How to copy a char array in C

Question

In C  I have two char arrays   char array1 18     abcdefg   char array2 18     How to copy the value of array1 to array2   Can I just do this  array2   array1

User · Answer

If your arrays are not string arrays  use  memcpy array2  array1  sizeof array2

User · Answer

c functions below only     c   you have to do char array then use a string copy then user the string tokenizor functions    c   made it a-lot harder to do anythng   include  lt iostream gt   include  lt fstream gt   include  lt cstring gt   define TRUE 1  define FALSE 0 typedef int Bool  using namespace std  Bool PalTrueFalse char str     int main void    char string 1000   ch  int i   0  cout lt  lt  Enter a message      while  ch   getchar         n     grab users input string untill                                    Enter is pressed     if   isspace ch   amp  amp   ispunct ch     Cstring functions checking for                                        spaces and punctuations of all kinds         string i    tolower ch            i            string i      0      hitting null deliminator once users input cout lt  lt  Your string    lt  lt string lt  lt endl   if PalTrueFalse string     the string i  user input is passed after                           being cleaned into the null function      cout lt  lt  is a   lt  lt  Palindrome n  lt  lt endl  else    cout lt  lt  Not a palindrome n  lt  lt endl  return 0     Bool PalTrueFalse char str      int left   0  int right   strlen str -1  while  left lt right       if str left     str right     comparing most outer values of string        return FALSE             to inner values     left       right--    return TRUE

User · Answer

You cannot assign arrays  the names are constants that cannot be changed   You can copy the contents  with   strcpy array2  array1     assuming the source is a valid string and that the destination is large enough  as in your example

User · Answer

for integer types   include  lt string h gt       int array1 10     0 1 2 3 4 5 6 7 8 9   int array2 10     memcpy array2 array1 sizeof array1       memcpy  destination   source   size

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None of the above was working for me   this works perfectly name here is char  name which is passed via the function  get length of char  name using strlen name  storing it in a const variable is important create same length size char array copy name  s content to temp using strcpy temp  name    use however you want  if you want original content back  strcpy name  temp   copy temp back to name and voila works perfectly     const int size   strlen name       char temp size       cout  lt  lt  size  lt  lt  endl      strcpy temp  name

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it should look like this   void cstringcpy char  src  char   dest        while   src              dest        src                dest     0            char src 6     Hello   char dest 6   cstringcpy src  dest

User · Answer

You cannot assign arrays to copy them  How you can copy the contents of one into another depends on multiple factors   For char arrays  if you know the source array is null terminated and destination array is large enough for the string in the source array  including the null terminator  use strcpy      include  lt string h gt   char array1 18     abcdefg   char array2 18         strcpy array2  array1     If you do not know if the destination array is large enough  but the source is a C string  and you want the destination to be a proper C string  use snprinf      include  lt stdio h gt   char array1      a longer string that might not fit   char array2 18         snprintf array2  sizeof array2    s   array1     If the source array is not necessarily null terminated  but you know both arrays have the same size  you can use memcpy    include  lt string h gt   char array1 28     a non null terminated string   char array2 28         memcpy array2  array1  sizeof array2

User · Answer

I recommend to use memcpy   for copying data  Also if we assign a buffer to another as array2   array1   both array have same memory and any change in the arrary1 deflects in array2 too   But we use memcpy  both buffer have different array  I recommend memcpy   because strcpy and related function do not copy NULL character

User · Answer

If you want to guard against non-terminated strings  which can cause all sorts of problems  copy your string like this   char array1 18      abcdefg    char array2 18    size t destination size   sizeof  array2    strncpy array2  array1  destination size   array2 destination size - 1      0     That last line is actually important  because strncpy   does not always null terminate strings   If the destination buffer is too small to contain the whole source string  sntrcpy   will not null terminate the destination string    The manpage for strncpy   even states  Warning  If there is no null byte among the first n bytes of src  the string placed in dest will not be null-terminated    The reason strncpy   behaves this somewhat odd way  is because it was not actually originally intended as a safe way to copy strings   Another way is to use snprintf   as a safe replacement for strcpy     snprintf array2  destination size    s   array1      Thanks jxh for the tip

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Well  techincally you can     typedef struct   char xx 18     arr wrap   char array1 18     abcdefg   char array2 18       arr wrap    array2       arr wrap    array1    printf   s n   array2           abcdefg       but it will not look very beautiful      Unless you use the C preprocessor      define CC MEMCPY DESTARR  SRCARR  ARRSIZE          struct  tmparrwrap    char xx ARRSIZE         struct  tmparrwrap     DESTARR       struct  tmparrwrap     SRCARR       You can then do   char array1 18     abcdefg   char array2 18    CC MEMCPY array2  array1  sizeof array1     printf   s n   array2           abcdefg       And it will work with any data type  not just char   int numbers1 3      1  2  3    int numbers2 3    CC MEMCPY numbers2  numbers1  sizeof numbers1     printf   d -  d -  d n   numbers2 0   numbers2 1   numbers2 2            abcdefg        Yes  the code above is granted to work always and it s portable

User · Answer

array2   array1    is not supported in c  You have to use functions like strcpy   to do it

User · Answer

You can t directly do array2   array1  because in this case you manipulate the addresses of the arrays  char    and not of their inner values  char    What you  conceptually  want is to do is iterate through all the chars of your source  array1  and copy them to the destination  array2   There are several ways to do this  For example you could write a simple for loop  or  use memcpy   That being said  the recommended way for strings is to use strncpy  It prevents common errors resulting in  for example  buffer overflows  which is especially dangerous if array1 is filled from user input  keyboard  network  etc   Like so      Will copy 18 characters from array1 to array2 strncpy array2  array1  18     As  Prof  Falken mentioned in a comment  strncpy can be evil  Make sure your target buffer is big enough to contain the source buffer  including the  0 at the end of the string

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As others have noted  strings are copied with strcpy   or its variants  In certain cases  you could use snprintf   as well   You can only assign arrays the way you want as part of a structure assignment   typedef struct   char a 18     array  array array1      abcdefg     array array2   array2   array1    If your arrays are passed to a function  it will appear that you are allowed to assign them  but this is just an accident of the semantics  In C  an array will decay to a pointer type with the value of the address of the first member of the array  and this pointer is what gets passed  So  your array parameter in your function is really just a pointer  The assignment is just a pointer assignment   void foo  char x 10   char y 10         x   y        pointer assignment         puts x       The array itself remains unchanged after returning from the function   This  decay to pointer value  semantic for arrays is the reason that the assignment doesn t work  The l-value has the array type  but the r-value is the decayed pointer type  so the assignment is between incompatible types   char array1 18     abcdefg   char array2 18   array2   array1     fails because array1 becomes a pointer type                      but array2 is still an array type      As to why the  decay to pointer value  semantic was introduced  this was to achieve a source code compatibility with the predecessor of C  You can read The Development of the C Language for details

[c] How to copy a char array in C?

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