp1);
PUSH(sp1, docal(t1, t2, POP(sp2)));
}
if(expr)
PUSH(sp2, *expr);
}

void eval2(char* expr)
{
double t1, t2;
while(TOP(sp2) != '(') {
t2 = POP(sp1);
t1 = POP(sp1);
PUSH(sp1, docal(t1, t2, POP(sp2)));
}
POP(sp2);
}

double parse(char* expr)
{
double val;
while(*expr) {
if(isdigit(*expr)) {
sscanf(expr, "%lf", &val);
PUSH(sp1, val);
while(isdigit(*expr) || *expr == '.')
++expr;
--expr;
} else if(isopt(*expr)) {
if(EMPTY2() || (isopt1(TOP(sp2)) && !isopt1(*expr)) || isbkt(TOP(sp2)))
PUSH(sp2, *expr);
else if(isopt(TOP(sp2)))
eval1(expr);
} else if(*expr == '(') {
PUSH(sp2, *expr);
}

Answer 2

float compvalue(char postexp[])
{

float d;
char ch;
int i=0;
st.top=-1;
ch=postexp[i];i++;
while(ch!='\0')
{
switch(ch)
{
case'+':
st.data[st.top-1]=st.data[st.top-1]+st.data[st.top];
st.top--;
break;

case'-':
st.data[st.top-1]=st.data[st.top-1]-st.data[st.top];
st.top--;
break;
case'*':
st.data[st.top-1]=st.data[st.top-1]*st.data[st.top];
st.top--;
break;
case'/':
if(st.data[st.top]!=0)
st.data[st.top-1]=st.data[st.top-1]/st.data[st.top];
else
{
printf("\n\t除零错误！\n");
exit(0);
}
st.top--;
break;

default:
d=0;
while(ch>='0'&&ch<='9')
{
d=10*d+ch-'0';
ch=postexp[i];
i++;
}
st.top++;
st.data[st.top]=d;
}
ch=postexp[i];
i++;
}
return st.data[st.top];

}

Answer 3

好像不是很容易！我再编下看看！