Can you please add a screenshot of the output in pep/9 Translate the following C program into Pep/9 assembly language.#include int minimum(int j1, int j2){if (jl < j2){return jl;}else{return j2;}}int main(){int n, m;scanf("%d %d", &n, &m);printf("Minimum: %d\n", minimum(n, m));return 0;}Draw the stack frame when the code is executing within the Function minimum()

Here i give the solution in pep/9 assembly language: DATA SEGMENT N1 DB ? N2 DB ?…

Translate the following C program into Pep/9 assembly language. #include int minimum(int j1, int j2) { if (jl < j2) { return j1; } else { return j2; } } int main() { int n, m; scanf("%d %d", &n, &m); printf("Minimum: %d\n", minimum(n, m)); return 0; } Draw the stack frame when the code is executing within the Function minimum()

Translate the following C program into Pep/9 assembly language. #include int minimum(int j1, int j2) { if (jl < j2) { return j1; } else { return j2; } } int main() { int n, m; scanf("%d %d", &n, &m); printf("Minimum: %d\n", minimum(n, m)); return 0; } Draw the stack frame when the code is executing within the Function minimum()

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

See similar textbooks

Similar questions

int func(int a, int b) { return (a
void fact(int num) { int k,f=1; for (k=1;k<=num; k++) f=f*k; cout<< f; int main() { int arri[3]={3,2,1} ; for(int i=0;i<3;i++) fact(arri[i]); } 261 O 621 O 7 462
int f(int &k){k++;return k * 2;}int main(){int i = 1, j = -1;int a, b, c;a = f(i) + i/2;b = j + f(j) + f(j);c = 2 * f(j);return 0;} What are the values of a, b and c id the operands in the expressions are evaluated from left to right and then what are the values when its evaluated right to left?
int main() { E(); C(); E(); } void B() {...} void C() {..} void D() {..} void E() { B(); D(); } A variable declared in function B and only in function B is accessible in?
debug c language main() { int m,n; for(i 0; i << 5; i++) { printf("\t\t\t\t"); for(j 0; j << 5; j+++) printf("*"); printf("/n"); }
#include using namespace std; bool isPalindrome(int x) { int n=0,val; val = x; while(x > 0) { n = n * 10 + x % 10; x = x / 10; } } int main() { int n; cin >>n; if(isPalindrome(n)) { cout <
IN THE C PROGRAMMING LANGUAGE j = 15; k = 9 // what are the values for n,m,and p after these operations: --j ; k++ ; n = j - ++k; m = j-- + k--; p = k + j;
int main() { E(); C(); E(); } void B() {...} void C() {...} void D() {...} void E() { B(); D(); } Which describes the order in which the functions are called when the program is run?
COnvert to C# Language only def Deposit(balance, pin): # Deposit functionp = int(input("Enter the PIN: "))if p == pin:amount = float(input("Enter deposit amount: "))balance += amountprint("You successfully deposited an amount of ", amount)else:print("Incorrect PIN!!")return balancedef Withdraw(balance, pin): # withdrawp = int(input("Enter the PIN: "))if p == pin:amount = float(input("Enter withdraw amount: "))if amount > balance:print("Insufficient Balance")else:balance -= amountprint("You successfully withdraw ", amount)else:print("Incorrect PIN!!")return balancedef Inquire(balance, pin): # Inquire functionp = int(input("Enter the PIN: "))if p == pin:print("The outstanding balance is =", balance)else:print("Incorrect PIN!!")# main program# taking initial amount from userinitial_amount = float(input("Enter initial amount: "))while True:pin = int(input("Enter a 4 digit PIN: "))if len(str(pin)) != 4:print("Invalid PIN!! PIN must be of 4 digit.")else:breakprint("\n**** SELECT…
CONVERT 1 & 2 from C++ to Python 1.NON RECURSIVE #include <iostream>using namespace std; int main(){ int days,buy_on_this_day ,sell_on_this_day; cout<<"Enter number of days: "; cin>>days; int stock_price[days]; for(int i=0;i<days;i++) { cin>>stock_price[i]; } int i=0; for(int i=0;i<days-1;i++) { //using this line we compare current price with the next day //price and find the local minima while(i<days-1 && stock_price[i+1]<=stock_price[i]) i++; if(i==days-1) break; buy_on_this_day =i++; while(i<days && stock_price[i]>= stock_price[i-1]) i++; sell_on_this_day =i-1; cout<<buy_on_this_day <<" : index of the change before which we buy"<<endl; cout<<sell_on_this_day<<" :index of the change before which we sell"<<endl; } int profit; int maxim =0; int…
// Return remainder after integer division // Global variables int a = 4; int b = 30; int main() { // Use registers for local variables int remainder; int divisor; } if (a > b) { } else { } remainder = a; divisor = b; } remainder = b; divisor = a; while (remainder > divisor) { remainder -= divisor; return remainder;
Describe the below c program. #include <stdio.h>#include <stdlib.h> int get_1st_weekday(int year){ int d;d = (((year - 1) * 365) + ((year - 1) / 4) - ((year - 1) / 100) + ((year) / 400) + 1) % 7;return d;} int main(){system("Color 3F");int year,month,day,daysInMonth,weekDay=0,startingDay;printf("\nEnter your desired year:");scanf("%d",&year); char *months[]={"January","February","March","April","May","June","July","August","September","October","November","December"};int monthDay[]={31,28,31,30,31,30,31,31,30,31,30,31}; if((year%4==0&&year%100!=0)||year%400==0)monthDay[1]=29; startingDay=get_1st_weekday(year); for(month=0;month<12;month++){ daysInMonth=monthDay[month];printf("\n\n---------------%s-------------------\n",months[month]);printf("\n Sun Mon Tue Wed Thurs Fri Sat\n"); for(weekDay=0;weekDay<startingDay;weekDay++)printf(" "); for(day=1;day<=daysInMonth;day++){printf("%5d",day); if(++weekDay>6){printf("\n");weekDay=0;}startingDay=weekDay;}…