//中缀表达式转后缀表达式
//(5*(((9+8)*(4*6))+7))
//？？？'*'与'+'没有按预想的输出
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
typedef struct
{
    char *base;
    char *top;
}SqStack;

void STACKpop(SqStack *S)
{
    

    printf("%c ",*--S->top);

}

void STACKpush(SqStack *S,char *c)
{
    *S->top=*c;
    S->top++;
}
main(int argc,char *argv[])
{
    char *a=argv[1];
    int i,N;
    SqStack S;
    N=strlen(a);
    S.base=(char *)malloc(N*sizeof(char));
    S.top=S.base;
    for (i=0;i<N;i++)
    {
        if (a[i]==')')
           STACKpop(&S);
        if ((a[i]=='+')||(a[i]=='*'))
           STACKpush(&S,&a[i]);
        if((a[i]>='0')&&(a[i])<='9')
           printf("%c ",a[i]);
    }
    printf("\n");
}

[[it] 本帖最后由 liyanhong 于 2008-10-25 19:15 编辑 [/it]]

这个小问题找了我好长时间……

恩

//后缀表达式求值
////598+46**7+*    结果不是2075  
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
typedef struct
{
    int *top;
    int *base;
}SqStack;

int STACKpop(SqStack *S)
{
    
    return *(--S->top);

}
void STACKpush(SqStack *S,int d)
{
    *S->top=d;
    S->top++;
}

main(int argc,char *argv[])
{
    char *a=argv[1];
    int i,N;
    SqStack S;
    N=strlen(a);
    S.base=(int *)malloc(N*sizeof(int));
    S.top=S.base;
    for(i=0;i<N;i++)
    {
        if(a[i]=='+')
      STACKpush(&S,(STACKpop(&S)+STACKpop(&S)));
       
       if(a[i]=='*')
           STACKpush(&S,(STACKpop(&S)*STACKpop(&S)));
        
       if((a[i]>='0')&&(a[i]<='9'));
     STACKpush(&S,a[i]-'0');
        
    }  
    printf("%d \n",STACKpop(&S));
}

[[it] 本帖最后由 liyanhong 于 2008-10-25 20:49 编辑 [/it]]