Question

Help me to fix this code in C language. This code converts infix expressions to postfix and then evaluate the expression. Right now, it works with single digits. I need to modify it so that it can evaluate expressions with also 2 digits, example: (60+82)%72. Additionally I need to display an error when the parenthesis don't match like (89+8(. I have muted some line that would print the postfix expression with a space like: 22 8 + when the input would have been (22+8). The problem is when I unmuted these lines, the evaluation doesn't work.

I'm using this online compiler: https://www.onlinegdb.com/online_c_compiler

#include <stdio.h>

#include <ctype.h>

#include <string.h>

#include <math.h>

#define SIZE 100

char s[SIZE];

int top=-1;

void infixToPostfix(char *infix, char *postfix);

void postfixEvaluation(char *postfix);

void push(char elem){

s[++top]=elem;

}

char pop(){

return(s[top--]);

}

int pr(char elem){ // Order of precedence

switch (elem)

{

case '(': return 1;

case '+': return 2;

case '-': return 2;

case '*': return 3;

case '/': return 3;

case '%': return 3; //modulo

case '^': return 4; //exponent

}

return -1;

}

int main ()

{

char infix[50];

  

printf("\nEnter an infix expression:\n");

scanf("%s", infix);

push('=');

  

char postfix[50];

infixToPostfix(infix, postfix);

  

postfixEvaluation(postfix);

  

return 0;

}

void infixToPostfix(char *infix, char *postfix){

  

char ch, elem;

int i=0, j=0;

  

for (i=0; infix[i] !='\0'; i++)

{

if (infix[i] == '(')

push(infix[i]);

  

else if (isalnum(infix[i]))

{

postfix[j++]=infix[i];

  

// if (infix[i+1] != '\0' && isalnum(infix[i+1]) !=0){ //This line check if there's another number aside

// postfix[j++]=infix[++i];

// }

// postfix[j++]=' '; //this line add an extra space

}

else if (infix[i] == ')')

{

while( s[top] != '(')

{

postfix[j++] = pop();

}

elem=pop();

}

else

{

while( pr(s[top]) >= pr(infix[i]) ){

postfix[j++] = pop();

}

push(infix[i]);

  

// postfix[j++]=' '; //This line separate the operators from operand

}

}

while ( s[top] != '=')

postfix[j++]=pop();

postfix[j]='\0';

  

printf("\n%s = %s", infix, postfix);

}

void postfixEvaluation(char *postfix){

  

int i=0, j=0, postfix_int[50];

  

while (top != -1)

pop();

  

int value=0, op1, op2;

  

for (i=0; postfix[i] != '\0'; i++)

{

  

if (isdigit(postfix[i]))

push(postfix[i]);

  

else {

op1 = pop() - '0';

op2 = pop() - '0';

switch (postfix[i]) {

case '+':

value = op2 + op1;

break;

case '-':

value = op2 - op1;

break;

case '*':

value = op2 * op1;

break;

case '/':

value = op2 / op1;

break;

case '%':

value = op2 % op1;

break;

case '^':

value = pow(op2, op1); // exponet math library

break;

}

push(value + '0');

}

}

printf(" = %d", pop() - '0');

}

Answer 1

PROBLEM IN THE CODE:

1.You have used a single dimensional char array and hence you can only store one char per a block in the array.

2.So,this requires including white spaces to differentiate multi digit numbers.But the problem in doing so is when we are creating a post fix expression,some times we pop operators out and again push another operator and sometimes we need to pop everything out until we get to a open bracket when we encounter a close bracket. In either of the cases,the spaces we used might get deleted and it is impossible to maintain uniformity in spaces in such cases.

3.So,using spaces is not helpful. Moreover,the code given doesn't work whenever it encounter a 3 digit number.So,it even does not work for single digit numbers. Eg: (9*9*9-7) gives an answer of -46 because 729 cannot simply be converted into a character by adding '0' as per the code.So,it does not work for such cases.

ALTERNATE SOLUTION:

1.Create an array of strings or create a 2 dimensional character array.and hence you can store each digit or operator in each line of 2 dimensional array or in each string of the string array.So,you don't need to use spaces and the rest process is same as the given code.And you can use the below code to convert a string to integer. Eg:'234' can be converted to 234.

int convert_int(char a[]) {
int c, sign, offset, n;

if (a[0] == '-') {
sign = -1;
}

if (sign == -1) {  
offset = 1;
}
else {
offset = 0;
}

n = 0;

for (c = offset; a[c] != '\0'; c++) {
n = n * 10 + a[c] - '0';
}

if (sign == -1) {
n = -n;
}

return n;
}

2.Create a data structure that consists of a string,a integer, a flag.

struct ele{
int no;
string value;
int flag;
};
and create a array of these structure.Use flag =0 to denote it is a integer and flag=1 to denote it is a character(for cases like +,-,*).In this way,you can simply do the required code.

And you can use the below code to take multiple digit input from a infix expression.The below code works in this way:

when it encounters a numbers it goes on until a non numerical character is encountered and store it in input and suing the above function it converts it to integer.Use the below code inside the for loop in your code.

char input[100];
int count1=0;
while(isalnum(infix[i+count1])){
input[count1]=infix[i+count1];
count1++;
}
i=i+count1-1;
input[count1]='\0';
printf("%s ",input);
int con_int=convert_int(input);
printf("%d\n",con_int);

To find error in matching brackets:This is simple.,use a for loop to go through the input and use a variable count(initially 0)and add 1 .whenever '(' is encountered and subtract by 1 if ')' is encountered.If the result is not 0,then there is an error.