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Finish function to complete code. #include <stdio.h> #include <stdlib.h> #include<string.h> #define Max_Size 20 void push(char S[],...

Finish function to complete code.

#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#define Max_Size 20
void push(char S[], int *p_top, char value);
char pop(char S[], int *p_top);
void printCurrentStack(char S[], int *p_top);
int validation(char infix[], char S[], int *p_top);
char *infix2postfix(char infix[], char postfix[], char S[], int *p_top);
int precedence(char symbol);
int main()
{
// int choice;
int top1=0; //top for S1 stack
int top2=0; //top for S2 stack
int *p_top1=&top1;
int *p_top2=&top2;
char infix[]="(2+3)*(4-3)"; //Stores infix string
int n=strlen(infix); //length of the string
char S1[Max_Size]; //Stack used for checking balanced expression
char S2[Max_Size]; //stack used for postfix conversion
char postfix[n]; //stores postfix string
char *p_postfix=postfix; //pointer to the string posfix (used for ease in passing between function) pointer points to the first element of postfix[]
if (validation(infix, S1, p_top1)==0){
printf("The expresion %s is NOT balanced \n", infix);
exit(1);
}
else
printf("The expresion %s is balanced\n", infix);


// printf("Infix Expression is %s\n", infix );
p_postfix=infix2postfix(infix, postfix, S2, p_top2);
// printf("Infix Expression is %s\n", infix );
printf("Postfix Expression is--->");
while (*p_postfix!='\0') {
printf("%c", *p_postfix++);
}
printf("\n" );

}


void push(char S[], int *p_top, char value) {
S[(*p_top)] = value;
(*p_top)++;
// printCurrentStack(s, p_top);
}
int isEmpty(char S[], int *p_top){
if (*p_top==0) {
return 1;
}
return 0;
}
char pop(char S[], int *p_top) {
if (isEmpty(S, p_top)==0) { //if not empty
(*p_top)--;
char x= S[(*p_top)];
// printCurrentStack(s, p_top);
return x;
}
}

char *infix2postfix(char infix[], char postfix[], char S2[], int *p_top2){

//write your code here

char * p_postfix=postfix;
return p_postfix;
}

void printCurrentStack(char s[], int *p_top) {
if ((*p_top) == 0) {
printf("The stack is empty.\n\n");
return;
}

printf("There are %d elements currently in the stack.\n", (*p_top));

for (int i = (*p_top) - 1; i >= 0; i--)
printf("%c \n", s[i]);
printf("\n");
}


int precedence (char symbol){
switch (symbol) {
case '(':
return 0; //this is special case as we do not pop it until we find a ')' hence 0
break;
case '^':
return 3;
break;
case '/':
case '*' :
return 2;
break;
case '+':
case '-':
return 1;
default:
printf("Error : not an operator\n");
return 0;
break;
}
}
int validation(char infix[], char S1[], int *p_top1){
char y;

//int n=sizeof(s)/sizeof(s[0]);
int i=0;
//printf("%c, %c, %c, %c, %c\n",s[i], s[i+1], s[i+2], s[i+3], s[i+9]);
while(infix[i]!='\0'){
switch (infix[i]) {
case '(':
case '{':
case '[':
push(S1, p_top1,infix[i] );
// printf("pushed %c\n",s[i] );
// printCurrentStack(s, p_top);
break;
case ')':
y=pop(S1, p_top1);
// printf("1. popped %c\n",y );
if(y=='(')
break;
else {
return 0;
}
break;
case '}':
y=pop(S1, p_top1);
//printf("2. popped %c\n",y );
if(y=='{')
break;
else {
return 0;
}
break;
case ']':
y=pop(S1, p_top1);
//printf("3. popped %c\n",y );
if(y=='[')
break;
else {
return 0;
}
break;
default :
break;
}
i++;
}
return 1;
}


the "char infix2postfix" function
engineering   Computer-Science   
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Homework Answers

Answer #1

Updated Code:

#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#define Max_Size 20
void push(char S[], int *p_top, char value);
char pop(char S[], int *p_top);
void printCurrentStack(char S[], int *p_top);
int validation(char infix[], char S[], int *p_top);
char *infix2postfix(char infix[], char postfix[], char S[], int *p_top);
int precedence(char symbol);
int main()
{
// int choice;
int top1=0; //top for S1 stack
int top2=0; //top for S2 stack
int *p_top1=&top1;
int *p_top2=&top2;
char infix[]="(2+3)*(4-3)"; //Stores infix string
int n=strlen(infix); //length of the string
char S1[Max_Size]; //Stack used for checking balanced expression
char S2[Max_Size]; //stack used for postfix conversion
char postfix[n]; //stores postfix string
char *p_postfix=postfix; //pointer to the string posfix (used for ease in passing between function) pointer points to the first element of postfix[]
if (validation(infix, S1, p_top1)==0){
printf("The expresion %s is NOT balanced \n", infix);
exit(1);
}
else
printf("The expresion %s is balanced\n", infix);


// printf("Infix Expression is %s\n", infix );
p_postfix=infix2postfix(infix, postfix, S2, p_top2);
// printf("Infix Expression is %s\n", infix );
printf("Postfix Expression is--->");
while (*p_postfix!='\0') {
printf("%c", *p_postfix++);
}
printf("\n" );

}


void push(char S[], int *p_top, char value) {
S[(*p_top)] = value;
(*p_top)++;
// printCurrentStack(s, p_top);
}
int isEmpty(char S[], int *p_top){
if (*p_top==0) {
return 1;
}
return 0;
}
char pop(char S[], int *p_top) {
if (isEmpty(S, p_top)==0) { //if not empty
(*p_top)--;
char x= S[(*p_top)];
// printCurrentStack(s, p_top);
return x;
}
}

char *infix2postfix(char infix[], char postfix[], char S2[], int *p_top2)
{
   // sz is used current size of postfix[]
   int sz=-1;
   // Iterate over all the elements in infix
for (int i = 0; infix[i]; ++i)
{
// Here test for the type of character and based on the type of character we perform task
       // if the infix[i] is operand
if (infix[i] != '+' && infix[i] != '-' && infix[i] != '/' && infix[i] != '*' && infix[i] != '^' && infix[i]!='(' && infix[i]!=')')
postfix[++sz] = infix[i];
  
// If the infix[i] is ( then push it to stack s2
else if (infix[i] == '(')
push(S2, p_top2,infix[i] ) ;
  
// If the infix[i] is an ‘)’, pop and output from the s2
// until an ‘(‘ is encountered.
else if (infix[i] == ')')
{
while(!isEmpty(S2,p_top2) && S2[(*p_top2)-1] != '(')
postfix[++sz] = pop(S2,p_top2);   
pop(S2,p_top2);
}
else // if infix i is operator
{
while (!isEmpty(S2,p_top2) && precedence(infix[i]) <= precedence(S2[(*p_top2)-1]))
postfix[++sz] = pop(S2,p_top2);
push(S2, p_top2,infix[i] ) ;
}
}
  
// at the end of iterating over infix we will simply pop all elements from stack which are remaining in stack
while (!isEmpty(S2,p_top2))
postfix[++sz] = pop(S2,p_top2);
  
// set the last character as null
   postfix[++sz] = '\0';
   char * p_postfix=postfix;
   // return the pointet to postfix
   return p_postfix;
}

void printCurrentStack(char s[], int *p_top) {
if ((*p_top) == 0) {
printf("The stack is empty.\n\n");
return;
}

printf("There are %d elements currently in the stack.\n", (*p_top));

for (int i = (*p_top) - 1; i >= 0; i--)
printf("%c \t", s[i]);
printf("\n");
}


int precedence (char symbol){
switch (symbol) {
case '(':
return 0; //this is special case as we do not pop it until we find a ')' hence 0
break;
case '^':
return 3;
break;
case '/':
case '*' :
return 2;
break;
case '+':
case '-':
return 1;
default:
printf("Error : not an operator\n");
return 0;
break;
}
}
int validation(char infix[], char S1[], int *p_top1){
char y;

//int n=sizeof(s)/sizeof(s[0]);
int i=0;
//printf("%c, %c, %c, %c, %c\n",s[i], s[i+1], s[i+2], s[i+3], s[i+9]);
while(infix[i]!='\0'){
switch (infix[i]) {
case '(':
case '{':
case '[':
push(S1, p_top1,infix[i] );
// printf("pushed %c\n",s[i] );
// printCurrentStack(s, p_top);
break;
case ')':
y=pop(S1, p_top1);
// printf("1. popped %c\n",y );
if(y=='(')
break;
else {
return 0;
}
break;
case '}':
y=pop(S1, p_top1);
//printf("2. popped %c\n",y );
if(y=='{')
break;
else {
return 0;
}
break;
case ']':
y=pop(S1, p_top1);
//printf("3. popped %c\n",y );
if(y=='[')
break;
else {
return 0;
}
break;
default :
break;
}
i++;
}
return 1;
}

Sample output:

The expresion (2+3)*(4-3) is balanced
Postfix Expression is--->23+43-*

Note: If you have any doubts regarding this solution please ask me in the comments section and please do not forget to upvote the answer.

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