中缀和前缀 [英] infix and prefix
本文介绍了中缀和前缀的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
亲爱的论坛，
我的代码有问题，输出不适用。我在我的智慧结束，我需要你明智的大师的建议。很多人
Dear forums,
i have problems with my code, where the output does not apppear. I am at my wits end, i need your advise from the wise gurus out ther. Thx a bunch
#include
#include
#include
#define MAX 10
#define EMPTY -1
struct stack
{
char data[MAX];
int top;
};
int isempty(struct stack *s)
{
return (s->top == EMPTY) ? 1 : 0;
}
void emptystack(struct stack* s)
{
s->top=EMPTY;
}
void push(struct stack* s,int item)
{
if(s->top == (MAX-1))
{
printf("\nSTACK FULL");
}
else
{
++s->top;
s->data[s->top]=item;
}
}
char pop(struct stack* s)
{
char ret=(char)EMPTY;
if(!isempty(s))
{
ret= s->data[s->top];
--s->top;
}
return ret;
}
void display(struct stack s)
{
while(s.top != EMPTY)
{
printf("\n%d",s.data[s.top]);
s.top--;
}
}
int isoperator(char e)
{
if(e == ''+'' || e == ''-'' || e == ''*'' || e == ''/'' || e == ''%'')
return 1;
else
return 0;
}
int priority(char e)
{
int pri = 0;
if(e == ''*'' || e == ''/'' || e ==''%'')
pri = 2;
else
{
if(e == ''+'' || e == ''-'')
pri = 1;
}
return pri;
}
void infix2postfix(char* infix, char * postfix, int insertspace)
{
char *i,*p;
struct stack X;
char n1;
emptystack(&X);
i = &infix[0];
p = &postfix[0];
while(*i)
{
while(*i == '' '' || *i == ''\t'')
{
i++;
}
if( isdigit(*i) || isalpha(*i) )
{
while( isdigit(*i) || isalpha(*i))
{
*p = *i;
p++;
i++;
}
/*SPACE CODE*/
if(insertspace)
{
*p = '' '';
p++;
}
/*END SPACE CODE*/
}
if( *i == ''('' )
{
push(&X,*i);
i++;
}
if( *i == '')'')
{
n1 = pop(&X);
while( n1 != ''('' )
{
*p = n1;
p++;
/*SPACE CODE*/
if(insertspace)
{
*p = '' '';
p++;
}
/*END SPACE CODE*/
n1 = pop(&X);
}
i++;
}
if( isoperator(*i) )
{
if(isempty(&X))
push(&X,*i);
else
{
n1 = pop(&X);
while(priority(n1) >= priority(*i))
{
*p = n1;
p++;
/*SPACE CODE*/
if(insertspace)
{
*p = '' '';
p++;
}
/*END SPACE CODE*/
n1 = pop(&X);
}
push(&X,n1);
push(&X,*i);
}
i++;
}
}
while(!isempty(&X))
{
n1 = pop(&X);
*p = n1;
p++;
/*SPACE CODE*/
if(insertspace)
{
*p = '' '';
p++;
}
/*END SPACE CODE*/
}
*p = ''\0'';
}
int evaluate(char eva)
{
char *p;
stack stack;
int op1,op2,result;
isempty(&stack);
p = &eva;
while(*p != ''\0'')
{
/* removes tabs and spaces */
while(*p == '' '' || *p == ''\t'')
{
p++;
}
/* if is digit */
if(isdigit(*p))
{
push(&stack,*p - 48);
}
else
{
/* it is an operator */
op1 = pop(&stack);
op2 = pop(&stack);
switch(*p)
{
case ''+'':
result = op2 + op1;
break;
case ''-'':
result = op2 - op1;
break;
case ''/'':
result = op2 / op1;
break;
case ''*'':
result = op2 * op1;
break;
default:
printf("\nInvalid Operator");
return 0;
}
push(&stack,result);
}
p++;
}
result = pop(&stack);
return result;
}
int main()
{
char in[50],post[50];
strcpy(&post[0],"");
printf("Enter Infix Expression : ");
fflush(stdin);
gets(in);
infix2postfix(&in[0],&post[0],1);
printf("Postfix Expression is : %s\n",&post[0]);
printf("Equal: %d\n\n",evaluate(&post));
return 0;
}
推荐答案
尝试以下代码（请注意，我的修复程序远非优雅......）。 br />
问题（您可以使用调试器轻松发现它们）：
- 您调用
isempty在evaluate函数的开头，而不是emptystack。- 你忘了检查字符串终止符在评估函数中跳过空格和制表符之后。
Try the following code (be aware, my fixes are far from being elegant...).
The problems (you could easily discover them with a debugger):
- You called
isemptyinstead ofemptystackat the begin ofevaluatefunction.- You forgot to check for string terminator after skipping spaces and tabulators in the evaluate function.
#define MAX 10
#define EMPTY -1
struct stack
{
char data[MAX];
int top;
};
int isempty(struct stack *s)
{
return (s->top == EMPTY) ? 1 : 0;
}
void emptystack(struct stack* s)
{
s->top=EMPTY;
}
void push(struct stack* s,int item)
{
if(s->top == (MAX-1))
{
printf("\nSTACK FULL");
}
else
{
++s->top;
s->data[s->top]=item;
}
}
char pop(struct stack* s)
{
char ret=(char)EMPTY;
if(!isempty(s))
{
ret= s->data[s->top];
--s->top;
}
return ret;
}
void display(struct stack s)
{
while(s.top != EMPTY)
{
printf("\n%d",s.data[s.top]);
s.top--;
}
}
int isoperator(char e)
{
if(e == '+' || e == '-' || e == '*' || e == '/' || e == '%')
return 1;
else
return 0;
}
int priority(char e)
{
int pri = 0;
if(e == '*' || e == '/' || e =='%')
pri = 2;
else
{
if(e == '+' || e == '-')
pri = 1;
}
return pri;
}
void infix2postfix(char* infix, char * postfix, int insertspace)
{
char *i,*p;
struct stack X;
char n1;
emptystack(&X);
i = infix;
p = postfix;
while(*i)
{
while(*i == ' ' || *i == '\t')
{
i++;
}
if( isdigit(*i) || isalpha(*i) )
{
while( isdigit(*i) || isalpha(*i))
{
*p = *i;
p++;
i++;
}
/*SPACE CODE*/
if(insertspace)
{
*p = ' ';
p++;
}
/*END SPACE CODE*/
}
if( *i == '(' )
{
push(&X,*i);
i++;
}
if( *i == ')')
{
n1 = pop(&X);
while( n1 != '(' )
{
*p = n1;
p++;
/*SPACE CODE*/
if(insertspace)
{
*p = ' ';
p++;
}
/*END SPACE CODE*/
n1 = pop(&X);
}
i++;
}
if( isoperator(*i) )
{
if(isempty(&X))
push(&X,*i);
else
{
n1 = pop(&X);
while(priority(n1) >= priority(*i))
{
*p = n1;
p++;
/*SPACE CODE*/
if(insertspace)
{
*p = ' ';
p++;
}
/*END SPACE CODE*/
n1 = pop(&X);
}
push(&X,n1);
push(&X,*i);
}
i++;
}
}
while(!isempty(&X))
{
n1 = pop(&X);
*p = n1;
p++;
/*SPACE CODE*/
if(insertspace)
{
*p = ' ';
p++;
}
/*END SPACE CODE*/
}
*p = '\0';
}
int evaluate(char *eva)
{
char *p;
stack stack;
int op1,op2,result;
emptystack(&stack);
p = eva;
while(*p != '\0')
{
/* removes tabs and spaces */
while(*p == ' ' || *p == '\t')
{
p++;
}
if (*p == '\0') break;
/* if is digit */
if(isdigit(*p))
{
push(&stack,*p - 48);
}
else
{
/* it is an operator */
op1 = pop(&stack);
op2 = pop(&stack);
switch(*p)
{
case '+':
result = op2 + op1;
break;
case '-':
result = op2 - op1;
break;
case '/':
result = op2 / op1;
break;
case '*':
result = op2 * op1;
break;
default:
printf("\nInvalid Operator");
return 0;
}
push(&stack,result);
}
p++;
}
result = pop(&stack);
return result;
}
int main()
{
char in[50],post[50];
strcpy(post,"");
printf("Enter Infix Expression : ");
fflush(stdin);
gets(in);
infix2postfix(in,post,1);
printf("Postfix Expression is : %s\n",post);
printf("Equal: %d\n\n",evaluate(post));
return 0;
}
非常感谢你的CPallini，你帮我很多了=）
Thank you so much CPallini, you helped me a lot =)
例如你的代码
3 + 12 * 2/2
后缀表达式为：3 12 2 * 2 / +
等于：3
3等于是错误
等于必须15
for example your code
3+12*2/2
postfix expression is : 3 12 2 * 2 / +
equal : 3
3 for equal is wrong
equal must 15
这篇关于中缀和前缀的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
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