What will be the content of memory location 250 in decimal after exclusion of the following program (here instructions are separated by semicolons, and numbers are represented in decimal)?: LDI R3, 248; LD R1, (R3); DEC R1, R1 INC R3, R3; LD R2, (R3); ADD R2, R2, R1 ; INC R3, R3; ST (R3), R2. (do not write any blank space) Data address memory ... 248 25 249 31 250
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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,…Write program segments that perform the operation C C+ A x B using each of the instruction classes indicated in Exercise 1 above. Assume that A, B, and C are memory 3. addresses. lleuing data Lin theWrite a program in C++ language to illustrates how a pointer variable works. Follow the instruction below; - Create main function and include your information details - Declare the pointer variable of p as an integer - Declare an integer variable of num1 and num2, and set the value of num1 = 5 and num2 = 8 - Store the address of num1 into p - Print the address of &num1 and value of p - Print the value of num1 and *p - Change the value of *p to 10 - Print the value of num1 and *p - Store the address of num2 into p - Print the address of &num2 and value of p - Print the value of num2 and *p - Multiply the value of *p by 2 - Print the value of num2 and *p
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