memory-address

In the 8086 architecture, the memory space is 1 MiB in size and divided into logical segments of up to 64 KiB each. i.e. it has 20 address lines thus the following method is used: That the data segment register is shifted left 4 bits then added to the offset register My question is: How we do the shift operation although all the registers are only 16 bits Address translation is done internally by a special unit without using the registers available to user code to store intermediate results - it just fetches 16-bit values and does the translation inside - it is not reflected anywhere where the

After searching all over the internet to find a way to symbolicate my crash logs I received from Apple, I finally figured out how to use the atos command in terminal to symbolicate the crash logs. I have the dSYM file, the .app file and the crash logs in the same folder, and using atos -arch armv7 -o APPNAME I have been able to enter memory addresses, and sometimes (but quite rarely) a method name has come up. To be perfectly honest, I don't have much experience with terminal, or crash logs. Attempting to symbolicate the crash logs from Xcode's organiser has unfortunately done absolutely

What forms of memory address spaces have been used? Today, a large flat virtual address space is common. Historically, more complicated address spaces have been used, such as a pair of a base address and an offset, a pair of a segment number and an offset, a word address plus some index for a byte or other sub-object, and so on. From time to time, various answers and comments assert that C/C++ pointers are essentially integers. That is an incorrect model for C/C++, since the variety of address spaces is undoubtedly the cause of some of the C rules about pointer operations. For example, not

Assume 32 Bit OS. One memory location in a computer stores how much data? Whats the basic unit of memory storage in a computer? For Example to a store a integer what will be the memory addresses required? If basic unit is BYTE the integer requires 4 bytes. So if I need to store a byte then if start putting in the 1st byte in memory location 0001 then will my integer end at 0003 memory location? Please correct me if am wrong? Most commonly, modern systems are what you call "byte-accessible" This means: One memory location stores 1 Byte (8 bits). The basic storage unit for memory is 1 byte. If

Given an STL container (you may also take boost::unordered_map and boost::multi_index_container into account) that is non-contiguous, is it guaranteed that the memory addresses of the elements inside the container never changes if no element is removed, (but new ones can be added)? e.g. class ABC { }; // //... // std::list<ABC> abclist; ABC abc; abclist.insert(abc); ABC * abc_ptr = &(*abclist.begin()); In other word will abc_ptr be pointed to abc throughout the execution, if I do not remove abc from abc_list . I am asking this because I am going to wrap the class ABC in C++/Cli, so I need

I'm trying to get the address of a member function, but I don't know how. I would appreciate if somebody could tell me what I'm doing wrong. As you can see in my example below, neither (long)&g nor (long)&this->g work and I can't figure out the correct syntax: /* Create a class that (redundantly) performs data member selection and a member function call using the this keyword (which refers to the address of the current object). */ #include<iostream> using namespace std; #define PR(STR) cout << #STR ": " << STR << endl; class test { public: int a, b; int c[10]; void g(); }; void f() { cout <<

I wrote the following Code. #include<stdio.h> int main() { int x = 1 ; int *j = &x ; int y = 2 ; int *t = &y ; printf("%p\n" , (void *)j); printf("%p" , (void *)t); } Output is 0028FF14 0028FF10 . The Point I want to make is that the difference between the addresses is `4'. Whereas in this case #include<stdio.h> int main() { char x = 't' ; char *j = &x ; char y = 'f' ; char *t = &y ; printf("%p\n" , (void *)j); printf("%p" , (void *)t); } Output is 0028FF17 0028FF16 The difference is 1 . Difference In First Case is 4 . Whereas in the second case it is 1 . Why is it so? And What will I get if I