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Convert the following c++ code into pep9 assembly language #include <iostream> using namespace std; void times(int& prod, int mpr, int mcand) { prod = 0; while (mpr != 0) { if (mpr % 2 == 1) prod = prod + mcand; mpr /= 2; mcand *= 2; } } int
* The output of the following code is #include <iostream> using namespace std: int prod (int x, int y=3){ int d; int a=x*y; return (a); int main ({ cout << prod (7) <<" "<<prod (6, 4); return 0; 21 12 O 49 24 O 21 24 O 21 18
Convert following code to implement linked list C++ language #include<iostream> using namespace std; int top = -1; //globally defining the value of top, as the stack is empty void push(int stack[], int x, int n) { if (top == -1) //if top position is the last of posiition of stack,means stack is full { cout << "Stack is full Overflow condition"; } else { top = top + 1; //incrementing top position stack[top] = x; //inserting element on incremented position...
How convert this c++ into C #include<iostream> #include <stdlib.h> using namespace std; void multiplication() { int num1; int c, num2, ans; cout<<"Enter difficulty level(1/2)\n"; cin>>c; if(c==1) { num1= rand() % 10; num2= rand() % 10; cout<<"what is"<<num1 <<"*"<<num2<<"?\n" ; cin>>ans while(ans!=(num1*num2)) { if(ans!=(num1*num2)) { cout<< "No. Please try again.\n"; cout<<"what is"<<num1 <<"*"<<num2<<"?\n" ; cin>>ans; } } } else if(c==2) { num1= rand() % 10+90; num2= rand() % 10+90; cout<<"what is"<<num1 <<"*"<<num2<<"?\n" ; cin>>ans; while(ans!=(num1*num2)) { if(ans!=(num1*num2)) { cout<< "No. Please...
How to convert C++ code to C #include<iostream> #include <stdlib.h> using namespace std; void multiplication() { int num1; int c, num2, ans; cout<<"Enter difficulty level(1/2)\n"; cin>>c; if(c==1) { num1= rand() % 10; num2= rand() % 10; cout<<"what is"<<num1 <<"*"<<num2<<"?\n" ; cin>>ans while(ans!=(num1*num2)) { if(ans!=(num1*num2)) { cout<< "No. Please try again.\n"; cout<<"what is"<<num1 <<"*"<<num2<<"?\n" ; cin>>ans; } } } else if(c==2) { num1= rand() % 10+90; num2= rand() % 10+90; cout<<"what is"<<num1 <<"*"<<num2<<"?\n" ; cin>>ans; while(ans!=(num1*num2)) { if(ans!=(num1*num2)) { cout<< "No....
Consider the following C++ program: #include <iostream> using namespace std; void f1(int); void f2(int); void f3(int); int main() { f1(10); return 0; } void f1(int n) { f2(n + 5); } void f2(int n) { f3(n - 2); } void f3(int n) { cout << n << endl; // LINE 1 } Just before the program statement marked with the comment "LINE 1" is executed, how many stack frames will be on the program call stack?
please write this in "MARIE assembly language" #include <iostream> using namespace std; int DivideByTwo(int, int); // Data section int Data[] = { 0x0102, 0x0105, 0x0106, 0x0108, 0x011A, 0x0120, 0x0225, 0x0230, 0x0231, 0x0238, 0x0339, 0x0350, 0x0459, 0x055F, 0x066A, 0x0790, 0x08AB, 0x09AF, 0x0AB9, 0x0BBD, 0x0CC1, 0x0DCA, 0x0EFE, 0x0FFE }; int main() { int* BAddr = &Data[0]; int* EAddr = &Data[23]; int Count = 24; // the number of Data int Ffff = 0xffff; // value for "not found" int num; // input...
#include using namespace std; void test(); int x = 0; int main(){ cout << x << endl; test(); cout << x << endl; } void test(){ int x = 100; cout << x << endl; for(int x = 0; x < 5; x++){ cout << x << endl; } } how many separate variables named ‘x’ are created?
Convert the following code to assembly MIPS void checki() { int i = 0; while(i < 10) { switch(i) { case 1: i += 5; break; case 2: i --; break; case 3: i += 2; break; default: i++; } } } void main() { checki(); }
Convert the below code into if else selection: #include <iostream> using namespace std; int main() { int num; sin. >> num; switch (num) { case 1: cout << "Casel: Value is: << num << endl; break; case 2: break; case 3: cout << "Case3: Value is: " << num << endl; break; default: cout << "Default: Value is: << num << endl; break; } return; }
#include <iostream> using namespace std; struct node { int base=0; int power=0; }; void insert(node ptr[],int basee,int powerr) { ptr[powerr].power=powerr; ptr[powerr].base=basee; } void subtract(node ptr1[],int size1,node ptr2[],int size2,node ptr3[]) { for(int j=0;j<=size1;j++) { if(ptr1[j].base!=0) { ptr3[j].base=(ptr3[j].base)+(ptr1[j].base); ptr3[j].power=ptr2[j].power; } } for(int j=0;j<=size2;j++) { if(ptr2[j].base!=0) { ptr3[j].base=(ptr3[j].base)-(ptr2[j].base); ptr3[j].power=ptr2[j].power; } } } void addition(node ptr1[],int size1,node ptr2[],int size2,node ptr3[]) { for(int j=0;j<=size1;j++) { if(ptr1[j].base!=0) { ptr3[j].base=(ptr3[j].base)+(ptr1[j].base); ptr3[j].power=ptr2[j].power; } } for(int j=0;j<=size2;j++) { if(ptr2[j].base!=0) { ptr3[j].base=(ptr3[j].base)+(ptr2[j].base); ptr3[j].power=ptr2[j].power; } } } void display(node ptr[],int size)...