求助大佬问题，谢谢：C语言，栈，有一块代码计算1+1，压进栈的时候我输出为什么是49

mogur

代码如下，代码我在别人中缀转后缀的代码上加了计算后缀的代码，不会调试我就在可能有数值改动的地方写个printf输出的数看看，

不知道为什么这个代码（188行）这个地方要减去48才可以算对
（不减去48的话在那几行增加个输出的代码的话，如果是1就会输出49）

[C] 纯文本查看 复制代码

Push_Stack(&s2,GetTop(&s1)-48);
                     Pop_Stack(&s1);

也没有大佬可以解答，谢谢。

[C] 纯文本查看 复制代码

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


//----------------------栈的实现--------------------------

typedef char ElemType;

#define STACKSIZE 50

typedef struct Stack   //定义栈的结构体
{
    ElemType  *data;
    ElemType  top;
    int stacksize;
}*StackPtr;

int Init_Stack(StackPtr stack)          //初始化
{
    if(stack == NULL)  exit(0);
    stack->data = (ElemType*)malloc(sizeof(ElemType)*STACKSIZE);
    stack->stacksize = STACKSIZE;
    stack->top = 0;
    return true;
}

int Push_Stack(StackPtr stack,ElemType val)      //入栈
{
    if(stack == NULL)  exit(0);
    stack->data[stack->top] = val;
    stack->top++;
    return true;
}

int Pop_Stack(StackPtr stack)          //出栈
{
    if(stack==NULL)  exit(0);
    stack->top--;
    return true;
}

int GetTop(StackPtr stack)          //获取栈顶
{
    if(stack == NULL)  exit(0);
    return stack->data[stack->top-1];
}

int Empty_Stack(StackPtr stack)          //判空
{
    if(stack == NULL)  exit(0);
    if(stack->top == 0)
    {
        return true;
    }
    else
    {
        return false;
    }
}

static int Apply_Stack(StackPtr stack)             //申请空间
{
    ElemType* p = (ElemType*)malloc(sizeof(ElemType)*stack->stacksize*2);
    if(p == NULL)  return false;
    for(int i=0;i<stack->top;i++)
    {
        p[i] = stack->data[i];
    }
    free(stack->data);
    stack->data = p;
    p = NULL;
    stack->stacksize*=2;
    return true;
}

static int Full_Stack(StackPtr stack)         //判满
{
    if(stack->stacksize == stack->top&&stack->top != 0)
    {
        Apply_Stack(stack);
        return true;
    }
    else
    {
        return false;
    }
}

int Clear_Stack(StackPtr stack)            //清空
{
    if(stack == NULL)  exit(0);
    stack->top = 0;
    return true;
}

int Destroy_Stack(StackPtr stack)          //销毁
{
    if(stack == NULL)  exit(0);
    free(stack->data);
    stack->data = NULL;
    stack->top = 0;
    stack->stacksize = 0;
    return true;
}

//------------------------------------------------------

//--------------------逆波兰式实现------------------------

int OperPre(char oper)         //处理优先级
{
    int tmp;
    switch(oper)
    {
    case '+':
    case '-':
        tmp = 1; break;
    case '*':
    case '/':
        tmp = 2;break;
    default:
        tmp = 3;
    }
    return tmp;
}

void RPNotation(char *buff)         //逆波兰式实现
{
    Stack s1,s2;                    //定义两个栈是s1，s2
    Init_Stack(&s1);                //初始化栈
    Init_Stack(&s2);

    while(1)                        //逆波兰实现
    {
        if(isdigit(*buff))           //将数字存入s2
        {
            Push_Stack(&s2,*buff);

        }

        if(!isdigit(*buff))
        {
            while(OperPre(*buff)<=OperPre(GetTop(&s1)))
            {
                if(s1.top == 0) break;          //若s1没数据则直接入栈
                if(GetTop(&s1)=='(')  break;    //处理‘（’问题
                Push_Stack(&s2,GetTop(&s1));    //将s1的栈顶压入s2
                Pop_Stack(&s1);                 //s1出栈
            }

            if(*buff != ')')                    //若出现‘）’则交给下一个if条件语句处理
            {
                Push_Stack(&s1,*buff);
            }
        }

        if(*buff == ')')                   //处理‘（’，‘）’问题
        {
            while(GetTop(&s1) != '(')      //将‘（’之前的运算符压入s2中
            {
                Push_Stack(&s2,GetTop(&s1));
                Pop_Stack(&s1);
            }
            Pop_Stack(&s1);                 //运算符处理完后，将‘（’出栈
        }

        buff++;
        if(*buff == '\n') break;            //结束条件
    }



   while(s2.top != 0)                     //将（栈）s2压入（栈）s1
        {
                Push_Stack(&s1,GetTop(&s2));
                Pop_Stack(&s2);
        }
    //==============================================================
      char op;

    while(s1.top != 0)

        {        op=GetTop(&s1);
                  int num1,num2,sum;
                 if(isdigit(op) )
                 {
                     Push_Stack(&s2,GetTop(&s1)-48);
                     Pop_Stack(&s1);



                 }
                 else
                 {
                   if  (op=='+')
               {
                    Pop_Stack(&s1);
                    num1 = GetTop(&s2);
                    Pop_Stack(&s2);

                    num2 = GetTop(&s2);
                    Pop_Stack(&s2);
                    sum=num1+num2;
                    Push_Stack(&s2,sum);


                }

               else if (op=='-')
               {
                  Pop_Stack(&s1);
                    num1 = GetTop(&s2);
                    Pop_Stack(&s2);
                    num2 = GetTop(&s2);
                    Pop_Stack(&s2);
                    sum=num1-num2;
                    Push_Stack(&s2,sum);



               }


               else if (op=='*')
               {
                    Pop_Stack(&s1);
                    num1 = GetTop(&s2);
                    Pop_Stack(&s2);
                    num2 = GetTop(&s2);
                    Pop_Stack(&s2);
                    sum=num1*num2;
                    Push_Stack(&s2,sum);


               }


              else if (op=='/')
               {
                    Pop_Stack(&s1);
                    num1 = GetTop(&s2);
                    Pop_Stack(&s2);
                    num2 = GetTop(&s2);
                    Pop_Stack(&s2);
                    sum=num1/num2;
                    Push_Stack(&s2,sum);


                }
                 }



        }

        while(s2.top != 0)
         {
        printf("%d",(int)GetTop(&s2));
        Pop_Stack(&s2);

         }

}

int main()
{
    char buff[128];
    printf("中缀表达式：\n");
    fgets(buff,127,stdin);
    RPNotation(buff);
    return 0;
}

DEATHTOUCH

学数据结构？

[C] 纯文本查看 复制代码

typedef char ElemType;
 
#define STACKSIZE 50

typedef struct Stack
{
    ElemType  *data;
    ElemType  top;
    int stacksize;
}*StackPtr;

这里你的类型是 char，你读取了字符串里的数字字符放进栈里，那栈里面的 '8' 其实转成 int 是 56
所以要减掉 48 才是真正的数字 8

nug

参考ascii对照表，一个是字符 1，一个是字符对应的整数

字符减去48 也是一种把字符转换整数的一种常见做法~