3. Write a code in AVR to load the value $15 into ocation $67 and add it to R19 five times and place he results in R19 as added. R19 should be zero pefore the addition starts. ! Write a code in AVR to load the value $15 into
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- You are required to make changes in the below programs and introduce the use of compaction where required. #include<stdio.h> #include<conio.h> main() { int ms, bs, nob, ef,n, mp[10],tif=0; int i,p=0; clrscr(); printf("Enter the total memory available (in Bytes) -- "); scanf("%d",&ms); printf("Enter the block size (in Bytes) -- "); scanf("%d", &bs); nob=ms/bs; ef=ms - nob*bs; printf("\nEnter the number of processes -- "); scanf("%d",&n); for(i=0;i<n;i++) { printf("Enter memory required for process %d (in Bytes)-- ",i+1); scanf("%d",&mp[i]); } printf("\nNo. of Blocks available in memory -- %d",nob); printf("\n\nPROCESS\tMEMORY REQUIRED\t ALLOCATED\tINTERNAL FRAGMENTATION"); for(i=0;i<n && p<nob;i++) { printf("\n %d\t\t%d",i+1,mp[i]); if(mp[i] > bs) printf("\t\tNO\t\t---"); else { printf("\t\tYES\t%d",bs-mp[i]);tif = tif + bs-mp[i]; p++; } } if(i<n) printf("\nMemory is Full, Remaining Processes cannot be accomodated"); printf("\n\nTotal…Need to create two MIPS functions: one that emulates DIV and one that emulates DIVU. I CANNOT USE DIV OR DIVU. It only takes the divisor and the dividend as arguments, so no need to load anything.write a statement that performs the specified task. Assume that double variables number1 and number2 have been declared and that number1 has been initialized to 7.3 . "Display the address stored in doublePtr . Is the address the same as that of number1 ?"
- Command line arguments are passed to int main(int argc, char** argv) as arguments argc and argv. You should assume that argc is at ebp+8 and argv is at ebp+12. 00000000 <what>: 0: push %ebp 1: mov %esp,%ebp 3: sub $0x10,%esp 6: mov 0x8(%ebp),%eax 9: add $0x4,%eax c: mov %eax,-0x4(%ebp) f: mov 0x8(%ebp),%eax 12: imul 0xc(%ebp),%eax 16: mov %eax,-0x8(%ebp) 19: mov 0x8(%ebp),%eax 1c: sub 0xc(%ebp),%eax 1f: mov %eax,-0xc(%ebp) 22: mov -0x4(%ebp),%edx 25: mov -0x8(%ebp),%eax 28: add %eax,%edx 2a: mov -0xc(%ebp),%eax 2d: add %edx,%eax 2f: leave 30: ret00000031 <main>: 31: lea 0x4(%esp),%ecx 35: and $0xfffffff0,%esp 38: pushl -0x4(%ecx) 3b: push %ebp 3c: mov %esp,%ebp 3e: push %ebx 3f: push %ecx 40: sub $0x10,%esp 43: mov %ecx,%ebx 45: mov 0x4(%ebx),%eax 48: add $0x4,%eax 4b: mov (%eax),%eax 4d: sub…After working this code, this is the error code that popped up resulted in half right. Traceback (most recent call last): File "main.py", line 292, in <module> print(double_pennies(starting_pennies, user_days)) File "main.py", line 264, in double_pennies total_pennies = double_pennies((num_pennies * 2), (num_days - 1)) File "main.py", line 264, in double_pennies total_pennies = double_pennies((num_pennies * 2), (num_days - 1)) File "main.py", line 264, in double_pennies total_pennies = double_pennies((num_pennies * 2), (num_days - 1)) [Previous line repeated 995 more times] File "main.py", line 260, in double_pennies if num_days == 1: RecursionError: maximum recursion depth exceeded in comparisonTo use dynamic memory allocation functions, which of the following header files must be included?a) stdlib.hb) stdio.hc) memory.hd) dos.h
- #1 A) Write a program to print all alphabets from A to Z. You will be using the JSR putchar command. B) Write a program to print the hexadecimal equivalents of all alphabets from A to Z. You will be using the JSR out2hex command in the previous question instead of JSR putchar.Question 10 Which statement of the following is the most appropriate? Group of answer choices In C++, the allo operator is used to allocate dynamic memory. The delete operator is used to free dynamic memory. In C++, the new operator is used to allocate dynamic memory. The delete operator is used to free dynamic memory. In C++, the new operator is used to allocate dynamic memory. The clean operator is used to free dynamic memory. In C++, the allo operator is used to allocate dynamic memory. The clean operator is used to free dynamic memory.In each of the following code snippets, data is copied from x to y. How many bytes of data arecopied? The answer should be a C expression. (e) int x=10, y=x;(f) int x[10]; int *y = x;(g) int x[10] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}; void f(int y[5]) { ... } int main() { f(x); }(h) int x[10], y[10]; ... memcpy(y, x, 5*sizeof(int));
- I need the code in C programing i am begging you please. There are several test cases. Each test case begins with a line containing a single integer n (1≤n≤1000). Each of the next n lines is either a type-1 command, or an integer 2 followed by an integer x. This means that executing the type-2 command returned the element x. The value of x is always a positive integer not larger than 100. Given a sequence of operations with return values, you’re going to guess the data structure. It is a stack (Last-In, First-Out), a queue (First-In, First-Out), a priority-queue (Always take out larger elements first), not sure(It can be more than one of the three data structures mentioned above) or impossible(It can’t be a stack, a queue or a priority queue). Sample Input Sample output 6 1 1 1 2 1 3 2 1 2 2 2 3 6 1 1 1 2 1 3 2 3 2 2 2 1 2 1 1 2 2 4 1 2 1 1 2 1 2 2 7 1 2 1 5 1 1 1 3 2 5 1 4 2 4 1 2 1 queue not sure impossible stack priority queue impossibleComplete the following sentences by filling the right word To check if two string variables string1 and string2 refer to the same memory address we use the command ………Hello I need help with the following question, I have provided my work that I have done and an example of completed work from the class. This is the question I need help with. A system currently accessing 3.8 G bytes of RAM and 10 M bytes of ROM. Compute the number of address lines required to access the designated storage. For a given # of lines how many more storage space is accessible? Currently out of 3.8 G, 3 G is used by the system programs, .4 G by an application. Can we expand the memory by another 890 M to support a new application? show total storage use and available. Show all your work.