# Evaluation of Prefix Expressions - GeeksforGeeks (2022)

Prefix and Postfix expressions can be evaluated faster than an infix expression. This is because we don’t need to process any brackets or follow operator precedence rule. In postfix and prefix expressions which ever operator comes before will be evaluated first, irrespective of its priority. Also, there are no brackets in these expressions. As long as we can guarantee that a valid prefix or postfix expression is used, it can be evaluated with correctness.

We can convert infix to postfix and can convert infix to prefix.
In this article, we will discuss how to evaluate an expression written in prefix notation. The method is similar to evaluating a postfix expression. Please read Evaluation of Postfix Expression to know how to evaluate postfix expressions

Algorithm:

`EVALUATE_PREFIX(STRING)Step 1: Put a pointer P at the end of the endStep 2: If character at P is an operand push it to StackStep 3: If the character at P is an operator pop two elements from the Stack. Operate on these elements according to the operator, and push the result back to the StackStep 4: Decrement P by 1 and go to Step 2 as long as there are characters left to be scanned in the expression.Step 5: The Result is stored at the top of the Stack, return itStep 6: End`

Example to demonstrate working of the algorithm

`Expression: +9*26Character | Stack | ExplanationScanned | (Front to | | Back) | -------------------------------------------6 6 6 is an operand, push to Stack2 6 2 2 is an operand, push to Stack* 12 (6*2) * is an operator, pop 6 and 2, multiply them and push result to Stack 9 12 9 9 is an operand, push to Stack+ 21 (12+9) + is an operator, pop 12 and 9 add them and push result to StackResult: 21`

Examples:

`Input : -+8/632Output : 8Input : -+7*45+20Output : 25`

Complexity The algorithm has linear complexity since we scan the expression once and perform at most O(N) push and pop operations which take constant time.
Implementation of the algorithm is given below.

Implementation:

## C++

`// C++ program to evaluate a prefix expression.`

`#include <bits/stdc++.h>`

`using` `namespace` `std;`

`bool` `isOperand(``char` `c)`

`{`

`// If the character is a digit then it must`

`// be an operand`

`return` `isdigit``(c);`

`}`

`double` `evaluatePrefix(string exprsn)`

`{`

`stack<``double``> Stack;`

`for` `(``int` `j = exprsn.size() - 1; j >= 0; j--) {`

`// Push operand to Stack`

`// To convert exprsn[j] to digit subtract`

`// '0' from exprsn[j].`

`if` `(isOperand(exprsn[j]))`

`Stack.push(exprsn[j] - ``'0'``);`

`else` `{`

`// Operator encountered`

`// Pop two elements from Stack`

`double` `o1 = Stack.top();`

`Stack.pop();`

`double` `o2 = Stack.top();`

`Stack.pop();`

`// Use switch case to operate on o1`

`// and o2 and perform o1 O o2.`

`switch` `(exprsn[j]) {`

`case` `'+'``:`

`Stack.push(o1 + o2);`

`break``;`

`case` `'-'``:`

`Stack.push(o1 - o2);`

`break``;`

`case` `'*'``:`

`Stack.push(o1 * o2);`

`break``;`

`case` `'/'``:`

`Stack.push(o1 / o2);`

`break``;`

`}`

`}`

`}`

`return` `Stack.top();`

`}`

`// Driver code`

`int` `main()`

`{`

`string exprsn = ``"+9*26"``;`

`cout << evaluatePrefix(exprsn) << endl;`

`return` `0;`

`}`

## Java

`// Java program to evaluate`

`// a prefix expression.`

`import` `java.io.*;`

`import` `java.util.*;`

`class` `GFG {`

`static` `Boolean isOperand(``char` `c)`

`{`

`// If the character is a digit`

`// then it must be an operand`

`if` `(c >= ``48` `&& c <= ``57``)`

`return` `true``;`

`else`

`return` `false``;`

`}`

`static` `double` `evaluatePrefix(String exprsn)`

`{`

`Stack<Double> Stack = ``new` `Stack<Double>();`

`for` `(``int` `j = exprsn.length() - ``1``; j >= ``0``; j--) {`

`// Push operand to Stack`

`// To convert exprsn[j] to digit subtract`

`// '0' from exprsn[j].`

`if` `(isOperand(exprsn.charAt(j)))`

`Stack.push((``double``)(exprsn.charAt(j) - ``48``));`

`else` `{`

`// Operator encountered`

`// Pop two elements from Stack`

`double` `o1 = Stack.peek();`

`Stack.pop();`

`double` `o2 = Stack.peek();`

`Stack.pop();`

`// Use switch case to operate on o1`

`// and o2 and perform o1 O o2.`

`switch` `(exprsn.charAt(j)) {`

`case` `'+'``:`

`Stack.push(o1 + o2);`

`break``;`

`case` `'-'``:`

`Stack.push(o1 - o2);`

`break``;`

`case` `'*'``:`

`Stack.push(o1 * o2);`

`break``;`

`case` `'/'``:`

`Stack.push(o1 / o2);`

`break``;`

`}`

`}`

`}`

`return` `Stack.peek();`

`}`

`/* Driver program to test above function */`

`public` `static` `void` `main(String[] args)`

`{`

`String exprsn = ``"+9*26"``;`

`System.out.println(evaluatePrefix(exprsn));`

`}`

`}`

`// This code is contributed by Gitanjali`

(Video) Evaluation of prefix | Evaluation Expression | Appliedcourse

## Python3

`"""`

`Python3 program to evaluate a prefix expression.`

`"""`

`def` `is_operand(c):`

`"""`

`Return True if the given char c is an operand, e.g. it is a number`

`"""`

`return` `c.isdigit()`

`def` `evaluate(expression):`

`"""`

`Evaluate a given expression in prefix notation.`

`Asserts that the given expression is valid.`

`"""`

`stack ``=` `[]`

`# iterate over the string in reverse order`

`for` `c ``in` `expression[::``-``1``]:`

`# push operand to stack`

`if` `is_operand(c):`

`stack.append(``int``(c))`

`else``:`

`# pop values from stack can calculate the result`

`# push the result onto the stack again`

`o1 ``=` `stack.pop()`

`o2 ``=` `stack.pop()`

`if` `c ``=``=` `'+'``:`

`stack.append(o1 ``+` `o2)`

`elif` `c ``=``=` `'-'``:`

`stack.append(o1 ``-` `o2)`

`elif` `c ``=``=` `'*'``:`

`stack.append(o1 ``*` `o2)`

`elif` `c ``=``=` `'/'``:`

`stack.append(o1 ``/` `o2)`

`return` `stack.pop()`

`# Driver code`

`if` `__name__ ``=``=` `"__main__"``:`

`test_expression ``=` `"+9*26"`

`print``(evaluate(test_expression))`

`# This code is contributed by Leon Morten Richter (GitHub: M0r13n)`

## C#

`// C# program to evaluate`

`// a prefix expression.`

`using` `System;`

`using` `System.Collections.Generic;`

`class` `GFG {`

`static` `Boolean isOperand(``char` `c)`

`{`

`// If the character is a digit`

`// then it must be an operand`

`if` `(c >= 48 && c <= 57)`

`return` `true``;`

`else`

`return` `false``;`

`}`

`static` `double` `evaluatePrefix(String exprsn)`

`{`

`Stack<Double> Stack = ``new` `Stack<Double>();`

`for` `(``int` `j = exprsn.Length - 1; j >= 0; j--) {`

`// Push operand to Stack`

`// To convert exprsn[j] to digit subtract`

`// '0' from exprsn[j].`

`if` `(isOperand(exprsn[j]))`

`Stack.Push((``double``)(exprsn[j] - 48));`

`else` `{`

`// Operator encountered`

`// Pop two elements from Stack`

`double` `o1 = Stack.Peek();`

`Stack.Pop();`

`double` `o2 = Stack.Peek();`

`Stack.Pop();`

`// Use switch case to operate on o1`

`// and o2 and perform o1 O o2.`

`switch` `(exprsn[j]) {`

`case` `'+'``:`

`Stack.Push(o1 + o2);`

`break``;`

`case` `'-'``:`

`Stack.Push(o1 - o2);`

`break``;`

`case` `'*'``:`

`Stack.Push(o1 * o2);`

`break``;`

`case` `'/'``:`

`Stack.Push(o1 / o2);`

`break``;`

`}`

`}`

`}`

`return` `Stack.Peek();`

`}`

`/* Driver code */`

`public` `static` `void` `Main(String[] args)`

`{`

`String exprsn = ``"+9*26"``;`

`Console.WriteLine(evaluatePrefix(exprsn));`

`}`

`}`

`/* This code contributed by PrinciRaj1992 */`

## Javascript

`<script>`

`// Javascript program to evaluate a prefix expression.`

`function` `isOperand(c)`

`{`

`// If the character is a digit`

`// then it must be an operand`

`if` `(c.charCodeAt() >= 48 && c.charCodeAt() <= 57)`

`return` `true``;`

`else`

`return` `false``;`

`}`

`function` `evaluatePrefix(exprsn)`

(Video) Evaluation of Expression Tree | GeeksforGeeks

`{`

`let Stack = [];`

`for` `(let j = exprsn.length - 1; j >= 0; j--) {`

`// Push operand to Stack`

`// To convert exprsn[j] to digit subtract`

`// '0' from exprsn[j].`

`if` `(isOperand(exprsn[j]))`

`Stack.push((exprsn[j].charCodeAt() - 48));`

`else` `{`

`// Operator encountered`

`// Pop two elements from Stack`

`let o1 = Stack[Stack.length - 1];`

`Stack.pop();`

`let o2 = Stack[Stack.length - 1];`

`Stack.pop();`

`// Use switch case to operate on o1`

`// and o2 and perform o1 O o2.`

`switch` `(exprsn[j]) {`

`case` `'+'``:`

`Stack.push(o1 + o2);`

`break``;`

`case` `'-'``:`

`Stack.push(o1 - o2);`

`break``;`

`case` `'*'``:`

`Stack.push(o1 * o2);`

`break``;`

`case` `'/'``:`

`Stack.push(o1 / o2);`

`break``;`

`}`

`}`

`}`

`return` `Stack[Stack.length - 1];`

`}`

`let exprsn = ``"+9*26"``;`

`document.write(evaluatePrefix(exprsn));`

`// This code is contributed by suresh07.`

`</script>`

Output

`21`

Note:
To perform more types of operations only the switch case table needs to be modified. This implementation works only for single digit operands. Multi-digit operands can be implemented if some character-like space is used to separate the operands and operators.

Below given is the extended program which allows operands to have multiple digits.

## C++

`// C++ program to evaluate a prefix expression.`

`#include <bits/stdc++.h>`

`using` `namespace` `std;`

`double` `evaluatePrefix(string exprsn)`

`{`

`stack<``double``> Stack;`

`for` `(``int` `j = exprsn.size() - 1; j >= 0; j--) {`

`// if jth character is the delimiter ( which is`

`// space in this case) then skip it`

`if` `(exprsn[j] == ``' '``)`

`continue``;`

`// Push operand to Stack`

`// To convert exprsn[j] to digit subtract`

`// '0' from exprsn[j].`

`if` `(``isdigit``(exprsn[j])) {`

`// there may be more than`

`// one digits in a number`

`double` `num = 0, i = j;`

`while` `(j < exprsn.size() && ``isdigit``(exprsn[j]))`

`j--;`

`j++;`

`// from [j, i] exprsn contains a number`

`for` `(``int` `k = j; k <= i; k++)`

`num = num * 10 + ``double``(exprsn[k] - ``'0'``);`

`Stack.push(num);`

`}`

`else` `{`

`// Operator encountered`

`// Pop two elements from Stack`

`double` `o1 = Stack.top();`

`Stack.pop();`

`double` `o2 = Stack.top();`

`Stack.pop();`

`// Use switch case to operate on o1`

`// and o2 and perform o1 O o2.`

`switch` `(exprsn[j]) {`

`case` `'+'``:`

`Stack.push(o1 + o2);`

`break``;`

`case` `'-'``:`

`Stack.push(o1 - o2);`

`break``;`

`case` `'*'``:`

`Stack.push(o1 * o2);`

`break``;`

`case` `'/'``:`

`Stack.push(o1 / o2);`

`break``;`

`}`

`}`

`}`

`return` `Stack.top();`

`}`

`// Driver code`

`int` `main()`

`{`

`string exprsn = ``"+ 9 * 12 6"``;`

`cout << evaluatePrefix(exprsn) << endl;`

`return` `0;`

`// this code is contributed by Mohd Shaad Khan`

`}`

## Java

`// Java program to evaluate a prefix expression.`

`import` `java.util.*;`

`public` `class` `Main`

`{`

`static` `boolean` `isdigit(``char` `ch)`

`{`

`if``(ch >= ``48` `&& ch <= ``57``)`

`{`

`return` `true``;`

`}`

`return` `false``;`

`}`

`static` `double` `evaluatePrefix(String exprsn)`

(Video) Length of longest balanced parentheses prefix | GeeksforGeeks

`{`

`Stack<Double> stack = ``new` `Stack<Double>();`

`for` `(``int` `j = exprsn.length() - ``1``; j >= ``0``; j--) {`

`// if jth character is the delimiter ( which is`

`// space in this case) then skip it`

`if` `(exprsn.charAt(j) == ``' '``)`

`continue``;`

`// Push operand to Stack`

`// To convert exprsn[j] to digit subtract`

`// '0' from exprsn[j].`

`if` `(isdigit(exprsn.charAt(j))) {`

`// there may be more than`

`// one digits in a number`

`double` `num = ``0``, i = j;`

`while` `(j < exprsn.length() && isdigit(exprsn.charAt(j)))`

`j--;`

`j++;`

`// from [j, i] exprsn contains a number`

`for` `(``int` `k = j; k <= i; k++)`

`{`

`num = num * ``10` `+ (``double``)(exprsn.charAt(k) - ``'0'``);`

`}`

`stack.push(num);`

`}`

`else` `{`

`// Operator encountered`

`// Pop two elements from Stack`

`double` `o1 = (``double``)stack.peek();`

`stack.pop();`

`double` `o2 = (``double``)stack.peek();`

`stack.pop();`

`// Use switch case to operate on o1`

`// and o2 and perform o1 O o2.`

`switch` `(exprsn.charAt(j)) {`

`case` `'+'``:`

`stack.push(o1 + o2);`

`break``;`

`case` `'-'``:`

`stack.push(o1 - o2);`

`break``;`

`case` `'*'``:`

`stack.push(o1 * o2);`

`break``;`

`case` `'/'``:`

`stack.push(o1 / o2);`

`break``;`

`}`

`}`

`}`

`return` `stack.peek();`

`}`

`// Driver code`

`public` `static` `void` `main(String[] args) {`

`String exprsn = ``"+ 9 * 12 6"``;`

`System.out.print((``int``)evaluatePrefix(exprsn));`

`}`

`}`

`// This code is contributed by mukesh07.`

## Python3

`# Python3 program to evaluate a prefix expression.`

`def` `isdigit(ch):`

`if``(``ord``(ch) >``=` `48` `and` `ord``(ch) <``=` `57``):`

`return` `True`

`return` `False`

`def` `evaluatePrefix(exprsn):`

`Stack ``=` `[]`

`for` `j ``in` `range``(``len``(exprsn) ``-` `1``, ``-``1``, ``-``1``):`

`# if jth character is the delimiter ( which is`

`# space in this case) then skip it`

`if` `(exprsn[j] ``=``=` `' '``):`

`continue`

`# Push operand to Stack`

`# To convert exprsn[j] to digit subtract`

`# '0' from exprsn[j].`

`if` `(isdigit(exprsn[j])):`

`# there may be more than`

`# one digits in a number`

`num, i ``=` `0``, j`

`while` `(j < ``len``(exprsn) ``and` `isdigit(exprsn[j])):`

`j``-``=``1`

`j``+``=``1`

`# from [j, i] exprsn contains a number`

`for` `k ``in` `range``(j, i ``+` `1``, ``1``):`

`num ``=` `num ``*` `10` `+` `(``ord``(exprsn[k]) ``-` `ord``(``'0'``))`

`Stack.append(num)`

`else``:`

`# Operator encountered`

`# Pop two elements from Stack`

`o1 ``=` `Stack[``-``1``]`

`Stack.pop()`

`o2 ``=` `Stack[``-``1``]`

`Stack.pop()`

`# Use switch case to operate on o1`

`# and o2 and perform o1 O o2.`

`if` `exprsn[j] ``=``=` `'+'``:`

`Stack.append(o1 ``+` `o2 ``+` `12``)`

`elif` `exprsn[j] ``=``=` `'-'``:`

`Stack.append(o1 ``-` `o2)`

`elif` `exprsn[j] ``=``=` `'*'``:`

`Stack.append(o1 ``*` `o2 ``*` `5` `)`

`elif` `exprsn[j] ``=``=` `'/'``:`

`Stack.append(o1 ``/` `o2)`

`return` `Stack[``-``1``]`

`exprsn ``=` `"+ 9 * 12 6"`

`print``(evaluatePrefix(exprsn))`

`# This code is contributed by divyesh072019.`

## C#

`// C# program to evaluate a prefix expression.`

`using` `System;`

`using` `System.Collections;`

`class` `GFG {`

`static` `bool` `isdigit(``char` `ch)`

`{`

`if``(ch >= 48 && ch <= 57)`

`{`

`return` `true``;`

`}`

`return` `false``;`

`}`

`static` `double` `evaluatePrefix(``string` `exprsn)`

(Video) Analysis of Prefix expression evaluation

`{`

`Stack stack = ``new` `Stack();`

`for` `(``int` `j = exprsn.Length - 1; j >= 0; j--) {`

`// if jth character is the delimiter ( which is`

`// space in this case) then skip it`

`if` `(exprsn[j] == ``' '``)`

`continue``;`

`// Push operand to Stack`

`// To convert exprsn[j] to digit subtract`

`// '0' from exprsn[j].`

`if` `(isdigit(exprsn[j])) {`

`// there may be more than`

`// one digits in a number`

`double` `num = 0, i = j;`

`while` `(j < exprsn.Length && isdigit(exprsn[j]))`

`j--;`

`j++;`

`// from [j, i] exprsn contains a number`

`for` `(``int` `k = j; k <= i; k++)`

`{`

`num = num * 10 + (``double``)(exprsn[k] - ``'0'``);`

`}`

`stack.Push(num);`

`}`

`else` `{`

`// Operator encountered`

`// Pop two elements from Stack`

`double` `o1 = (``double``)stack.Peek();`

`stack.Pop();`

`double` `o2 = (``double``)stack.Peek();`

`stack.Pop();`

`// Use switch case to operate on o1`

`// and o2 and perform o1 O o2.`

`switch` `(exprsn[j]) {`

`case` `'+'``:`

`stack.Push(o1 + o2);`

`break``;`

`case` `'-'``:`

`stack.Push(o1 - o2);`

`break``;`

`case` `'*'``:`

`stack.Push(o1 * o2);`

`break``;`

`case` `'/'``:`

`stack.Push(o1 / o2);`

`break``;`

`}`

`}`

`}`

`return` `(``double``)stack.Peek();`

`}`

`static` `void` `Main() {`

`string` `exprsn = ``"+ 9 * 12 6"``;`

`Console.Write(evaluatePrefix(exprsn));`

`}`

`}`

`// This code is contributed by divyeshrabadiya07.`

## Javascript

`<script>`

`// Javascript program to evaluate a prefix expression.`

`function` `isdigit(ch)`

`{`

`if``(ch.charCodeAt() >= 48 && ch.charCodeAt() <= 57)`

`{`

`return` `true``;`

`}`

`return` `false``;`

`}`

`function` `evaluatePrefix(exprsn)`

`{`

`let Stack = [];`

`for` `(let j = exprsn.length - 1; j >= 0; j--) {`

`// if jth character is the delimiter ( which is`

`// space in this case) then skip it`

`if` `(exprsn[j] == ``' '``)`

`continue``;`

`// Push operand to Stack`

`// To convert exprsn[j] to digit subtract`

`// '0' from exprsn[j].`

`if` `(isdigit(exprsn[j])) {`

`// there may be more than`

`// one digits in a number`

`let num = 0, i = j;`

`while` `(j < exprsn.length && isdigit(exprsn[j]))`

`j--;`

`j++;`

`// from [j, i] exprsn contains a number`

`for` `(let k = j; k <= i; k++)`

`num = num * 10 + (exprsn[k].charCodeAt() - ``'0'``.charCodeAt());`

`Stack.push(num);`

`}`

`else` `{`

`// Operator encountered`

`// Pop two elements from Stack`

`let o1 = Stack[Stack.length - 1];`

`Stack.pop();`

`let o2 = Stack[Stack.length - 1];`

`Stack.pop();`

`// Use switch case to operate on o1`

`// and o2 and perform o1 O o2.`

`switch` `(exprsn[j]) {`

`case` `'+'``:`

`Stack.push(o1 + o2);`

`break``;`

`case` `'-'``:`

`Stack.push(o1 - o2);`

`break``;`

`case` `'*'``:`

`Stack.push(o1 * o2);`

`break``;`

`case` `'/'``:`

`Stack.push(o1 / o2);`

`break``;`

`}`

`}`

`}`

`return` `Stack[Stack.length - 1];`

`}`

`let exprsn = ``"+ 9 * 12 6"``;`

`document.write(evaluatePrefix(exprsn));`

`// This code is contributed by rameshtravel07.`

`</script>`

Output

`81`

Time Complexity: O(n)
Space Complexity: O(n)

RecommendedSolve DSA problems on GfG Practice.Solve Problems

My Personal Notesarrow_drop_up

## FAQs

### What is evaluation of prefix expression? ›

The evaluation of prefix expression requires a stack data structure. We will push the operators in the stack and then solve the expression. We will visit each element of the expression one by one. If the current element is an operand, we will push it to the stack.

### What determines order of evaluation of a prefix expression? ›

What determines the order of evaluation of a prefix expression? Explanation: Precedence is a very important factor in determining the order of evaluation. If two operators have the same precedence, associativity comes into action.

### How do you find the prefix of an expression? ›

Prefix: An expression is called the prefix expression if the operator appears in the expression before the operands. Simply of the form (operator operand1 operand2). Postfix: An expression is called the postfix expression if the operator appears in the expression after the operands.

### What is evaluation of expression in data structure? ›

For example, in expression a + b − c, both + and – have the same precedence, then which part of the expression will be evaluated first, is determined by associativity of those operators. Here, both + and − are left associative, so the expression will be evaluated as (a + b) − c.

### How do you evaluate the prefix infix? ›

We use the same to convert Infix to Prefix.
1. Step 1: Reverse the infix expression i.e A+B*C will become C*B+A. Note while reversing each '(' will become ')' and each ')' becomes '('.
2. Step 2: Obtain the “nearly” postfix expression of the modified expression i.e CB*A+.
3. Step 3: Reverse the postfix expression.
12 Aug 2022

### How do you evaluate a postfix expression? ›

Evaluation of Postfix Expression Using Stack:
1. Create a stack to store operands (or values).
2. Scan the given expression from left to right and do the following for every scanned element. If the element is a number, push it into the stack. ...
3. When the expression is ended, the number in the stack is the final answer.
22 Aug 2022

### What is the time complexity of evaluating a prefix expression? ›

Explanation: The time complexity of evaluation of infix, prefix and postfix expressions is O (N).

### Which direction of scanning is suitable for the evaluation of a prefix expression? ›

Right to Left direction of scanning is suitable for evaluation of prefix expression - Data Structure.

### Why prefix and postfix expression is required? ›

Need of Prefix and Postfix Notations:

Prefix notations are needed when we require operators before the operands while postfix notations are needed when we require operators after the operands. Prefix notations are used in many programming languages like LISP.

### How do you evaluate a prefix or postfix expression using stack? ›

Prefix notation

Reverse the given expression. Start scanning from left to right. If the current value is an operand, push it onto the stack. If the current is an operator, pop two elements from the stack, apply the at-hand operator on those two popped operands, and push the result onto the stack.

### How do you evaluate expressions using stack? ›

3.9 Evaluation of Prefix and Postfix expressions using stack - YouTube

### How do you convert expressions into prefix and postfix? ›

Prefix expression notation requires that all operators precede the two operands that they work on. Postfix, on the other hand, requires that its operators come after the corresponding operands. A few more examples should help to make this a bit clearer (see Table 2). A + B * C would be written as + A * B C in prefix.

### How do you evaluate an expression explain with example? ›

To evaluate an algebraic expression, you have to substitute a number for each variable and perform the arithmetic operations. In the example above, the variable x is equal to 6 since 6 + 6 = 12. If we know the value of our variables, we can replace the variables with their values and then evaluate the expression.

### What is evaluation of expression in DBMS? ›

In this method, the given expression evaluates one relational operation at a time. Also, each operation is evaluated in an appropriate sequence or order. After evaluating all the operations, the outputs are materialized in a temporary relation for their subsequent uses.

### What are the three different methods of evaluating arithmetic expressions? ›

Arithmetic Expressions can be written in one of three forms: Infix Notation: Operators are written between the operands they operate on, e.g. 3 + 4. Prefix Notation: Operators are written before the operands, e.g + 3 4. Postfix Notation: Operators are written after operands.

### How do you read the prefix notation? ›

Prefix notation puts the operator before all of the operands. This means that the precedence is always clear. We are going to compare the Infix notation to this Prefix notation. so to derive the prefix equivalent we simply take the operator then move from left to right reading off the operands in left-to-right order.

### How do you convert infix expressions into prefix expressions? ›

Rules for the conversion of infix to prefix expression:

First, reverse the infix expression given in the problem. Scan the expression from left to right. Whenever the operands arrive, print them. If the operator arrives and the stack is found to be empty, then simply push the operator into the stack.

### What is prefix notation? ›

Prefix Notation

In this notation, operator is prefixed to operands, i.e. operator is written ahead of operands. For example, +ab. This is equivalent to its infix notation a + b. Prefix notation is also known as Polish Notation.

### What is postfix evaluation in C? ›

Evaluation of postfix expression

If it is a digit then, push it on to the stack. If it is an operator then, pop out the top most two contents from the stack and apply the operator on them. Later on, push the result on to stack. If the input symbol is '\0', empty the stack.

### What is the value of the postfix expression 6 324 +-? ›

1 Answer. The best explanation: Postfix Expression is (6*(3-(2+4))) which results -18 as output.

### Which expression is most suitable for evaluation using stack? ›

Detailed Solution. Stack data structure is suitable for evaluating postfix expression.

### What is meant by postfix expression? ›

In a postfix expression, • an operator is written after its operands. • the infix expression 2+3 is 23+ in postfix notation. • For postfix expressions, operations are performed in the order in which they are written (left to right).

### What is the use of postfix expression? ›

Postfix notation, also known as RPN, is very easy to process left-to-right. An operand is pushed onto a stack; an operator pops its operand(s) from the stack and pushes the result. Little or no parsing is necessary. It's used by Forth and by some calculators (HP calculators are noted for using RPN).

### Why prefix and postfix expression is required? ›

Need of Prefix and Postfix Notations:

Prefix notations are needed when we require operators before the operands while postfix notations are needed when we require operators after the operands. Prefix notations are used in many programming languages like LISP.

### What will be the postfix expression for the given prefix expression? ›

Prefix expression notation requires that all operators precede the two operands that they work on. Postfix, on the other hand, requires that its operators come after the corresponding operands.
...
2.9. Infix, Prefix and Postfix Expressions.
Infix ExpressionPrefix ExpressionPostfix Expression
A + B+ A BA B +
A + B * C+ A * B CA B C * +

### What is other name for postfix expression? ›

What is the other name for a postfix expression? Explanation: Reverse polish Notation is the other name for a postfix expression whereas Polish Notation, Warsaw notation are the other names for a prefix expression.

### What is infix and postfix expression? ›

Infix expression: The expression of the form a op b. When an operator is in-between every pair of operands. Postfix expression: The expression of the form a b op. When an operator is followed for every pair of operands.

### What is prefix postfix and infix with example? ›

For example, The above expression can be written in the prefix form as / * A + B C D. This type of expression cannot be simply decoded as infix expressions.
...
Definition of Infix, Postfix, and Prefix.
InfixPrefixPostfix
A+B+ABAB+
A+B-C-+ABCAB+C-
(A+B)*C-D-*+ABCDAB+C*D-
6 Sept 2021

### What is the use of prefix expression? ›

Prefix and Postfix expressions can be evaluated faster than an infix expression. This is because we don't need to process any brackets or follow operator precedence rule. In postfix and prefix expressions which ever operator comes before will be evaluated first, irrespective of its priority.

### What is the advantage of postfix expression? ›

Postfix has a number of advantages over infix for expressing algebraic formulas. First, any formula can be expressed without parenthesis. Second, it is very convenient for evaluating formulas on computers with stacks. Third, infix operators have precedence.

### Why do we use infix and postfix expression? ›

Infix expressions are readable and solvable by humans. We can easily distinguish the order of operators, and also can use the parenthesis to solve that part first during solving mathematical expressions. The computer cannot differentiate the operators and parenthesis easily, that's why postfix conversion is needed.

### What is the advantage of using postfix or prefix expression over infix expression? ›

Explanation: Postfix expression is simple to execute as a comparison to the infix expression it required more operation to execute. In the postfix expression, the overhead of brackets is not there while in the infix expression the overhead of brackets is there.

### What is prefix notation? ›

Prefix Notation

In this notation, operator is prefixed to operands, i.e. operator is written ahead of operands. For example, +ab. This is equivalent to its infix notation a + b. Prefix notation is also known as Polish Notation.

### What do you mean by prefix notation? ›

Definitions of prefix notation. a parenthesis-free notation for forming mathematical expressions in which each operator precedes its operands.

### How do you read the prefix notation? ›

Prefix notation puts the operator before all of the operands. This means that the precedence is always clear. We are going to compare the Infix notation to this Prefix notation. so to derive the prefix equivalent we simply take the operator then move from left to right reading off the operands in left-to-right order.

### How do I convert infix to postfix and prefix manually? ›

To convert an infix expression to postfix, you simply place each operator in the infix expression immediately to the right of its respective right parenthesis. Then you rewrite the expression in the new order, and what you get is the same expression in prefix notation.

## Videos

1. Evaluation of Prefix Expression using Stack in Data structure & Algorithms Hello World Hindi
(Hello World)
2. Evaluate an array expression with numbers, + and – | GeeksforGeeks
(GeeksforGeeks)
3. Evaluation of Prefix expression (step by step using stack) | Data Structure
(Emmy's Lectures)
4. Stack | Set 4 (Evaluation of Postfix Expression) | GeeksforGeeks
(GeeksforGeeks)
5. Evaluate Prefix Expression
(Mrunal Fatangare)
6. Evaluation of posfix expression | Love Babbar DSA sheet

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