1. Convert infix to postfix: Infix: * b + (c - d) Postfix: a
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- Q: Write the code in c++ of Stack By Link List and Perform MultipleTask which are given below. Add all the given tasks in the code 1. Push value in stack2. Check that Stack is Empty or Not.3. POP values From Stack.4. Search any values from stack.5. Update any value from stack. NOTE: STACK BY LINKLISTGiven the following infix expression: (A + B) / (C + (D - E)) Convert from infix to postfix using stack. You must show all states of the stack during the conversion process. Infix expression STACK (head to the left) Output (A + B) / (C + (D - E))stack permutation of the numbers 0, 1, ..., n-1 is a permutation obtained as the sequence of numbers output by a stack after the original numbers have been processed left-to-right through a stack (via legal push and pop operations). For example, 0, 3, 2, 1 is a stack permutation of 0, 1, 2, 3 obtained by the sequence of operations: push, pop, push, push, push, pop, pop, pop. However, the permutation 1, 3, 0, 2 is not a stack permutation of 0, 1, 2, 3 Which of the following sequences are not stack permutations of 0,1,2,3,4,5,6,7,8,9? There could be more than one answer: select all those that you think are correct. There is a penalty for incorrect choices. Group of answer choices 4, 6, 8, 7, 5, 3, 2, 9, 0, 1 4, 3, 2, 1, 0, 9, 8, 7, 6, 5 2, 5, 6, 7, 4, 8, 9, 3, 1, 0 0, 4, 6, 5, 3, 8, 1, 7, 2, 9
- Q: Write the code in c++ of Stack By Link List and Perform MultipleTask which are given below. 1. Push value in stack2. Check that Stack is Empty or Not.3. POP values From Stack.4. Search any values from stack.5. Update any value from stack. NOTE: STACK BY LINKLISTPostfix using StacksThe rules to convert an infix expression into an equivalent postfix expression are as follows: Suppose infx represents the expression and pfx represents the postfix expression. The rules to convert infx into pfx are as follows: 1. Initialize pfx to an empty expression and also initialize the stack.2. Get the next symbol, sym, from infx. a. if sym is an operand, append sym to pfx b. if sym is (. push sym into the stack. c. if dym is ). pop snd append all the symbols from the stack until the most recent left parenthesis. Pop and discard the left parentheses. d. If sym is an operator: i. Pop and append all of the operators from the stack to pfx that are above the most recent left parenthesis and have precededence greater than or equal to sym ii. Push sym onto the stack. 3. After processing infx, some operators might be left in the stack. Pop and append to pfx everything from the stack. In this program, you will…Q: Write the code in c++ of Stack By Link List and Perform MultipleTask which are given below. 1. Push value in stack2. Check that Stack is Empty or Not.3. POP values From Stack.4. Search any values from stack.5. Update any value from stack.
- Passing Arguments to a FunctionThe sample program in Example 5.4 shows how the stack frame is used to passarguments to a function.The code simply creates some local stack variables, fillsthem with values, and passes them to a function called callex().The callex() function takes the supplied arguments and prints them to the screen.Example 5.4 Stack and Passing Parameters to a Function/* stack2.c */#include <stdlib.h>#include <stdio.h>int callex(char *buffer, int int1, int int2){/* This prints the input variables to the screen:*/printf("%s %d %d\n",buffer,int1, int2);return 1;}int main(int argc, char **argv){Stack Overflows • Chapter 5 173 char buffer[15]="Hello Buffer"; /* a 15-byte character bufferwith12 characters filled/*int int1=1, int2=2; /* two four-byte integers */callex(buffer,int1,int2); /*call our function*/return 1; /*leaves the main function*/}You need to compile this example in MSVC in a console application inRelease mode or in GCC without optimizations. Example…Question 11 mah.c- Write a program to perform the following operations on a stack data structure. c.1 Create push and pop functions and display the items [12,9,3,25,19,0,5,30,9] after performing push and then pop operation. c.2 Write a function to convert an infix expression (2 – (1 / 5 ) * 6 + 12 ) * 33 % 9) to postfix expression and display the result. c.3 Applying best practices for writing code comments I want the code in c++ language Full explain this question and text typing work only We should answer our question within 2 hours takes more time then we will reduce Rating Dont ignore this line56. What is the stack pop operation? Group of answer choices The pop operation reads and discards the next character from the input stream. An istream processes characters in the order they are received. The pop operation reads and discards the next character from the input stream. A stack processes characters in the LIFO fashion. The pop operation reads and discards the first character from the input stream. A stack processes characters in the LIFO fashion. The pop operation reads and discards the last character from the input stream. An istream processes characters in the order they are received. The stack top operation reads and discards the next character from the input stream. An istream processes characters in the order they are received.
- Write the code in assebly language code that can be read in LC3 simulators! .ORIG x3000 ; Set up the user stack: LD R6, STKBASE ; Sets up the stack by loading the base address into R6 ; Prompt for the first string: LEA R0, PRMPT1 ; Loads the address of PRMPT1 into R0 PUTS ; Call GETS to get first string: LEA R0, STRNG1 ; Loads the address of STRNG1 ADD R6, R6, #-1 ; Push the address to store the string at STR R0, R6, #0 JSR GETSP ; Call GETSP ADD R6, R6, #2 ; Clean up (pop parameter & return value) ; Prompt for second string: LEA R0, PRMPT2 ; Loads the address of PRMPT2 PUTS ; Call GETS to get second string: LEA R0, STRNG2 ; Loads the address of STRNG2 ADD R6, R6, #-1 ; Push the second address to store the string at STR R0, R6, #0 JSR GETSP ; Call GETSP ADD R6, R6, #2 ; Clean up (pop parameter & return value) ; Output both strings: LEA R0, OUT1 ; First string... PUTS LEA R0, STRNG1 PUTSP LD R0, LF ;…54. What is the difference between the top and pop operations of a stack? Group of answer choices The top operation removes the top item of the stack. The pop operation returns a reference to the top item on the stack and remove it. The top operation removes the top item of the stack. The pop operation returns a reference to the top item on the stack without removing it. The pop operation removes the top item of the stack. The top operation returns a reference to the top item on the stack without removing it. The pop and top operations are doing the same thing which removes the top item of the stack. The pop operation removes the top item of the stack. The top operation returns a reference to the top item on the stack and remove it.Given the following sequence of characters: DA&T!A ST*R&U!CT!&UR*ES. Consider the stack data structure, supporting two operations push and pop. Suppose that for the above sequence, each letter (such as DATA STRUCTURES) corresponds to a push of that letter onto the stack, each exclamation mark(!) corresponds one pop operation on the stack, each ampersand (&) corresponds one peep operation on the stack and each asterisk (*) corresponds two pop operations on the stack. Show the sequence of values returned by the pop operations. Also, display the remaining elements in the stack after all the push and pop operations.