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以下C代码中的item=infix_exp[i++];是什么意思？第21行。它用于后缀转换。据我所知，这里i是数组索引。但为什么它没有任何循环递增？在下面的C代码中，“item = infix_exp [i ++];”是什么意思？
这是代码
#include<stdio.h>
#include<conio.h>
#define SIZE 100
int top = -1;
char stack[SIZE];
void push(char item);
char pop();
int is_operator(char symbol);
int precedence(char symbol);
void main()
{
int i;
int j;
char infix_exp[SIZE], postfix_exp[SIZE];
char item;
char x;
printf("\nEnter Infix expression in parentheses: \n");
gets(infix_exp);
i=0;
j=0;
item=infix_exp[i++]; /* HERE */
while(item != '\0')
{
if(item == '(')
{
push(item);
}
else if((item >= 'A' && item <= 'Z') ||
(item >= 'a' && item <= 'z'))
{
postfix_exp[j++] = item;
}
else if(is_operator(item) == 1)
{
x=pop();
while(is_operator(x) == 1 && precedence(x)
>= precedence(item))
{
postfix_exp[j++] = x;
x = pop();
}
push(x);
push(item);
}
else if(item == ')')
{
x = pop();
while(x != '(')
{
postfix_exp[j++] = x;
x = pop();
}
}
else
{
printf("\nInvalid Arithmetic Expression.\n");
getch();
}
item = infix_exp[i++];
}
postfix_exp[j++] = '\0';
printf("\nArithmetic expression in Postfix notation: ");
puts(postfix_exp);
getch();
}
void push(char item)
{
if(top >= SIZE-1)
{
printf("\nStack Overflow. Push not possible.\n");
}
else
{
top = top+1;
stack[top] = item;
}
}
char pop()
{
char item = NULL;
if(top <= -1)
{
printf("\nStack Underflow. Pop not possible.\n");
}
else
{
item = stack[top];
stack[top] = NULL;
top = top-1;
}
return(item);
}
int is_operator(char symbol)
{
if(symbol == '^' || symbol == '*' || symbol == '/' ||
symbol == '+' || symbol == '-')
{
return 1;
}
else
{
return 0;
}
}
int precedence(char symbol)
{
if(symbol == '^')
{
return(3);
}
else if(symbol == '*' || symbol == '/')
{
return(2);
}
else if(symbol == '+' || symbol == '-')
{
return(1);
}
else
{
return(0);
}
}