What is the MIPS machine language code for the following instruction? Write opcode, rs, rt, rd, shamt, and funct in is of R-type, write opcode, rs, rt, offset/const/address in decimal if it is of I-type, and write opcode and const/addre decimal if it is of J-type. In the answer fields below, if any field is not used, enter exactly as written here inside of t "N/A". bne $t3, Szero, label where label = Ox1000 and PC = Ox1012 opcode (in decimal)
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- AIM- Write an 8085 sequence to check whether the first set of reading is higher than the second one or not. PROBLEM STATEMENT- The pressure of two boilers is monitored and controlled by a microcomputer works based on microprocessor programming. A set of 6 readings of first boiler, recorded by six pressure sensors, which are stored in the memory location starting from 2050H. A corresponding set of 6 reading from the second boiler is stored at the memory location starting from 2060H. Each reading from the first set is expected to be higher than the corresponding position in the second set of readings. Write an 8085 sequence to check whether the first set of reading is higher than the second one or not. If all the readings of first set is higher than the second set, store 00 in the 'D' register. If any one of the readings is lower than the corresponding reading of second set, stop the process and store FF in the register 'D'. Data (H): First set: 78, 89, 6A, 80, 90, 85 Second Set:71, 78,…Q5 - Where is the operand (data) found in each of the following addressing modes? Example: d = direct (the operand specifier is the address of the operand) i.e.; the operand is found in the memory location whose address is given in the second and third bytes of the instruction. (DO NOT USE Mem[ OprndSpec ] as given in the text but rather write it out as shown) i = s = sf = x = sx = n =Give solution in C ++ Language. Write logic also which is applicable in this question. First read the instruction and hints then solve the question. Hints and Instructions : First read the number of rows and numbers of columns for the grid of information. Store them in Array (2D) These values are then followed by the elevation values, in row order. The maximum size of the grid is 5 rows by 5 columns. Then extend it to 10x10. Forget the file handling part. Just randomly fill in data in 2d array. And then try to find maximum of the array while keeping in mind the conditions mentioned. Don't store date or file in grid.text.
- 6. Given single-character codes for the colored bands that mark a resistor, compute its resistance. The color codes are as follows: Color Code Character Code Black 'B' Brown 'N' 1 bandı Red 2 | banda bands Orange Yellow 3 4 band4 Green Blue 6 Violet Gray White 7 'V' 'A' 9 "W' Figure 1: Resistor Color Bands Table 1: Resistor Color Codes The resistance (2) value of a resistor can be found using the integer value for each color band as follows: R = (bandl * 10 + band2) * 10band3 Write C++ program that finds the resistor value from its color code and output the answer as a numerical format. Remember to use function(s)! P.S.: Use the character code in shown in Table 1 order to represent each color. Sample output Input three color codes: YVO Resistance is: 47000 Ohms.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<< "" <<Assignment for Computer Architecture Instructions: The assignment is to create a program that adds the number 1/2 to itself a large number of times and adds the number 1/3 to itself a large number of times separately first using type float and then type double. It is to then compare the values of adding the numbers to multiplying 1/2 time the number of times added to compute the “same sum” in a different way. The program will also multiply 1/3 times the number of times 1/3 was added to itself to compute the “same sum” in a different way. The program will then compare these two methods at arrive for the same value and output the difference. Hint, the value for the ½’s will be the same for the smaller numbers of times, the 1/3’s will never be the same. The output from your program is to be to a *.txt file which you are to turn in along with your code. The program must first add the ½’s and 1/3’s using type float and compare to the value obtain using multiplication instead of addition.…IN C PROGRAMMING LANGUAGE AND COMMENT EVERY LINE PLEASE SO I CAN UNDERSTAND EVERY STEP , Write a program that accept two (3 x 3) matrices from the user and then ask the user to select one operation to do: a. Multiplication. b. Addition. c. Subtraction. d. Transpose. The program should print the output as a (3 x 3) matrix.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 .ENDIf the value of ESP register before running line number 9 is 2000. What will be its value at line 15? Note that the "sum" function is defined in the second picture below. extern sum, print_sint, print_uint, print_hex global main segment .text main: push 1 push-2 call sum add esp, 8 push eax call print_sint call print_uint call print_hex add esp, 4 Imov eax, 1 int 0x80 #include int sum(int a, int b) { return a+b; } void print_sint(int a) { printf("%d\n", a); } void print_uint (int a) { printf("%u\n", a); void print_hex(int a) { printf("%x\n", a); } A. 1996 B. 2012 C. 1988 D. 2008Overall Requirements Write two programs encode.toy and decode.toy. Each TOY instruction must have corresponding pseudocode. (This is auto-generated by Visual X-Toy – see below.) It's also good practice to add line breaks between logically related "sections" of TOY code and write a comment above each "section" explaining what that code does. encode.toy Write a TOY program encode.toy to encode a binary message using the scheme described above. Repeatedly read four (4) bits m1, m2, m3, m4 from TOY standard input and write the seven (7) bits m1, m2, m3, m4, p1, p2, p3 to TOY standard output. Stop upon reading FFFF from standard input. p1 = m1 ^ m2 ^ m4 p2 = m1 ^ m3 ^ m4 p3 = m2 ^ m3 ^ m4 Recall that ^ is the exclusive or operator in Java and TOY. This captures the parity concept described above. decode.toy Write a TOY program decode.toy to decode and correct a Hamming encoded message. Repeatedly read seven (7) bits m1, m2, m3, m4, p1, p2, p3 from TOY standard input and write the four…it is said that it should use POINTER ARITHMETIC SOLUTIONS ONLYNote: 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, -2SEE MORE QUESTIONS