Infix to Prefix Conversion using Stack in C
Converting Infix to Prefix Expression
On this page we will discuss about Infix to Prefix Conversion using Stack in C . As we know for the compiler it may be difficult to read an infix expression as it can’t start operations and assignments until it has read the whole expression and know about the precedence of various operations in the expression.
While prefix expression is easier for the compiler, as it is already sorted in accordance with precedence and compiler, can start assignments and operations, without caring about precedence.
What is Infix to Prefix Conversion using Stack ?
- Infix: Expressions of format
(A + B)
are called as infix expressions, these are just like mathematical expressions- Example –
((a / b) + c) - (d + (e * f ))
- Example –
- Prefix: Expressions wherein the operator comes before the operands are prefix expression like – Infix:
(A + B)
can be expressed as+AB
- Example – Prefix result would be :
-+/abc+d*ef
- Example – Prefix result would be :
- Postfix: Expression operator comes after the operands are prefix expression like – Infix:
(A + B)
can be expressed asAB+
- Example – Prefix result would be:
ab/c+def*+-
- Example – Prefix result would be:
Methods Discussed
Firstly we are going to discuss the manual way of calculating prefix. We will use this to understand the algorithm of the code. Then we are going to code, the coding methods that we have used are –- Method 1 – Using Array
- Method 2 – Dynamically created Stack
Steps to convert
- Any infix
op1 oper op2
can be written asop1 op2 oper
- Where op1 = Operand 1
- op2 = Operand2
- oper = Operation
- Example
a + b
can be written asab+
in prefix
Problem (This is how to convert manually for MCQ Question in the exam)
- Infix:
(a / b + c) - ( d + e * f)
can be written as((a / b) + c) - ( d + (e * f))
- Now, we have done the above according to associativity
- Solving and converting innermost bracket to prefix
- Step 1 –
(/ab + c) - ( d + *ef)
- Step 2 – Consider
/ab
and*ef
as separate operandx
andy
- the innermost bracket now looks like
(x + c) - (d + y)
- Applying prefix it looks like –
(+xc - +dy)
replacing x and y here(+/abc - +d*ef)
- Applying prefix it looks like –
- Step 3 – Considering
+/abc
and+d*ef
as separate operand z and w, the final bracket looks like –
(z - w)
the result would be-zw
- replacing z and w value =
-+/abc+d*ef
- replacing z and w value =
Alert
The above may give wrong results sometimes, which is why its always safer to use below algorithm for both coding and manual calculation -Also note below algorithm is given wrong on Geeks4Geek website, only refer from here.(As most codes are made by interns and PrepInsta pages are made by Ph.D Teachers)
Algorithm for Prefix
Given Infix - ((a/b)+c)-(d+(e*f))
- Step 1: Reverse the infix string. Note that while reversing the string you must interchange left and right parentheses.
- Step 2: Obtain the postfix expression of the infix expression Step 1.
- Step 3: Reverse the postfix expression to get the prefix expression
This is how you convert manually for theory question in the exam
- String after reversal – ))f*e(+d(-)c+)b/a((
- String after interchanging right and left parenthesis – ((f*e)+d)-(c+(b/a))
- Apply postfix – Below is postfix (On this page you check infix to postfix conversion rule)
- Reverse Postfix Expression (Given After the table below)
Sr. no. | Expression | Stack | Prefix |
---|---|---|---|
0 | ( | (( | |
1 | ( | ((( | |
2 | f | ((( | f |
3 | * | (((* | f |
4 | e | (((* | fe |
5 | ) | (( | fe* |
6 | + | ((+ | fe* |
7 | d | ((+ | fe*d |
8 | ) | ( | fe*d+ |
9 | – | (- | fe*d+ |
10 | ( | (-( | fe*d+ |
11 | c | (-( | fe*d+c |
12 | + | (-(+ | fe*d+c |
13 | ( | (-(+( | fe*d+c |
14 | b | (-(+( | fe*d+cb |
15 | / | (-(+(/ | fe*d+cb |
16 | a | (-(+(/ | fe*d+cba |
17 | ) | (-(+ | fe*d+cba/ |
18 | ) | (- | fe*d+cba/+ |
19 | ) | fe*d+cba/+- |
Final prefix: -+/abc+d*ef
Method 1 (Using Array)
Let us have a look at this method below –#include<stdio.h> #include<string.h> #include<limits.h> #include<stdlib.h> #define MAX 100 int top = -1; char stack[MAX]; // checking if stack is full int isFull () { return top == MAX - 1; } // checking is stack is empty int isEmpty () { return top == -1; } void push (char item) { if (isFull ()) return; top++; stack[top] = item; } // Function to remove an item from stack. It decreases top by 1 int pop () { if (isEmpty ()) return INT_MIN; // decrements top and returns what has been popped return stack[top--]; } // Function to return the top from stack without removing it int peek () { if (isEmpty ()) return INT_MIN; return stack[top]; } // A utility function to check if the given character is operand int checkIfOperand (char ch) { return (ch >= 'a' && ch <= 'z') || (ch >= 'A' && ch <= 'Z'); } // Fucntion to compare precedence // If we return larger value means higher precedence int precedence (char ch) { switch (ch) { case '+': case '-': return 1; case '*': case '/': return 2; case '^': return 3; } return -1; } // The driver function for infix to postfix conversion int getPostfix (char *expression) { int i, j; for (i = 0, j = -1; expression[i]; ++i) { if (checkIfOperand (expression[i])) expression[++j] = expression[i]; else if (expression[i] == '(') push (expression[i]); else if (expression[i] == ')') { while (!isEmpty (stack) && peek (stack) != '(') expression[++j] = pop (stack); if (!isEmpty (stack) && peek (stack) != '(') return -1; // invalid expression else pop (stack); } else // if an opertor { while (!isEmpty (stack) && precedence (expression[i]) <= precedence (peek (stack))) expression[++j] = pop (stack); push (expression[i]); } } // Once all inital expression characters are traversed // adding all left elements from stack to exp while (!isEmpty (stack)) expression[++j] = pop (stack); expression[++j] = '\0'; } void reverse (char *exp) { int size = strlen (exp); int j = size, i = 0; char temp[size]; temp[j--] = '\0'; while (exp[i] != '\0') { temp[j] = exp[i]; j--; i++; } strcpy (exp, temp); } void brackets (char *exp) { int i = 0; while (exp[i] != '\0') { if (exp[i] == '(') exp[i] = ')'; else if (exp[i] == ')') exp[i] = '('; i++; } } void InfixtoPrefix (char *exp) { int size = strlen (exp); // reverse string reverse (exp); //change brackets brackets (exp); //get postfix getPostfix (exp); // reverse string again reverse (exp); } int main () { printf ("The infix is: "); char expression[] = "((a/b)+c)-(d+(e*f))"; printf ("%s\n", expression); InfixtoPrefix (expression); printf ("The prefix is: "); printf ("%s\n", expression); return 0; }
Output
The infix is: ((a/b)+c)-(d+(e*f)) The prefix is: -+/abc+d*ef
Method 2 (Dynamically created Stack)
Let us have a look on this method below –#include<string.h> #include<limits.h> #include<stdio.h> #include<stdlib.h> #define MAX 100 // A structure to represent a stack struct Stack { int top; int maxSize; int *array; }; struct Stack *create (int max) { struct Stack *stack = (struct Stack *) malloc (sizeof (struct Stack)); stack->maxSize = max; stack->top = -1; stack->array = (int *) malloc (stack->maxSize * sizeof (int)); return stack; } // Checking with this function is stack is full or not // Will return true is stack is full else false //Stack is full when top is equal to the last index int isFull (struct Stack *stack) { if (stack->top == stack->maxSize - 1) { printf ("Will not be able to push maxSize reached\n"); } // Since array starts from 0, and maxSize starts from 1 return stack->top == stack->maxSize - 1; } // By definition the Stack is empty when top is equal to -1 // Will return true if top is -1 int isEmpty (struct Stack *stack) { return stack->top == -1; } // Push function here, inserts value in stack and increments stack top by 1 void push (struct Stack *stack, char item) { if (isFull (stack)) return; stack->array[++stack->top] = item; } // Function to remove an item from stack. It decreases top by 1 int pop (struct Stack *stack) { if (isEmpty (stack)) return INT_MIN; return stack->array[stack->top--]; } // Function to return the top from stack without removing it int peek (struct Stack *stack) { if (isEmpty (stack)) return INT_MIN; return stack->array[stack->top]; } // A utility function to check if the given character is operand int checkIfOperand (char ch) { return (ch >= 'a' && ch <= 'z') || (ch >= 'A' && ch <= 'Z'); } // Fucntion to compare precedence // If we return larger value means higher precedence int precedence (char ch) { switch (ch) { case '+': case '-': return 1; case '*': case '/': return 2; case '^': return 3; } return -1; } // The driver function for infix to postfix conversion int getPostfix (char *expression) { int i, j; // Stack size should be equal to expression size for safety struct Stack *stack = create (strlen (expression)); if (!stack) // just checking is stack was created or not return -1; for (i = 0, j = -1; expression[i]; ++i) { if (checkIfOperand (expression[i])) expression[++j] = expression[i]; else if (expression[i] == '(') push (stack, expression[i]); else if (expression[i] == ')') { while (!isEmpty (stack) && peek (stack) != '(') expression[++j] = pop (stack); if (!isEmpty (stack) && peek (stack) != '(') return -1; // invalid expression else pop (stack); } else // if an opertor { while (!isEmpty (stack) && precedence (expression[i]) <= precedence (peek (stack))) expression[++j] = pop (stack); push (stack, expression[i]); } } // Once all inital expression characters are traversed // adding all left elements from stack to exp while (!isEmpty (stack)) expression[++j] = pop (stack); expression[++j] = '\0'; } void reverse (char *exp) { int size = strlen (exp); int j = size, i = 0; char temp[size]; temp[j--] = '\0'; while (exp[i] != '\0') { temp[j] = exp[i]; j--; i++; } strcpy (exp, temp); } void brackets (char *exp) { int i = 0; while (exp[i] != '\0') { if (exp[i] == '(') exp[i] = ')'; else if (exp[i] == ')') exp[i] = '('; i++; } } void InfixtoPrefix (char *exp) { int size = strlen (exp); // reverse string reverse (exp); //change brackets brackets (exp); //get postfix getPostfix (exp); // reverse string again reverse (exp); } int main () { printf ("The infix is: "); char expression[] = "((a/b)+c)-(d+(e*f))"; printf ("%s\n", expression); InfixtoPrefix (expression); printf ("The prefix is: "); printf ("%s\n", expression); return 0; }
Output
The infix is: ((a/b)+c)-(d+(e*f)) The prefix is: -+/abc+d*ef
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Queues
- Queues in Data Structures (Introduction)
Click Here - Queues Program in C and implementation
Click Here - Implementation of Queues using Arrays | C Program
Click Here - Types of Queues in Data Structure
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Click Here - Insertion in Queues Program (Enqueuing) –
C | C++ | Java - Deletion (Removal) in Queues Program(Dequeuing) –
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