Please solve and show work. Thank you. What is the MIPS instructions corresponding to this C code: Int leaf_example (int g, h, i, j) { int f; f = (g - h) - (i + j); f +=2 return f; } Arguments g, …, j in $a0, …, $a3, f in $s0 (hence, need to save $s0 on stack), Result in $v0
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Please solve and show work. Thank you.
- What is the MIPS instructions corresponding to this C code:
Int leaf_example (int g, h, i, j)
{ int f;
f = (g - h) - (i + j);
f +=2
return f;
}
Arguments g, …, j in $a0, …, $a3, f in $s0 (hence, need to save $s0 on stack), Result in $v0
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- 5. Evaluate the following postfix expression. (a + b) * c - (d - e) / f The postfix will be a. abcde f + * - -/ b. a b +c *de-f/- Write step by step procedure of solving this problem. Your Answer is (a or b) Use the above stack and fill (push) in operators, +, -, *, /, and 'C'. Type "pop" beside the operators or '(' when you need to pop it out. Finally, convert the expression from infix to the postfix. Submit Type here to searchQuestion 2: Given a balanced expression that can contain opening and closing parenthesis, check if it contains any duplicate parenthesis or not. Example: Input: ((x+y))+z it should give an error Note: for above questions implement stack on your own rather than built in stack. C++Complete the following ():a. A stack is used by the system when a function call is madeb. A stack can become full during program execution
- Draw the sequence of stack configurations in the evaluation of the following postfix expression: 3 5 5 * + 4 7 + / 6 *SOLVE THE MAZE FUNCTION USING STACK.PLEASE ONLY C CODE,DONT USE C++ AND C#. /* R\C 0 1 2 3 4 5 6 7 *//* 0 */ {0, 0, 0, 1, 0, 0, 0, 0},/* 1 */ {0, 1, 1, 1, 0, 0, 1, 0},/* 2 */ {0, 1, 0, 1, 0, 0, 1, 0},/* 3 */ {0, 1, 0, 1, 1, 1, 1, 0},/* 4 */ {0, 1, 0, 0, 0, 0, 1, 1},/* 5 */ {0, 1, 1, 0, 1, 1, 1, 0},/* 6 */ {0, 0, 0, 0, 1, 0, 0, 0},/* 7 */ {0, 0, 0, 0, 1, 0, 0, 0}, #include <stdio.h>#include <stdlib.h>#include <string.h> typedef struct LINKED_STACK_NODE_s *LINKED_STACK_NODE; typedef struct LINKED_STACK_NODE_s{LINKED_STACK_NODE next;void *data;} LINKED_STACK_NODE_t[1]; typedef struct LINKED_STACK_s{LINKED_STACK_NODE head;int count;} LINKED_STACK_t[1], *LINKED_STACK; typedef struct{int R;int C;} POS_t[1], *POS; LINKED_STACK stack_init();void stack_free(LINKED_STACK stack);void stack_push(LINKED_STACK stack, void *data);void *stack_pop(LINKED_STACK stack);void *stack_top(LINKED_STACK stack);int is_empty(LINKED_STACK stack); int is_empty(LINKED_STACK stack){return…The following is the postfix evaluation code for your reference. Explain the stack status for evaluating the following postfix expression 3 4 5 + *
- Solve the maze function using stack.PLEASE ONLY C CODE, DONT USE C++ AND C# . /* R\C 0 1 2 3 4 5 6 7 *//* 0 */ {0, 0, 0, 1, 0, 0, 0, 0},/* 1 */ {0, 1, 1, 1, 0, 0, 1, 0},/* 2 */ {0, 1, 0, 1, 0, 0, 1, 0},/* 3 */ {0, 1, 0, 1, 1, 1, 1, 0},/* 4 */ {0, 1, 0, 0, 0, 0, 1, 1},/* 5 */ {0, 1, 1, 0, 1, 1, 1, 0},/* 6 */ {0, 0, 0, 0, 1, 0, 0, 0},/* 7 */ {0, 0, 0, 0, 1, 0, 0, 0}, void solve_maze(){int M[8][8] = {/* R\C 0 1 2 3 4 5 6 7 *//* 0 */ {0, 0, 0, 1, 0, 0, 0, 0},/* 1 */ {0, 1, 1, 1, 0, 0, 1, 0},/* 2 */ {0, 1, 0, 1, 0, 0, 1, 0},/* 3 */ {0, 1, 0, 1, 1, 1, 1, 0},/* 4 */ {0, 1, 0, 0, 0, 0, 1, 1},/* 5 */ {0, 1, 1, 0, 1, 1, 1, 0},/* 6 */ {0, 0, 0, 0, 1, 0, 0, 0},/* 7 */ {0, 0, 0, 0, 1, 0, 0, 0},};int I, J, R = 0, C = 3;LINKED_STACK stack;POS pos = (POS)malloc(sizeof(POS));int flag;stack = stack_init(); do{// TODO: Fill this block.} while (R != 7 && C != 7 && R != 0 && C != 0);}int main(){solve_maze();return 0;}Requirements:1. Evaluate infix expressions below2. Convert each infix expression to a postfix expression and evaluate it, find the result.3. Show Stack operations.Given: infix expressions3. 18 != 12 < 22 % 3 || !(55 == 5 * 5)Do not copy from chegg and do it in c++ Write a program to convert an infix expression to a postfix expression. Input: an infix expression Output: the corresponding postfix expression Note: You must use the stack to implement this task. And you can use any one of the above stack implementations. Test your program using the following expressions: (a + b) * c (a - b) / (c +d) (a*b+c/d)/e
- C++ Programming, Stack Queues and Deque EXPLAIN THE FLOW OF THIS MAIN.CPP CODE DO NOT USE FLOW CHART. The code is already correct you just need to explain the flow if the code (See attached photo for the problem) int main(int argc, char** argv) { SLLStack* stack = new SLLStack(); int test; int length; string str; char top; bool flag = true; cin >> test; switch (test) { case 0: getline(cin, str); length = str.length(); for(int i = 0; i < length; i++){ if(str[i] == '{' || str[i] == '(' || str[i] == '['){ stack->push(str[i]); } else if (str[i] == '}' || str[i] == ')' || str[i] == ']'){ if(!stack->isEmpty()){ top = stack->top(); if(top == '{' && str[i] == '}' || top == '(' && str[i] == ')' || top == '[' && str[i] == ']'){…Do not copy from chegg and do it in c++ Write a program to evaluate a postfix expression. Input: a postfix expression. E.g. 3 5 + Output: the result of the expression. E.g. the result for the above input is 8. Note: You must use the stack to implement this task. And please give the stack solutions for both the following implementations: Implement the array-based stack Implement the pointer-based stack Test your program using the following expressions: 3 5 + 3 5 + 6 * 3 5 6 + * 3 5 6 * 8 – 2 / + 12 20 + 16 /Evaluate the following statement using stack : Z = mod ( 10 , 3 ) ; Where mod ( ) defined as follows : int mod ( int x , int y ) { int m ; if ( x < y ) return x ; else { m = mod ( x - y , y ); return 0 + m ; } }