Char array in multiple test cases - c

Question

#include<stdio.h>
#include<stdlib.h>
#include<ctype.h>
#include<string.h>
#define SIZE 100
char stack[SIZE];
int top = -1;
void push(char item)
{
if(top >= SIZE-1)
{
printf("\nStack Overflow.");
}
else
{
top = top+1;
stack[top] = item;
}
}
char pop()
{
char item;
if(top <0)
{
printf("stack under flow: invalid infix expression");
}
else
{
item = stack[top];
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(5);
}
else if(symbol == '/')
{
return(4);
}
else if(symbol == '*')
{
return(3);
}
else if(symbol == '+' )
{
return(2);
}
else if(symbol == '-' )
{
return(1);
}
else
{
return(0);
}
}
void InfixToPostfix(char infix_exp[], char postfix_exp[])
{
int i, j;
char item;
char x;
push('(');
strcat(infix_exp,")");
i=0;
j=0;
item=infix_exp[i];
while(item != '\0')
{
if(item == '(')
{
push(item);
}
else if( isdigit(item) || isalpha(item))
{
postfix_exp[j] = item; /* add operand symbol to postfix expr */
j++;
}
else if(is_operator(item) == 1) /* means symbol is operator */
{
x=pop();
while(is_operator(x) == 1 && precedence(x)>= precedence(item))
{
postfix_exp[j] = x; /* so pop all higher precendence operator and */
j++;
x = pop(); /* add them to postfix expresion */
}
push(x);
push(item);
}
else if(item == ')')
{
x = pop();
while(x != '(')
{
postfix_exp[j] = x;
j++;
x = pop();
}
}
else
{
printf("\nInvalid infix Expression.\n");
}
i++;
item = infix_exp[i];
}
if(top>0)
{
printf("\nInvalid infix Expression.\n");
}
if(top>0)
{
printf("\nInvalid infix Expression.\n");
}
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
char infix[SIZE], postfix[SIZE]; /* declare infix string and postfix string */
scanf("\n%[^\n]s",infix);
InfixToPostfix(infix,postfix); /* call to convert */
//printf("Postfix Expression: ");
printf("%s\n",postfix);
/* print postfix expression */
}
return 0;
}
This is the code to convert infix to postfix expression I am using multiple test cases the problem is the second time(second test case the postfix_exp is printing more than it should)
Input
2
a+b*(c^d-e)^(f+gh)-i
a(b+c)/d
Output
abcd^e-fgh*+^*+i-
abc+d/fgh+^*+i-
Can you tell me what is wrong because of which "fgh+^*+i-" the extra expression from the previous expression is being concatenated at the end of the output?
The postfix_exp seems to be overwritten but I was expecting a new assigned postfix expression. Why is it that earlier data(data from previous test case) is saved?
Please ask any further query in the comments section

Answer 1

You did not closed postfix_exp with the terminating '\0'. After the first iteration postfix will look like something like this
XXXXXXXXXXX0
After the shorter 2nd input it will look something like
YYYYYYYXXXX0
Where X represent the characters written in the first iteration and Y the ones which in the second. So when you print it it will contain the last iteration results too.
Add this postfix_exp[j] = '\0'; before the if (top>0) line.
Note that this is still undefined behavior since postfix array is unitialized and there is no guarantee that it will be zeroed out.

Answer 2

you should clos postfix_exp with the terminating '\0' by adding
Add this postfix_exp[j] = '\0';
before the if (top>0) line.