What is the difference between "x86" and "x86_64"? Does "i386" mean "x86"?
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x86 refers to Intel's (and compatibles like AMD Athlon, and VIA C3) line of processors ending in the value 86 since the release of the original 8086.  x86_64 indicates an x86 processor that supports 64-bit extensions (ironically developed by AMD and adopted by Intel).\n\ni386 derives from Windows NT.  It is the default directory containing all the cab files used to install the O/S.  Since Microsoft developed Windows NT for several different processor platforms (MIPS, Alpha, PowerPC - way back when) this was used to distinguish which platform it was intended for.  Now, NT derivatives like Win2000 and XP are x86 only but the i386 stuck around.
518024
There wont be much of a difference between 384 and 320. I think the display will say 320, but your computer is actually using all of 384. You dont need to worry about it.
41411
It sounds like you are running into 'versioning' problems with .NET.  First of all, to register your assembly in the Global Assembly Cache (GAC) you have to sign your assembly with a strong name.  Your applications that use that assembly then reference the assembly.\n\nThe problem comes in if you do not specifically designate the version of your assembly.  Look in the AssemblyInfo.vb, most specifically at the AssemblyVersion attribute.  If this is set to something like "1.0.*", that will direct the compiler to update the version number each time you compile.\n\nChange the version to a specific number, like "1.0.0.1" and leave out the '*'.  Even if you have a strongly named assembly with the same key, it will not work with other applications that have been compiled against a different version, which is changing every time you compile!
397669
8 bits is a byte... 4 bits is a byte... whether your binary looks like 10011000, 11111111, or 00000001 or 00000000, it's all a byte. If you have 1001, the machine will automatically make it 00001001. It will just add zero's in front of it to make it a byte.
453934
You don't use binary in this way, and compilator embedded into IDE (interface of development environment) does it for you. \n\nAny program is the combination of bytes, and each byte (in majority of systems) consists of 8 bits of information. Each bit is 0 or 1.\n\nActually, you can write in hex code (almost binary format) and it's hard and inconvenient, or in assembler, or more advanced programming languages.\n\nFor example, In assembler, command, say,  "JMP 280000"\nfor example, means go to address 280000, and translated into\ncombination of bytes.\n\nIt's very easy to transform the program written in C or assembler, into binary format, automatically.\nAlso, if you already have somebody's else binary, you could desassemble it into assembler or C code, with some limitations because there is no 1 to 1 mapping all the time.\n\nTherefore, binary is not a programming language.
730566
Doesn't sound good. Search Google or microsoft.com with "0x00000024."
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What os u means? specify correctely. I m not getting.
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intel\nOne entry found for intel.\n \n\nMain Entry: intel\nFunction: abbreviation\nintelligence
302941
translated in assembler this addition and store statement would be written as follows:\nMOVE J, R0 (1 Instr. fetch + 1 Operand fetch)\nADD  K, R0 (1 Instr. fetch + 1 Operand fetch + 1 Addition)\nMOVE R0, I (1 Instr. fetch + 1 Operand store)\n\nwith a speed of 200MHz 1 cycle is 5ns (nano seconds)\nlets assume an Instr. fetch is 1 cycle, an Operand fetch/store is 3 cycles and the Addtion is 1 cycle then we get 3 + 9 + 1 = 13 cycles, 13 * 5 ns = 65ns approx.
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16 and 32bit programs (OS, programming) can be considered in many issues;\nas C, C++, assembly, ... (not high level languages) programmer basically you can address different size of memory in each schemes:\n1) In 16bit programs, every "int" variable is 16bit long, so 2^16 bytes of memory can be accessed. (there were different memory models those-days: huge, far)\n2) in the same way every "int" in a 32bit programs is 32bit long and thus 2^32 bytes of memory handled.\n\nalso your program in runtime is tightly dependent of the OS and hardware architecture (they could be 16, 32 or even 64bit)\n\nOther programming issues (in practice) rise up when you want to switch between 16 and 32bit programming, eg Multithreading & synchronization, I/O addressing, ... which are OS and hardware dependent.
1021350
direct addressing (which means specifying the effective address within the assembly instruction) is pretty limited - it means that you have to know the absolute address when you write the code, which is at most times impractical since programs and their data are loaded into a different addresses in memory each time they are loaded. \n\nindirect addressing (which means that the effective address is stored in a register and the assembly instruction points to that register) is much more useful - for example when a subroutine is called, the caller may store the return address in a register, and when the subroutine finishes, it jumps to that return address using indirect addressing.
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64. or 72, depending on what type
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Any 64-bit version if you want to take full advatage of your 64-bit CPU. Note that any 32-bit version will work as well.
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everything that interprets 1s and 0s into something you can use is an operating system, the difference is the architechture, ex:windows is x86
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C:&#92;WINDOWS&#92;system32
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You need to use CPUID assembler instruction to get it. To learn more, follow the link below
622118
16 bit consists of instruction code that are 16 bit in length. And 32 bit contails 32 bit. \n\nSome problems occur when we are migrating from 16 bit to 32 bit. Example...\n\nA programmer was using a pointer (16 bit). He needed to increment the pointer so that the pointer now points to next word. So he wrote the expression like\n\niptr = iptr + 2 (2 bytes means 16 bits).\n\nBut when the compiler design was changed to 32 bit, the program crased. Because the expression that he wrote now made the pointer jump in between the word.\n\nInclude this point dude.