A program and its data are both stored in memory in the von Neumann architecture. It is thus possible for a program to modify itself accidentally (or on purpose) by mistaking a memory location for a piece of data when it actually contains a program instruction. What are the ramifications for you as a programmer?
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A program and its data are both stored in memory in the von Neumann architecture. It is thus possible for a program to modify itself accidentally (or on purpose) by mistaking a memory location for a piece of data when it actually contains a program instruction. What are the ramifications for you as a programmer?
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- It is necessary to convert mnemonics to binary machine code because the central processing units (CPUs) of microprocessors do not understand them in their original form.The use of transistors in the construction of RAM and ROM leads me to believe that there is no need for cache memory.The term "temporary storage" may also be thought of as "random access memory" (RAM) that is momentarily vacant. Imagine a machine that only had one kind of memory—is it even possible?Mnemonics must be converted to binary machine code for microprocessors' CPUs to interpret.
- A general register, a single accumulator, or a stack are the three potential configurations that might be chosen for the central processing unit (CPU) of a computer. The stack is the most common of these three options. Each choice comes with its own individual combination of benefits and drawbacks. It is up to you to provide a response, and the response that you provide may be accurate or it may be erroneous.Does the memory fetch procedure always happen instantly when a computer wants data? So what is the point of it?memory management using segmentation.... need in c language with output screenshot
- Suppose a program’s 15-th logical instruction (counting starts from zeroth) is at physical address 1234ABDE16 in the RAM. (contents of any logical address fits into any physical address) What is the physical address in the RAM where the program has been loaded? What is the physical address in the RAM of the last instruction of the program if it has 37 logical instructions altogether?The central processing unit (CPU) of a computer may be designed in one of three different ways: with a general register, a single accumulator, or a stack. There are benefits and drawbacks to each option. Your answer, if you choose to provide one, might be right or wrong; it's up to you to decide.Answer the given question with a proper explanation and step-by-step solution. Write an assembly language program which will multiply x by y. The final answer should be stored into memory location A000H.You may assume that memory location B000H is holding the value of x, and memory location B001H is holding the value of y.Your program design must include a loop. You may assume that your program will begin execution at line 0000H.You must enter a NOP command at the end of your program to make the program stop.Enter one command per line.
- 4. A portion of a computer program consists of a sequence of calculations where the results are stored in variables, like this (with inputs a, b and outputs d, g, h): Step Calculation 1 C = a + b d = a * C e = c + 3 4 f = C e g = a + f 6 h = f + 1 A computer can perform such calculations most quickly if the value of each variable is stored in a register, a chunk of very fast memory inside the microprocessor. Compilers face the problem of assigning each variable in a program to a register. Computers usually have few registers, however, so they must be used wisely and reused often. This is called the register allocation problem. In the example above, variables a and b must be assigned different registers, because they hold distinct input values. Furthermore, c and d must be assigned different registers; if they used the same one, then the value of c would be overwritten in the second step and wed get the wrong answer in the third step. On the other hand, variables b and d may use the…All microprocessors have same number of address lines but different number of data lines. Select one: True FalseA digital computer has a memory unit with 24 bits per word. The instruction set consists of 150 different operations. All instructions have an operation code part (opcode) and an address part (allowing for only one address). Each instruction is stored in one word ofmemory.a. How many bits are needed for the opcode?b. How many bits are left for the address part of the instruction?c. What is the maximum allowable size for memory?d. What is the largest unsigned binary number that can be accommodated in one word of memory?