问题遇到的现象和发生背景
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#define ERROR 0
#define TRUE 1
#define FALSE 0
#define OK 1
#define max 100
typedef struct STACK
{
int date[max];
int top;
}stack;
stack* initStack()//创建线性存储结构的栈结构
{
stack* s;
s = (stack*)malloc(sizeof(stack));
s->top = -1;
return s;
}
void destroyStack(stack* s)//摧毁栈
{
free(s);
printf("清空成功！");
}
stack* clearStack(stack* s)//清空栈
{
s->top = -1;
printf("清理成功！");
return s;
}
int stackEmpty(stack* s)//判断链表是否为空
{
if (s->top >= max)
return FALSE;
if (s->top == -1)
return TRUE;
else
return FALSE;
}
int getTop(stack* s, int* e)//获取栈顶元素
{
if (s->top > -1 && s->top < max)
{
*e = s->date[s->top];
return OK;
}
else
return FALSE;
}
int push(stack* s, int e)//进栈操作
{
if (s->top == max - 1)
{
return ERROR;
}
s->top++;
s->date[s->top] = e;
return OK;
}
int pop(stack* s, int* e)//出栈操作
{
if (s->top == -1)
return ERROR;
*e = s->date[s->top];
s->top--;
return OK;
}
int stackLength(stack* s)//获取链表长度
{
return s->top++;
}
int main()
{
char str[100];
int i = 0, a, b, c, sum;
stack* s;
s = initStack();
gets_s(str);
while (str[i] != '\0')
{
if (str[i] >= '0' && str[i] <= '9')//1. 将字符串型的数字转换为整型（或者浮点型）。
{
sum = 0;
sum = str[i] - '0';
i++;
while (str[i] >= '0' && str[i] <= '9')
{
sum = sum * 10 + str[i] - '0';
printf("%d\n",sum);
i++;
}
push(s, sum);
}
if (str[i] == '+' || str[i] == '-' || str[i] == '' || str[i] == '/')//3. 遇到运算符（#），则将栈中的前两个数字（a和b）出栈，并将b#a的值压入栈中。
{
pop(s, &a);
pop(s, &b);
if (str[i] == '+')
push(s, b + a);
else if (str[i] == '-')
push(s, b - a);
else if (str[i] == '')
push(s, b * a);
else if (str[i] == '/')
push(s, b / a);
else
printf("error\n");
}
i++;
}
pop(s, &c);//栈顶的数字出栈
printf("%s=%d\n", str, c);
return 0;
}
这是怎么变的 sum是整形 我不应该 用sum等于 str{i}-0 就行了吗 为什么sum*10还要加str
