Infix Expression: In infix expression, an operator is placed between the two operands. Example: x + y, here operator + is placed between operands x and y.
Postfix Expression: In postfix expression, an operator is placed after the operands. Example: xy+, here operator + is placed after the operands x and y.
Algorithm for the conversion from infix to postfix
- Read the expression from user.
- Scan the expression character by character, if the character is alphabet or number print it to the console
- If the expression is operator then,
- If the precedence of current character operator is greater than top of the stack (or the stack is empty) then push an operator into the stack.
- If the precedence of current character operator is less than the top of stack, pop all elements from the stack whose precedence is greater than or equal to the current element.
- If the operator is ‘(‘, push it on stack.
- If the operator is ‘) ‘, pop all elements from stack until we get ‘(‘and also remove ‘(‘and ‘) ‘operator from the stack.
- Repeat the steps from 3 to 6 until the expression becomes empty.
- Print the output to screen
- Pop all elements from the stack and print the output to console.
Consider the below code:
int top = -1;
void push(char x)
stack[++top] = x;
if(top == -1)
int priority(char x)
if(x == '(')
if(x == '+' || x == '-')
if(x == '*' || x == '/')
char *e, x;
printf("Enter the expression : ");
e = exp;
printf("Postfix expression : ");
while(*e != '\0')
else if(*e == '(')
else if(*e == ')')
while((x = pop()) != '(')
printf("%c ", x);
while(priority(stack[top]) >= priority(*e))
while(top != -1)
Enter the expression: a*(b+c)-d
Postfix expression: a b c + * d –
Explanation of Code:
- As the program execution starts from the main(), in main() firstly we are reading expression in exp character array. Assigning this char array to pointer variable e.
- We are scanning the expression character by character and checking if the character is alphabet or a number then print that character to the screen.
- Next, we are checking if the character is ‘(‘then push it into the stack.
- If the character is ‘) ‘, then pop all elements from the stack and print them to the screen until we get ‘) ‘and remove the both parenthesis ‘(‘and ‘) ‘.
- Next checking is if the character is operator then check if priority of current operator is greater than top of stack element then push the current character into the stack. Else if, the priority of current operator is less than top of stack then pop all elements with higher priority from the stack, print them and then push current operator into the stack.
- Move to the next character and repeat the steps from 2 to 5 until the expression ends.
- Pop all elements from the stack and print them to console. This is the functionality of main().
- As we are using stack in this solution, we had to write the code for stack functionalities. Stack works on last in first out property and top is the pointer which points to the top element of stack.
- In push(), insert the element into the stack and increment the top by 1.
- In pop(), return the top most element and decrement the top by 1.
- Next priority(), returns the priority of operators.