memcpy vs memmove

Question

I am trying to understand the difference between memcpy   and memmove    and I have read the text that memcpy   doesn t take care of the overlapping source and destination whereas memmove   does   However  when I execute these two functions on overlapping memory blocks  they both give the same result  For instance  take the following MSDN example on the memmove   help page -  Is there a better example to understand the drawbacks of memcpy and how memmove solves it      crt memcpy c    Illustrate overlapping copy  memmove always handles it correctly  memcpy may handle    it correctly    include  lt memory h gt   include  lt string h gt   include  lt stdio h gt   char str1 7     aabbcc    int main  void         printf   The string   s n   str1        memcpy  str1   2  str1  4        printf   New string   s n   str1         strcpy s  str1  sizeof str1    aabbcc          reset string      printf   The string   s n   str1        memmove  str1   2  str1  4        printf   New string   s n   str1        Output   The string  aabbcc New string  aaaabb The string  aabbcc New string  aaaabb

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The memory in memcpy cannot overlap or you risk undefined behaviour  while the memory in memmove can overlap   char a 16   char b 16    memcpy a b 16                valid memmove a b 16               Also valid  but slower than memcpy  memcpy  amp a 0    amp a 1  10       Not valid since it overlaps  memmove  amp a 0    amp a 1  10      valid     Some implementations of memcpy might still work for overlapping inputs but you cannot count of that behaviour  While memmove must allow for overlapping

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The difference between memcpy and memmove is that   in memmove  the source memory of specified size is copied into buffer and then moved     to destination  So if the memory is overlapping  there are no side effects  in case of memcpy    there is no extra buffer taken for source memory  The copying     is done directly on the memory so that when there is memory overlap  we get unexpected results    These can be observed by the following code     include string h  stdio h  stdlib h int main      char a    hare rama hare rama      char b    hare rama hare rama      memmove a 5 a 20     puts a      memcpy b 5 b 20     puts b       Output is   hare hare rama hare rama hare hare hare hare hare hare rama hare rama

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I m not entirely surprised that your example exhibits no strange behaviour  Try copying str1 to str1 2 instead and see what happens then   May not actually make a difference  depends on compiler libraries    In general  memcpy is implemented in a simple  but fast  manner  Simplistically  it just loops over the data  in order   copying from one location to the other  This can result in the source being overwritten while it s being read   Memmove does more work to ensure it handles the overlap correctly   EDIT    Unfortunately  I can t find decent examples  but these will do   Contrast the memcpy and memmove implementations shown here  memcpy just loops  while memmove performs a test to determine which direction to loop in to avoid corrupting the data  These implementations are rather simple  Most high-performance implementations are more complicated  involving copying word-size blocks at a time rather than bytes

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Your demo didn t expose memcpy drawbacks because of  bad  compiler  it does you a favor in Debug version  A release version  however  gives you the same output  but because of optimization       memcpy str1   2  str1  4   00241013  mov         eax dword ptr  str1  243018h       load 4 bytes from source string     printf  New string   s n   str1   00241018  push        offset str1  243018h   0024101D  push        offset string  New string   s n   242104h   00241022  mov         dword ptr  str1 2  24301Ah   eax     put 4 bytes to destination 00241027  call        esi     The register  eax here plays as a temporary storage  which  elegantly  fixes overlap issue   The drawback emerges when copying 6 bytes  well  at least part of it   char str1 9     aabbccdd    int main  void         printf  The string   s n   str1       memcpy str1   2  str1  6       printf  New string   s n   str1        strcpy s str1  sizeof str1    aabbccdd         reset string      printf  The string   s n   str1       memmove str1   2  str1  6       printf  New string   s n   str1       Output   The string  aabbccdd New string  aaaabbbb The string  aabbccdd New string  aaaabbcc   Looks weird  it s caused by optimization  too       memcpy str1   2  str1  6   00341013  mov         eax dword ptr  str1  343018h    00341018  mov         dword ptr  str1 2  34301Ah   eax    put 4 bytes to destination  earlier than the above example 0034101D  mov         cx word ptr  str1 4  34301Ch       HA  new register  Holding a word  which is exactly the left 2 bytes  after 4 bytes loaded to  eax      printf  New string   s n   str1   00341024  push        offset str1  343018h   00341029  push        offset string  New string   s n   342104h   0034102E  mov         word ptr  str1 6  34301Eh   cx     Again  pulling the stored word back from the new register 00341035  call        esi     This is why I always choose memmove when trying to copy 2 overlapped memory blocks

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As already pointed out in other answers  memmove is more sophisticated than memcpy such that it accounts for memory overlaps  The result of memmove is defined as if the src was copied into a buffer and then buffer copied into dst  This does NOT mean that the actual implementation uses any buffer  but probably does some pointer arithmetic

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The code given in the links http   clc-wiki net wiki memcpy for memcpy seems to confuse me a bit  as it does not give the same output when I implemented it using the below example    include  lt memory h gt   include  lt string h gt   include  lt stdio h gt   char str1 11     abcdefghij    void  memcpyCustom void  dest  const void  src  size t n        char  dp    char   dest      const char  sp    char   src      while  n--           dp      sp        return dest     void  memmoveCustom void  dest  const void  src  size t n        unsigned char  pd    unsigned char   dest      const unsigned char  ps    unsigned char   src      if   ps  lt  pd           for  pd    n  ps    n  n--                --pd    --ps      else         while n--               pd      ps        return dest     int main  void         printf   The string   s n   str1        memcpy  str1   1  str1  9        printf   Actual memcpy output   s n   str1         strcpy s  str1  sizeof str1    abcdefghij          reset string      memcpyCustom  str1   1  str1  9        printf   Implemented memcpy output   s n   str1         strcpy s  str1  sizeof str1    abcdefghij          reset string      memmoveCustom  str1   1  str1  9        printf   Implemented memmove output   s n   str1        getchar        Output     The string  abcdefghij Actual memcpy output  aabcdefghi Implemented memcpy output  aaaaaaaaaa Implemented memmove output  aabcdefghi   But you can now understand why memmove will take care of overlapping issue

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compiler could optimize memcpy  for example   int x  memcpy  amp x  some pointer  sizeof int      This memcpy may be optimized as  x     int  some pointer

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C11 standard draft  The C11 N1570 standard draft says   7 24 2 1  The memcpy function       2 The memcpy function copies n characters from the object pointed to by s2 into the   object pointed to by s1  If copying takes place between objects that overlap  the behavior   is undefined    7 24 2 2  The memmove function       2   The memmove function copies n characters from the object pointed to by s2 into the   object pointed to by s1  Copying takes place as if the n characters from the object   pointed to by s2 are first copied into a temporary array of n characters that does not   overlap the objects pointed to by s1 and s2  and then the n characters from the   temporary array are copied into the object pointed to by s1    Therefore  any overlap on memcpy leads to undefined behavior  and anything can happen  bad  nothing or even good  Good is rare though  -   memmove however clearly says that everything happens as if an intermediate buffer is used  so clearly overlaps are OK   C   std  copy is more forgiving however  and allows overlaps  Does std  copy handle overlapping ranges

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Just because memcpy doesn t have to deal with overlapping regions  doesn t mean it doesn t deal with them correctly  The call with overlapping regions produces undefined behavior  Undefined behavior can work entirely as you expect on one platform  that doesn t mean it s correct or valid

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I have tried to run same program using eclipse and it shows clear difference between memcpy and memmove  memcpy   doesn t care about overlapping of memory location which results in corruption of data  while memmove   will copy data to temporary variable first and then copy into actual memory location   While trying to copy data from location str1 to str1 2  output of memcpy is  aaaaaa   The question would be how  memcpy   will copy one byte at a time from left to right  As shown in your program  aabbcc  then  all copying will take place as below    aabbcc - gt  aaabcc aaabcc - gt  aaaacc aaaacc - gt  aaaaac aaaaac - gt  aaaaaa   memmove   will copy data to temporary variable first and then copy to actual memory location    aabbcc actual  - gt  aabbcc temp  aabbcc temp  - gt  aaabcc act  aabbcc temp  - gt  aaaacc act  aabbcc temp  - gt  aaaabc act  aabbcc temp  - gt  aaaabb act    Output is  memcpy    aaaaaa  memmove   aaaabb

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Both memcpy and memove do similar things   But to sight out one difference    include  lt memory h gt   include  lt string h gt   include  lt stdio h gt   char str1 7     abcdef    int main         printf   The string   s n   str1       memcpy   str1 6   str1  10       printf   New string   s n   str1        strcpy s  str1  sizeof str1    aabbcc          reset string      printf   nstr1   s n   str1      printf   The string   s n   str1       memmove   str1 6   str1  10       printf   New string   s n   str1         gives   The string  abcdef New string  abcdefabcdefabcd The string  abcdef New string  abcdefabcdef

[c] memcpy() vs memmove()

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Examples related to memcpy

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