Question: Show the execution of each line of the given MIPS Code for $s0 = 0xA starting from“Main” till the code reaches “End”.Main: ble $s0, $0, Endsub $a0, $s0, $0Jal FuncJ EndFunc: add $v0, $0, $0Loop: addi $v0, $v0, 1srl $a0, $a0, 1bgt $a0, $0, LoopRet: jr $raEnd:
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“Main” till the code reaches “End”.
Main: ble $s0, $0, End
sub $a0, $s0, $0
Jal Func
J End
Func: add $v0, $0, $0
Loop: addi $v0, $v0, 1
srl $a0, $a0, 1
bgt $a0, $0, Loop
Ret: jr $ra
End:
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- In execution of the following code with the following bytes contents: START: CLC MOV CX, 0002 MOV SI, 0300 MOV DI, 0302 LOOP: MOV AL, (DI) ADC (SI), AL INC SI INC DI DEC CX JNZ LOOP HLT (0300H) = 1AH, (0301H) = 2BH, (0302H) = 3CH, (0303H) = 4DH. %3D %3D %3D The result will be:Rewrite the code with local variables only and add trace tags to the code. BR main // branch to main loop x: .BLOCK 2 // storage for x Y: .BLOCK 2 // storage for y Temp: EQUATE 0 // set Temp=0 local variable #2d Main: LDWA 3,i // input i=3 STWA x,d //Store x LDWA 8,i // i=8 STWA y,d //store y SUBSP 2,i // subtract from stack pointer STWA temp,s //store word to stack LDWA x,d // Load word accumulator d to x STWA y,d // Store word accumulator d to y LDWA temp,S // Load word accumulator temp to stack STWA x,d // store d to x ADDSP 2,i // add to stack pointer STOP // stop execution .ENDBelow C-code has a for loop which will repeat 1000 times of the operation. This code will be running on very simple machine. Change the code to reduce the number of branch instructions. You can have up to 5 lines (statements) within the for loop. What is the number of branch operations before and after? for (i=0; i<1000; i++){ y[i] = w * x[i] + b[i]; }
- Use select to avoid deadlock goroutines and write go codeSession: 3 Topic: Introduction to Computer Programming-1 C++ Languages. Aim: Write a C+ program to check the entered number is within the 0- 10. Tool: Turbo C++ PROGRAM LOGIC: • Read any numbers by the users during execution. • Check entered number is within 0-10 which is entered by the user • Print the output Code: TURBO C++ /* Check entered number is within 0-10 */ #include #include void main () int num; cout> num; if (0 Compile (Alt+F9) -> Run (Ctrl +F9) 10Write the program code that compares the data at the address that will come over PO with the data at the address 70h and sends the larger value to P2 in the solution field given in Question 2. The box below for the 1st question will be left blank. Yanıt: Soru 1 için gerekli olan program kodunu aşağıda verilen alana yazınız. Write the program code required for Question 1 in the space provided below.
- :Write the result of AL and CL after executing syscall MOV AL, 07H · MOV CL, 02H. SAL AL, CL syscallSo I ONLY need help with #1, so I was attempting to calculate Trotation & Ttransfer for #1 and NEED to know if I did my calculations correctly..........Trotation: (1/2)*(1/(10000/60))*1000 = 3ms & Ttransfer:Implement 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
- Please write the lines that the code will print out. Assume the following addresses: - a in main has the address 1000-b in main has the address 1004 c in main has the address 1008 - d in main has the address 1012 - b in fooA has the address 2000 - a in fooA has the address 2008 int fooA (int* b, int& c, int a) { (*b) ++; c+ = 2; a + = 4; cout <<<<<<' <<<&Q<<< ""«a« ""<< b<< "" <<In executing the following code: MOV CX, 0002 START: MOV SI, 0300 MOV DI, 0302 LOOP: XOR (DI), AL INC SI MOV AL, (SI) INC DI DEC CX JNZ LOOP HLT with the following byte contents: (0300H) = 0AH, (0301H) = OBH, (0302H) = 11H, (0303H) = 21H. The result will be: Select one: a. (0300H) = 0AH, (0301H) = OBH, (0302H) = 1AH, (0303H) = 2BH (0300H) = 0AH, (0301H) = OBH, (0302H) = 10H, (0303H) = 20H b. c. (0300H) = 0AH, (0301H) = OBH, (0302H) = 1AH, (0303H) = 20H d. None of the options given here e. (0300H) = 0AH, (0301H) = OBH, (0302H) = 1BH, (0303H) = 2AHIn executing the following code: START: MOV CX, 0002 MOV SI, 0300 MOV DI, 0302 LOOP: MOV AL, (SI) XOR (DI), AL INC SI INC DI DEC CX JNZ LOOP HLT with the following byte contents: (0300H) = OAH, (0301H) = OBH, (0302H) = 11H, (0303H) = 21H. The result will be:SEE MORE QUESTIONS