1. A Simple Conversion

1. Input: 3 + 4
2. Push 3 to the output queue (whenever a number is read it is pushed to the output)
3. Push + (or its ID) onto the operator stack
4. Push 4 to the output queue
5. After reading the expression, pop the operators off the stack and add them to the output.

   In this case there is only one, "+".

6. Output: 3 4 +

This already shows a couple of rules:

• All numbers are pushed to the output when they are read.
• At the end of reading the expression, pop all operators off the stack and onto the output.

2. Graphical Illustration

Shunting yard. https://handwiki.org/wiki/index.php?curid=2019399

Graphical illustration of algorithm, using a three-way railroad junction. The input is processed one symbol at a time: if a variable or number is found, it is copied directly to the output a), c), e), h). If the symbol is an operator, it is pushed onto the operator stack b), d), f). If the operator's precedence is lower than that of the operators at the top of the stack or the precedences are equal and the operator is left associative, then that operator is popped off the stack and added to the output g). Finally, any remaining operators are popped off the stack and added to the output i).

3. The Algorithm in Detail

/* This implementation does not implement composite functions, functions with variable number of arguments, and unary operators. */ while there are tokens to be read: read a token if the token is: - a number: put it into the output queue - a function: push it onto the operator stack - an operator o1: while ( there is an operator o2 other than the left parenthesis at the top of the operator stack, and (o2 has greater precedence than o1 or they have the same precedence and o1 is left-associative) ): pop o2 from the operator stack into the output queue push o1 onto the operator stack - a left parenthesis (i.e. "("): push it onto the operator stack - a right parenthesis (i.e. ")"): while the operator at the top of the operator stack is not a left parenthesis: {assert the operator stack is not empty} /* If the stack runs out without finding a left parenthesis, then there are mismatched parentheses. */ pop the operator from the operator stack into the output queue {assert there is a left parenthesis at the top of the operator stack} pop the left parenthesis from the operator stack and discard it if there is a function token at the top of the operator stack, then: pop the function from the operator stack into the output queue /* After the while loop, pop the remaining items from the operator stack into the output queue. */ while there are tokens on the operator stack: /* If the operator token on the top of the stack is a parenthesis, then there are mismatched parentheses. */ {assert the operator on top of the stack is not a (left) parenthesis} pop the operator from the operator stack onto the output queue 

To analyze the running time complexity of this algorithm, one has only to note that each token will be read once, each number, function, or operator will be printed once, and each function, operator, or parenthesis will be pushed onto the stack and popped off the stack once—therefore, there are at most a constant number of operations executed per token, and the running time is thus O(n)—linear in the size of the input.

The shunting yard algorithm can also be applied to produce prefix notation (also known as Polish notation). To do this one would simply start from the end of a string of tokens to be parsed and work backwards, reverse the output queue (therefore making the output queue an output stack), and flip the left and right parenthesis behavior (remembering that the now-left parenthesis behavior should pop until it finds a now-right parenthesis). And changing the associativity condition to right.

4. Detailed Example

Input: 3 + 4 × 2 ÷ ( 1 − 5 ) ^ 2 ^ 3

Operator	Precedence	Associativity
^	4	Right
×	3	Left
÷	3	Left
+	2	Left
−	2	Left

The symbol ^ represents the power operator.

Token	Action	Output (in RPN)	Operator stack	Notes
3	Add token to output	3
+	Push token to stack	3	+
4	Add token to output	3 4	+
×	Push token to stack	3 4	× +	× has higher precedence than +
2	Add token to output	3 4 2	× +
÷	Pop stack to output	3 4 2 ×	+	÷ and × have same precedence
	Push token to stack	3 4 2 ×	÷ +	÷ has higher precedence than +
(	Push token to stack	3 4 2 ×	( ÷ +
1	Add token to output	3 4 2 × 1	( ÷ +
−	Push token to stack	3 4 2 × 1	− ( ÷ +
5	Add token to output	3 4 2 × 1 5	− ( ÷ +
)	Pop stack to output	3 4 2 × 1 5 −	( ÷ +	Repeated until "(" found
	Pop stack	3 4 2 × 1 5 −	÷ +	Discard matching parenthesis
^	Push token to stack	3 4 2 × 1 5 −	^ ÷ +	^ has higher precedence than ÷
2	Add token to output	3 4 2 × 1 5 − 2	^ ÷ +
^	Push token to stack	3 4 2 × 1 5 − 2	^ ^ ÷ +	^ is evaluated right-to-left
3	Add token to output	3 4 2 × 1 5 − 2 3	^ ^ ÷ +
end	Pop entire stack to output	3 4 2 × 1 5 − 2 3 ^ ^ ÷ +

Input: sin ( max ( 2, 3 ) ÷ 3 × π )

Token	Action	Output = (in RPN)	Operator stack	Notes
sin	Push token to stack		sin
(	Push token to stack		( sin
max	Push token to stack		max ( sin
(	Push token to stack		( max ( sin
2	Add token to output	2	( max ( sin
,	Ignore	2	( max ( sin
3	Add token to output	2 3	( max ( sin
)	Pop stack to output	2 3	( max ( sin	Repeated until "(" is at the top of the stack
Pop stack	2 3	max ( sin	Discarding matching parentheses
Pop stack to output	2 3 max	( sin	Function at top of the stack
÷	Push token to stack	2 3 max	÷ ( sin
3	Add token to output	2 3 max 3	÷ ( sin
×	Pop stack to output	2 3 max 3 ÷	( sin
	Push token to stack	2 3 max 3 ÷	× ( sin
π	Add token to output	2 3 max 3 ÷ π	× ( sin
)	Pop stack to output	2 3 max 3 ÷ π ×	( sin	Repeated until "(" is at the top of the stack
	Pop stack	2 3 max 3 ÷ π ×	sin	Discarding matching parentheses
	Pop stack to output	2 3 max 3 ÷ π × sin		Function at top of the stack
end	Pop entire stack to output	2 3 max 3 ÷ π × sin

The content is sourced from: https://handwiki.org/wiki/Shunting-yard_algorithm