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The subject you're describing often goes under the names computer architecture, computer systems, computer organization and design, and the like. One example is Elements of computing systems, based on a course From Nand to Tetris originating in the Hebrew University. The Amazon page on the book links to other popular textbooks on the topic. Generally ...


2

The page tables are not kept in the MMU. In the MMU is a cache called the TLB which holds translated addresses. The TLB holds virtual addresses to physical addresses translations. For example, the TLB on intel core i5 has 64 entries for pages of 4KB. Since the 10 upper bits are used as the offset in the page directory on 32 bits architecture, there are 2^10 =...


1

if there are only 28 address lines and each carries one bit how there can be any more data? I don't read 28 address lines all told, but 28 address lines going into one specific memory device. where are the 4 address lines to carry [the 4 bits out of 32 that are used to activate the memory device]? There are two answers on the conceptual level: A> Those ...


1

The program is not really split up into pages. The program is unaware of where it lands in physical memory. This is all the essence of paging. For example, if you take Linux on x86 CPUs and the C++ language, you will have the compiler generating machine code which will be linked into a final executable which has a .ELF extension. ELF files support virtual ...


1

This question illustrates a lack of understanding of fundamentals. What does moving mean!? The RAM stores data in transistors. Do you want a little robot to pick up the specific transistors you want out of the RAM chip and deliver them to the CPU? I hope it is obvious why that is not possible. Instead, an electronic circuit (which those transistors are part ...


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It depends. A program can store 1 ASCII character in each 64-bit word, or 8 ASCII characters in each 64-bit word. It's up to each individual program to decide how it wants to store and format its data in memory. The latter would probably be more typical.


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First of all you need to specify whether your partitions are fixed size or variable size. In case they are using dynamic partitioning(variable size) the internal fragmentation will always be 0 since its defined only for fixed size partitioning. Now assuming your partitioning is static, total internal fragmentation will be the sum of all internal ...


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