While (Operand1 < Operand2) { Operand1++; If (operand1==operand2) X=2; Else
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Consider the following program and implement it’s equivalent in assembly language.
Assume, Operand1=0; Operand2=10 initially,
While (Operand1 < Operand2)
{
Operand1++;
If (operand1==operand2)
X=2;
Else
X=3;
}
Step by step
Solved in 2 steps with 1 images
- Note: assembly language(8086 microprocessor) Write a program that takes a number N (2 digit decimal) as input. Then the user will enter that many numeric values ranging from 0 to 9 as input. Here the user may enter a negative number too. The program then finds a pair of values whose summation is closest to zero.Explanation: Here at first the user enters 10, which is a 2 digit decimal number. That means the user now has to enter 10 negative or positive digits ranging from 0 to 9. Then the program will calculate the sum of all possible pairs and find the pair which gives the minimum sum. Here from the given example, we see that the sum of -2 and -4 which is -6 is the least sum.Input:10 -2, 3, 7, 4, -4, 7, -8, 0, 9, 9 Output: Pair of values which has the smallest sum = -4, -2In C code This is what I have so far #include <stdio.h> int main () { unsigned int x = 0x76543210; char *c = (char*) &x; printf ("*c is: 0x%x\n", *c); if (*c == 0x10) { printf ("Architecture is little endian. \n"); } else { printf ("Architecture is big endian. \n"); } return 0; }Consider the following C code: long func (1long x, long m, long n) { long result = 0; int p = 1; int g = 10 if ((x > m) && (x < n)) { p = p + m; 9 = g + n; } else p = p n; result = return result; } • Rewrite the code and replace the if-else with 'goto' statements. • Write the assembly code that corresponds to the C goto' version. • Compile the C code to assembly using the gcc compiler and see how your code compares with the compiler version.
- Consider the C code given below: volatile static int sum = 0 for (int i=0; i<8; i++) { if (i<4) sum = sum + 2*i; else sum = sum + 3*i; } ; // assume sum is a memory variableTranslate the following C program to Pep/9 assembly language. #include int inpNum, sum, count, min; int main() { sum = 0; count = 0; min = 32767; printf("? "); scanf("%d", &inpNum); while (inpNum != 0 ) { sum += inpNum; count++; if (inpNum < min) { min = inpNum; } printf("? "); scanf("%d", &inpNum); } // end while printf("Sum printf("Count printf("Min } // end of main = = %d\n", sum); = %d\n", count); %d\n", min);For this question you must design and implement a precondition loop in Python that will display the following numbers (one on each line) to the terminal: 79, 92, 105, 118, 131, 144, 157, 170, 183, 196, 209, 222, 235, 248, 261, and 274. Although a counter-controlled loop is ideal for this task, you are not permitted to use a FOR loop in your program - you must use a precondition loop that has been implemented using WHILE instead.
- Translate the following C program into Pep/9 assembly language:#include <stdio.h>int main () {int num1;int num2;scanf("%d", &num1);num2 = -num1;printf("num1 = %d\n", num1);printf("num2 = %d\n", num2);return 0;} please explain and provide cPlease can you convert this part of a c program to pep 9 assembly language void findLog ( int *result, int num, int b ) { int k; num--; *result = 0; while (num > 0) { *result = *result + 1; num = divide(num, b); } // end while } // end of findLog( )In python, Problem DescriptionProfessor Santos has decided to hide a secret formula for a new type of biofuel. She has, however, left a sequence of coded instructions for her assistant. Each instruction is a sequence of five digits which represents a direction to turn and the number of steps to take. The first two digits represent the direction to turn: • If their sum is odd, then the direction to turn is left.• If their sum is even and not zero, then the direction to turn is right.• If their sum is zero, then the direction to turn is the same as the previous instruction. The remaining three digits represent the number of steps to take which will always be at least 100. Your job is to decode the instructions so the assistant can use them to find the secret formula. Input SpecificationThere will be at least two lines of input. Each line except the last line will contain exactly five digits representing an instruction. The first line will not begin with 00. The last line will contain…
- please modify the c language program into a pep 9 assembly language #include <stdio.h> int num, answer; int divide ( int numer, int denom ) { int quotient, remain; remain = numer; quotient = 0; while ( remain >= denom ) { remain -= denom; quotient++; } // end for return quotient; } // end of divide( ) int modulus ( int n, int d ) { int quot, rem; rem = n; quot = 0; while ( rem >= d ) { rem -= d; quot++; } // end for return rem; } // end of modulus( ) int main() { printf("? "); scanf("%d", &num); answer = divide(num, 20); printf("%d / 20 = %d\n", num, answer); answer = modulus(num, 20); printf("%d mod 20 = %d\n", num, answer); answer = divide(num, 35); printf("%d / 35 = %d\n", num, answer); answer = modulus(num, 35); printf("%d mod 35 = %d\n", num, answer); return 0; } // end of mainThe C programming language supports left-shift and right-shift operators. This Arduino sketch tests both shift operators: void setup () { Serial.begin (9600); 1 void loop (0 I Serial.println ("Test shift operations byte a = 0xl; Serial.println (a, BIN); a = a > 2; // right shift Serial.print.In (a, BIN); delay (10000); ") ; 6 10 11 12 13 The AVR instruction set supports five shift operations: logical shift left, logical shift right, arithmetic shift right, rotate left, rotate right. 1sl 1sr asr rol rorImplement the counter increment, and branch back to the start of the loop. When you havecompleted these steps, the basic structure of the counter control loop has been implemented,and your code should look similar to the following:.text li $s0, 0 lw $s1, n start_loop: sle $t1, $s0, $s1 beqz $t1, end_loop # code block addi $s0, $s0, 1 b start_loop end_loop:.data n: .word 5