In C# does a dataset occupy more RAM memory if passed by value, reference or out?
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In all programming languages passing by reference will take less memory than others
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Here are some general data structures:\n\n- Set\n- List\n- Map\n- Tree\n- Table\n- Stack\n- Queue
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The data is not earsed from the disk but the pointers (memory address) to the data (located in the FAT?) gets erased. The actual data stays there until you do 'Disk fragmentation'.
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no where.  Just the data pointers are deleted.
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How it works exactly is compiler dependent. It may also depend on whether or not you are building with debugging information turned on.\n\nfree is going to release memory that you've allocated and return it to the heap for reallocation. Generally, the memory manager as a linked list of blocks of memory that it has allocated. It is going to search the list for your block of memory, remove the link from the allocated list and insert it in the list of free blocks of memory. It may also attempt to merge you block with adjacent blocks of free memory to maximize the size of available blocks.\n\nIf you are running Microsoft C++, the debug memory manager will check a couple of bytes that are added to the front and back of the block to be sure that you haven't overflowed when you used it. It will also write a specific byte pattern to the block so that it is easily recognized as deallocated if you use a pointer to it after it has been released.
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Static RAM and Dynamic RAM
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alternative 1)\nUse int/short instead. Make sure you mask potential values with "& 0xff" so they don't overflow.\n\nalternative 2)\nUse (signed) byte for storage, but when doing comparisons, assignment to other integer types, cast to int/short and mask with 0xff
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By using "Call by Reference", we can modify values that passsed to a function. If we use "Call by Value", we can not modify value of parameters, because actually we just modify values that stored in stack, not in the memory.\n\nHere is the example. In the main program, we pass a variabel contains number 1, and functions try to change it. Excuse me, my English is bad.\n\nvoid test_one(int a) {\n  a = 2;\n};\n\nvoid test_two(int *a) {\n  *a = 2;\n};\n\n// C++\nvoid test_three(int &a) {\n  a = 2;\n};\n\nvoid main(void) {\n  int a;\n\n  a = 1;\n  test_one(a);\n  printf("%d&#92;n", a);  // Result: 1;\n\n  a = 1;\n  test_two(&a);\n  printf("%d&#92;n", a);  // Result: 2;\n\n  a = 1;\n  test_three(a);\n  printf("%d&#92;n", a);  // Result: 2;\n};\n\n"&" sign means: Address of this variable in memory.\n"*" sign means: Address that pointed by this memory address.\n\nThe best way to understand this, is by examine the Assembly Code of this program. You can use Turbo Debugger or just DEBUG.COM.\n\nIf you think my answer is the best answer, would you please do rate mine? Thank you.\n\nGreetings from Tangerang, Indonesia.
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There are a number places in the C++ spec that say that the method of implementation is left up to the designer of the compiler as long as they meet certain requirement.\n\nFor example, the rules require that the storage size of a long be at least as large as that for an int.\nMicrosoft C++ makes both ints and longs 32 bits. A program that works with 32 bit longs could very easily fail with 16 bit longs and vice versa. \n\nIt's also possible that you are breaking a rule that you can get away with on with one compiler, but not the other. For example, if you are running a program built with MS C++ using debug heap allocation, when you free the memory, it gets overwritten with a specific pattern to aid in debugging. If you use a stale pointer, you'll be pointing to trash. The same code built using another compiler might still find valid values in the memory, even though you released it.
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In windows there are many forms of memory.. programs request memory as required and any memory that cannot be used from available ram is spooled to disk (pagefile).. As requests to/from the pagefile are rather slow (compared to memory) the more ram that you have the less dependant you are upon the pagefile.  Some programs actually will request all available memory while they operate to speed up their operation. \n\nWindows usually holds .dll's in memory while the using program wants them and slowly releases this memory after the program terminates. (you can change this setting by editing your registry to unload .dlls directly).. If you have sufficient memory (256MB +) you can keep the core parts of the operating system in memory (non-paged) to speed up system calls dramatically.  \n\nSome programs use what is called shared memory (either paged or non-paged) to transfer information from one process to another or only 1 copy of the .dll is required in memory and  memory used to hold variables is allocated seperately.\n\nWhen programs load they can either load entirely or just load pieces and request the additional pieces from disk as required and unload unneeded pieces as required but this is not the norm in windows xp. \n\nAs programs load/operate and unload the physical ram and pagefile dynamically go up and down.. The more ram you have (I cannot stress this enough) the less the pagefile will be used for system mrmory/disk cache, etc)) the better the user experience will be.\nWith 32 bit processors the maximum ram that can be accessed is 4Gb but in normal operation more than 2G of ram doesn't seem to result in much of a performance gain (unless you are using programs that use a LOT of memory i.e. adobe photoshop, video editors, etc).. Another item to consider is that some motherboards that allow the maximum amount of memory 4G cannot really use more than 3G as the extra gig is  tied up by the bios and other hardware operations.\n\nOf course, once power is lost what you have in ram is lost forever as it is volatile memory.
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Here's a few things you might want to think about:\n\nPress {windows} and {pause break} and see if it's reporting your full memory amount.  \n\nWatch the performance tab in task manager and watch your memory usage as you copy and paste.\n\nAre you managing your swap file manually or do you let Windows do it?  \n\nAlso, try running memtest to check your memory for errors:\n\nhttp://hcidesign.com/memtest/MemTest.zip
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C# is actually a lot more like Java than it is C++.  They are both object-oriented languages, but C#, like Java, is designed to be compiled to "byte code" rather than an native application (that is, one that will run without the .Net framework).  Most C++ compilers create native applications.\n\nC# also handles memory management issues for you, so you are not required to keep track of what memory you have allocated.  However, it does not guarantee for you when exactly your objects will be destroyed, which can be a pain in some circumstances.
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A program sets aside an allocated chunk of memory, called a buffer.  The size of the buffer is typically calculated to allow some time to pass before incoming data fills it.  If the data comes in faster than anticipated, or the buffer is too small, than the buffer "overflows" and data is lost.
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Why are you passing large amounts of data between programs?\n\nIt would be better to pass a pointer to the working storage location.\nYou may have to call another program to get the address of the storage.
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The two calls use structures called localtime and localtime_s. They  have different storage sizes.  The compiler is simply telling you that 5 bytes are being put in a 4 byte space, you'll loose data, for example.  Try looking up the time function that you are using and see if there is a secure version from the Win32s (as opposed to Win32) API.  I'd bet there is and you'll need to use it in place of what you have now.  This happens a lot, older code examples sometimes use the Win32 functions instead of the Win32 API current versions as a minimum.  The secure versions are required by Windows 2000, Windows 2003 and of course, Windows XP.  Whether you are writing this yourself or the older calls are still in your library and headers paths, this can hold you up until you find and set things right.
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It doesn't look like any memory error I have seen but errors which resemble this are indicating that a particular program cannot utilize a certain required memory address.
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That depends on how you run your Java programs.  Since you say "compilation", I assume you're actually building them into executable form.\n\nMemory is allocated in real-time, that is, when an object is instantiated, that is when the memory is allocated by the system.  Say you have an object that is created only after you enter your name.  When the program starts, the object is not in memory.  Once you enter your name, the object will create, and the system will allocate the necessary memory.\n\nRemember that with Java, when you "delete" an object, it doesn't always fall immediately out of the memory stack.  The Garbage Collection routine will occasionally filter through and "de-allocate" all those homeless memory pointers.