Returning an array using C

Question

I am relatively new to C and I need some help with methods dealing with arrays  Coming from Java programming  I am used to being able to say int    method  in order to return an array  However  I have found out that with C you have to use pointers for arrays when you return them  Being a new programmer  I really do not understand this at all  even with the many forums I have looked through    Basically  I am trying to write a method that returns a char array in C  I will provide the method  lets call it returnArray  with an array  It will create a new array from the previous array and return a pointer to it  I just need some help on how to get this started and how to read the pointer once it is sent out of the array  Any help explaining this is appreciated   Proposed Code Format for Array Returning Function  char  returnArray char array       char returned  10      methods to pull values from array  interpret them  and then create new array  return  amp  returned 0      is this correct       Caller of the Function  int main     int i 0   char array     1 0 0 0 0 1 1    char arrayCount 0   char  returnedArray   returnArray  amp arrayCount      is this correct   for  i 0  i lt 10 i      printf  d       returnedArray i       is this correctly formatted      I have not tested this yet as my C compiler is not working at the moment but I would like to figure this out

User · Answer

I am not saying that this is the best solution or a preferred solution to the given problem. However, it may be useful to remember that functions can return structs. Although functions cannot return arrays, arrays can be wrapped in structs and the function can return the struct thereby carrying the array with it. This works for fixed length arrays.

    #include <stdio.h>
    #include <stdlib.h>
    #include <string.h>

    typedef
    struct 
    {
        char v[10];
    } CHAR_ARRAY;



    CHAR_ARRAY returnArray(CHAR_ARRAY array_in, int size)
    {
        CHAR_ARRAY returned;

        /*
        . . . methods to pull values from array, interpret them, and then create new array
        */

        for (int i = 0;  i < size; i++ )
            returned.v[i] = array_in.v[i] + 1;

        return returned; // Works!
    } 




    int main(int argc, char * argv[])
    {
        CHAR_ARRAY array = {1,0,0,0,0,1,1};

        char arrayCount = 7;

        CHAR_ARRAY returnedArray = returnArray(array, arrayCount); 

        for (int i = 0; i < arrayCount; i++)
            printf("%d, ", returnedArray.v[i]);  //is this correctly formatted?

        getchar();
        return 0;
    }

I invite comments on the strengths and weaknesses of this technique. I have not bothered to do so.

User · Answer

Your method will return a local stack variable that will fail badly   To return an array  create one outside the function  pass it by address into the function  then modify it  or create an array on the heap and return that variable   Both will work  but the first doesn t require any dynamic memory allocation to get it working correctly   void returnArray int size  char  retArray         work directly with retArray or memcpy into it from elsewhere like      memcpy retArray  localArray  size        define ARRAY SIZE 20  int main void      char foo ARRAY SIZE     returnArray ARRAY SIZE  foo

User · Answer

How about this deliciously evil implementation   array h   define IMPORT ARRAY TYPE             struct TYPE  Array            TYPE  contents           size t size                     struct TYPE  Array new   TYPE  Array              struct TYPE  Array a           a contents   NULL           a size   0           return a                    void array add struct TYPE  Array  o  TYPE value             TYPE  a   malloc  o- gt size   1    sizeof TYPE             TYPE i           for i   0  i  lt  o- gt size    i                 a i    o- gt contents i                          o- gt size            a o- gt size - 1    value           free o- gt contents            o- gt contents   a              void array destroy struct TYPE  Array  o             free o- gt contents               TYPE  array begin struct TYPE  Array  o             return o- gt contents              TYPE  array end struct TYPE  Array  o             return o- gt contents   o- gt size           main c   include  lt stdlib h gt   include  array h   IMPORT ARRAY int    struct intArray return an array         struct intArray a      a   new intArray        array add  amp a  1       array add  amp a  2       array add  amp a  3       return a     int main         struct intArray a      int  it      int  begin      int  end      a   return an array        begin   array begin  amp a       end   array end  amp a       for it   begin  it    end    it            printf   d     it             array destroy  amp a       getchar        return 0

User · Answer

C s treatment of arrays is very different from Java s  and you ll have to adjust your thinking accordingly   Arrays in C are not first-class objects  that is  an array expression does not retain it s  array-ness  in most contexts    In C  an expression of type  N-element array of T  will be implicitly converted   decay   to an expression of type  pointer to T   except when the array expression is an operand of the sizeof or unary  amp  operators  or if the array expression is a string literal being used to initialize another array in a declaration     Among other things  this means that you cannot pass an array expression to a function and have it received as an array type  the function actually receives a pointer type   void foo char  a  size t asize         do something with a    int bar void      char str 6     Hello     foo str  sizeof str       In the call to foo  the expression str is converted from type char  6  to char    which is why the first parameter of foo is declared char  a instead of char a 6    In sizeof str  since the array expression is an operand of the sizeof operator  it s not converted to a pointer type  so you get the number of bytes in the array  6      If you re really interested  you can read Dennis Ritchie s The Development of the C Language to understand where this treatment comes from    The upshot is that functions cannot return array types  which is fine since array expressions cannot be the target of an assignment  either     The safest method is for the caller to define the array  and pass its address and size to the function that s supposed to write to it   void returnArray const char  srcArray  size t srcSize  char  dstArray  char dstSize            dstArray i    some value derived from srcArray i             int main void      char src      This is a test     char dst sizeof src           returnArray src  sizeof src  dst  sizeof dst             Another method is for the function to allocate the array dynamically and return the pointer and size   char  returnArray const char  srcArray  size t srcSize  size t  dstSize      char  dstArray   malloc srcSize     if  dstArray           dstSize   srcSize                return dstArray     int main void      char src      This is a test     char  dst    size t dstSize     dst   returnArray src  sizeof src   amp dstSize           free dst             In this case  the caller is responsible for deallocating the array with the free library function     Note that dst in the above code is a simple pointer to char  not a pointer to an array of char   C s pointer and array semantics are such that you can apply the subscript operator    to either an expression of array type or pointer type  both src i  and dst i  will access the i th element of the array  even though only src has array type    You can declare a pointer to an N-element array of T and do something similar   char   returnArray const char  srcArr  size t srcSize   SOME SIZE      char   dstArr  SOME SIZE    malloc sizeof  dstArr     if  dstArr                    dstArr  i                       return dstArr     int main void      char src      This is a test     char   dst  SOME SIZE           dst   returnArray src  sizeof src           printf   c     dst  j              Several drawbacks with the above   First of all  older versions of C expect SOME SIZE to be a compile-time constant  meaning that function will only ever work with one array size   Secondly  you have to dereference the pointer before applying the subscript  which clutters the code   Pointers to arrays work better when you re dealing with multi-dimensional arrays

User · Answer

You can use code like this   char  MyFunction some arguments           char  pointer   malloc size for the new array       if   pointer          An error occurred  abort or do something about the error      return pointer     Return address of memory to the caller      When you do this  the memory should later be freed  by passing the address to free   There are other options  A routine might return a pointer to an array  or portion of an array  that is part of some existing structure  The caller might pass an array  and the routine merely writes into the array  rather than allocating space for a new array

User · Answer

You can t return arrays from functions in C   You also can t  shouldn t  do this   char  returnArray char array       char returned  10      methods to pull values from array  interpret them  and then create new array  return  amp  returned 0      is this correct       returned is created with automatic storage duration and references to it will become invalid once it leaves its declaring scope  i e   when the function returns   You will need to dynamically allocate the memory inside of the function or fill a preallocated buffer provided by the caller   Option 1    dynamically allocate the memory inside of the function  caller responsible for deallocating ret   char  foo int count        char  ret   malloc count       if  ret          return NULL       for int i   0  i  lt  count    i           ret i    i       return ret      Call it like so   int main         char  p   foo 10       if p               do stuff with p         free p              return 0      Option 2    fill a preallocated buffer provided by the caller  caller allocates buf and passes to the function   void foo char  buf  int count        for int i   0  i  lt  count    i          buf i    i      And call it like so   int main         char arr 10     0       foo arr  10          No need to deallocate because we allocated         arr with automatic storage duration         If we had dynamically allocated it         i e  malloc or some variant  then we         would need to call free arr

User · Answer

In your case  you are creating an array on the stack and once you leave the function scope  the array will be deallocated  Instead  create a dynamically allocated array and return a pointer to it   char   returnArray char  arr  int size        char  new arr   malloc sizeof char    size       for int i   0  i  lt  size    i            new arr i    arr i             return new arr     int main          char arr 7    1 0 0 0 0 1 1       char  new arr   returnArray arr  7           don t forget to free the memory after you re done with the array     free new arr

User · Answer

You can do it using heap memory  through malloc   invocation  like other answers reported here  but you must always manage the memory  use free   function everytime you call your function   You can also do it with a static array   char  returnArrayPointer      static char array SIZE       do something in your array here  return array       You can than use it without worrying about memory management   int main      char  myArray   returnArrayPointer       use your array here       don t worry to free memory here        In this example you must use static keyword in array definition to set to application-long the array lifetime  so it will not destroyed after return statement  Of course  in this way  you occupy SIZE bytes in your memory for the entire application life  so size it properly

[c] Returning an array using C

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