# Evaluation of Prefix Expressions

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 end
Step 2: If character at P is an operand push it to Stack
Step 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 Stack
Step 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 it
Step 6: End
```

Example to demonstrate working of the algorithm

```Expression: +9*26

Character | Stack       |  Explanation
Scanned   | (Front to   |
|  Back)      |
-------------------------------------------
6           6             6 is an operand,
push to Stack
2           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 Stack

Result: 21
```

Examples:

```Input : -+8/632
Output : 8

Input : -+7*45+20
Output : 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.

## CPP

 `// CPP program to evaluate a prefix expression. ` `#include ` `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 Stack = ``new` `Stack(); ` ` `  `        ``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 `

## 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 Stack = ``new` `Stack(); ` ` `  `        ``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 */`

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.

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