Passing an array by reference in C

Question

How can I pass an array of structs by reference in C   As an example   struct Coordinate      int X     int Y     SomeMethod Coordinate  Coordinates          Do Something with the array   int main        Coordinate Coordinates 10      SomeMethod  amp Coordinates

User · Answer

The C language does not support pass by reference of any type. The closest equivalent is to pass a pointer to the type.

Here is a contrived example in both languages

C++ style API

void UpdateValue(int& i) {
  i = 42;
}

Closest C equivalent

void UpdateValue(int *i) {
  *i = 42;
}

User · Answer

In plain C you can use a pointer size combination in your API   void doSomething MyStruct  mystruct  size t numElements        for  size t i   0  i  lt  numElements    i                MyStruct current   mystruct i           handleElement current             Using pointers is the closest to call-by-reference available in C

User · Answer

Hey guys here is a simple test program that shows how to allocate and pass an array using new or malloc  Just cut  paste and run it   Have fun   struct Coordinate       int x y      void resize  int   p  int size        free   p        p    int   malloc  size   sizeof int        void resizeCoord  struct Coordinate   p  int size        free   p        p    Coordinate   malloc  size   sizeof Coordinate        void resizeCoordWithNew  struct Coordinate   p  int size        delete     p      p    struct Coordinate   new struct Coordinate size      void SomeMethod Coordinate Coordinates          Coordinates 0  x        Coordinates 0  y   6     void SomeOtherMethod Coordinate Coordinates    int size        for  int i 0  i lt size  i                  Coordinates i  x   i          Coordinates i  y   i 2           int main           static array     Coordinate tenCoordinates 10       tenCoordinates 0  x 0      SomeMethod tenCoordinates       SomeMethod  amp  tenCoordinates 0         if tenCoordinates 0  x - 2     0                printf  test1 coord change successful n              else               printf  test1 coord change unsuccessful n                 dynamic int    int  p    int   malloc  10   sizeof int        resize   amp p  20          dynamic struct with malloc    int myresize   20     int initSize   10     struct Coordinate  pcoord    struct Coordinate   malloc  initSize   sizeof struct Coordinate       resizeCoord  amp pcoord  myresize       SomeOtherMethod pcoord  myresize      bool pass   true     for  int i 0  i lt myresize  i                if      pcoord i  x    i   amp  amp   pcoord i  y    i 2                                printf  Error dynamic Coord struct   d  failed with   d  d  n  i pcoord i  x pcoord i  y              pass   false                   if  pass              printf  test2 coords for dynamic struct allocated with malloc worked correctly n                dynamic struct with new    myresize   20     initSize   10     struct Coordinate  pcoord2    struct Coordinate   new struct Coordinate initSize      resizeCoordWithNew  amp pcoord2  myresize       SomeOtherMethod pcoord2  myresize      pass   true     for  int i 0  i lt myresize  i                if      pcoord2 i  x    i   amp  amp   pcoord2 i  y    i 2                                printf  Error dynamic Coord struct   d  failed with   d  d  n  i pcoord2 i  x pcoord2 i  y              pass   false                   if  pass              printf  test3 coords for dynamic struct with new worked correctly n              return 0

User · Answer

To expand a little bit on some of the answers here     In C  when an array identifier appears in a context other than as an operand to either  amp  or sizeof  the type of the identifier is implicitly converted from  N-element array of T  to  pointer to T   and its value is implicitly set to the address of the first element in the array  which is the same as the address of the array itself    That s why when you just pass the array identifier as an argument to a function  the function receives a pointer to the base type  rather than an array   Since you can t tell how big an array is just by looking at the pointer to the first element  you have to pass the size in as a separate parameter     struct Coordinate   int x  int y     void SomeMethod struct Coordinate  coordinates  size t numCoordinates                coordinates i  x            coordinates i  y                   int main  void        struct Coordinate coordinates 10               SomeMethod  coordinates  sizeof coordinates   sizeof  coordinates               There are a couple of alternate ways of passing arrays to functions     There is such a thing as a pointer to an array of T  as opposed to a pointer to T   You would declare such a pointer as  T   p  N     In this case  p is a pointer to an N-element array of T  as opposed to T  p N   where p is an N-element array of pointer to T    So you could pass a pointer to the array as opposed to a pointer to the first element   struct Coordinate   int x  int y     void SomeMethod struct Coordinate   coordinates  10                   coordinates  i  x              coordinates  i  y                   int main void        struct Coordinate coordinates 10               SomeMethod  amp coordinates               The disadvantage of this method is that the array size is fixed  since a pointer to a 10-element array of T is a different type from a pointer to a 20-element array of T      A third method is to wrap the array in a struct   struct Coordinate   int x  int y     struct CoordinateWrapper   struct Coordinate coordinates 10      void SomeMethod struct CoordinateWrapper wrapper                wrapper coordinates i  x            wrapper coordinates i  y                  int main void        struct CoordinateWrapper wrapper              SomeMethod wrapper               The advantage of this method is that you aren t mucking around with pointers   The disadvantage is that the array size is fixed  again  a 10-element array of T is a different type from a 20-element array of T

User · Answer

also be aware that if you are creating a array within a method  you cannot return it  If you return a pointer to it  it would have been removed from the stack when the function returns  you must allocate memory onto the heap and return a pointer to that  eg     this is bad char  getname       char name 100     return name       this is better char  getname       char  name   malloc 100     return name      remember to free name

User · Answer

In C arrays are passed as a pointer to the first element  They are the only element that is not really passed by value  the pointer is passed by value  but the array is not copied   That allows the called function to modify the contents   void reset  int  array  int size       memset array 0 size   sizeof  array      int main        int array 10      reset  array  10       sets all elements to 0     Now  if what you want is changing the array itself  number of elements     you cannot do it with stack or global arrays  only with dynamically allocated memory in the heap  In that case   if you want to change the pointer you must pass a pointer to it   void resize  int   p  int size        free   p        p   malloc  size   sizeof int       int main        int  p   malloc  10   sizeof int        resize   amp p  20        In the question edit you ask specifically about passing an array of structs  You have two solutions there  declare a typedef  or make explicit that you are passing an struct   struct Coordinate      int x     int y     void f  struct Coordinate coordinates    int size    typedef struct Coordinate Coordinate      generate a type alias  Coordinate  that is equivalent to struct Coordinate void g  Coordinate coordinates    int size       uses typedef ed Coordinate   You can typedef the type as you declare it  and it is a common idiom in C    typedef struct Coordinate      int x     int y    Coordinate

User · Answer

Arrays are effectively passed by reference by default  Actually the value of the pointer to the first element is passed  Therefore the function or method receiving this can modify the values in the array   void SomeMethod Coordinate Coordinates    Coordinates 0  x      int main      Coordinate tenCoordinates 10     tenCoordinates 0  x 0    SomeMethod tenCoordinates       SomeMethod  amp tenCoordinates 0      if 0  tenCoordinates 0  x - 2        exit 0         exit -1       The two calls are equivalent  and the exit value should be 0

[c] Passing an array by reference in C?

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