Passing an array as an argument to a function in C

Question

I wrote a function containing array as argument  and call it by passing value of array as follows   void arraytest int a             changed the array a     a 0  a 0  a 1       a 1  a 0 -a 1       a 0  a 0 -a 1      void main         int arr    1 2       printf   d  t  d  arr 0  arr 1        arraytest arr       printf   n After calling fun arr contains   d t  d  arr 0  arr 1        What I found is though I am calling arraytest   function by passing values  the original copy of int arr   is changed   Can you please explain why

User · Accepted Answer

When passing an array as a parameter  this  void arraytest int a      means exactly the same as  void arraytest int  a    so you are modifying the values in main   For historical reasons  arrays are not first class citizens and cannot be passed by value

User · Answer

You are passing the address of the first element of the array

User · Answer

Arrays in C are converted  in most of the cases  to a pointer to the first element of the array itself  And more in detail arrays passed into functions are always converted into pointers    Here a quote from K amp R2nd      When an array name is passed to a function  what is passed is the   location of the initial element  Within the called function  this   argument is a local variable  and so an array name parameter is a   pointer  that is  a variable containing an address     Writing   void arraytest int a      has the same meaning as writing   void arraytest int  a    So despite you are not writing it explicitly it is as you are passing a pointer and so you are modifying the values in the main   For more I really suggest reading this   Moreover  you can find other answers on SO here

User · Answer

You are not passing the array as copy  It is only a pointer pointing to the address where the first element of the array is in memory

User · Answer

1  Standard array usage in C with natural type decay from array to ptr  Bo Persson correctly states in his great answer here   When passing an array as a parameter  this void arraytest int a     means exactly the same as void arraytest int  a    However  let me add also that the above two forms also   mean exactly the same as  void arraytest int a 0     which means exactly the same as  void arraytest int a 1     which means exactly the same as  void arraytest int a 2     which means exactly the same as  void arraytest int a 1000     etc    In every single one of the array examples above  and as shown in the example calls in the code just below  the input parameter type decays to an int    and can be called with no warnings and no errors  even with build options -Wall -Wextra -Werror turned on  see my repo here for details on these 3 build options   like this  int array1 2   int   array2   array1      works fine because  array1  automatically decays from an array type    to  int    arraytest array1      works fine because  array2  is already an  int     arraytest array2    As a matter of fact  the  quot size quot  value   0    1    2    1000   etc   inside the array parameter here is apparently just for aesthetic self-documentation purposes  and can be any positive integer  size t type I think  you want  In practice  however  you should use it to specify the minimum size of the array you expect the function to receive  so that when writing code it s easy for you to track and verify  The MISRA-C-2012 standard  buy download the 236-pg 2012-version PDF of the standard for   15 00 here  goes so far as to state  emphasis added    Rule 17 5 The function argument corresponding to a parameter declared to have an array type shall have an appropriate number of elements      If a parameter is declared as an array with a specified size  the corresponding argument in each function call should point into an object that has at least as many elements as the array      The use of an array declarator for a function parameter specifies the function interface more clearly than using a pointer  The minimum number of elements expected by the function is explicitly stated  whereas this is not possible with a pointer   In other words  they recommend using the explicit size format  even though the C standard technically doesn t enforce it--it at least helps clarify to you as a developer  and to others using the code  what size array the function is expecting you to pass in   2  Forcing type safety on arrays in C  Not recommended  but possible  See my brief argument against doing this at the end   As  Winger Sendon points out in a comment below my answer  we can force C to treat an array type to be different based on the array size  First  you must recognize that in my example just above  using the int array1 2   like this  arraytest array1   causes array1 to automatically decay into an int    HOWEVER  if you take the address of array1 instead and call arraytest  amp array1   you get completely different behavior  Now  it does NOT decay into an int    Instead  the type of  amp array1 is int     2   which means  quot pointer to an array of size 2 of int quot   or  quot pointer to an array of size 2 of type int quot   or said also as  quot pointer to an array of 2 ints quot   So  you can FORCE C to check for type safety on an array  like this  void arraytest int   a  2            my function here    This syntax is hard to read  but similar to that of a function pointer  The online tool  cdecl  tells us that int   a  2  means   quot declare a as pointer to array 2 of int quot   pointer to array of 2 ints   Do NOT confuse this with the version withOUT parenthesis  int   a 2   which means   quot declare a as array 2 of pointer to int quot   AKA  array of 2 pointers to int  AKA  array of 2 int s   Now  this function REQUIRES you to call it with the address operator   amp   like this  using as an input parameter a POINTER TO AN ARRAY OF THE CORRECT SIZE   int array1 2       ok  since the type of  array1  is  int     2    ptr to array of     2 ints  arraytest  amp array1      you must use the  amp  operator here to prevent                         array1  from otherwise automatically decaying                        into  int     which is the WRONG input type here   This  however  will produce a warning  int array1 2       WARNING  Wrong type since the type of  array1  decays to  int             main c 32 15  warning  passing argument 1 of    arraytest    from          incompatible pointer type  -Wincompatible-pointer-types                                                                      main c 22 6  note  expected    int     2     but argument is of type    int      arraytest array1       missing  amp  operator   You may test this code here  To force the C compiler to turn this warning into an error  so that you MUST always call arraytest  amp array1   using only an input array of the corrrect size and type  int array1 2   in this case   add -Werror to your build options  If running the test code above on onlinegdb com  do this by clicking the gear icon in the top-right and click on  quot Extra Compiler Flags quot  to type this option in  Now  this warning   main c 34 15  warning  passing argument 1 of    arraytest    from incompatible pointer type  -Wincompatible-pointer-types                                                              main c 24 6  note  expected    int     2     but argument is of type    int            will turn into this build error   main c  In function    main     main c 34 15  error  passing argument 1 of    arraytest    from incompatible pointer type  -Werror incompatible-pointer-types       arraytest array1      warning                        main c 24 6  note  expected    int     2     but argument is of type    int       void arraytest int   a  2                   cc1  all warnings being treated as errors    Note that you can also create  quot type safe quot  pointers to arrays of a given size  like this  int array 2       quot type safe quot  ptr to array of size 2 of int  int   array p  2     amp array      but I do NOT necessarily recommend this  using these  quot type safe quot  arrays in C   as it reminds me a lot of the C   antics used to force type safety everywhere  at the exceptionally high cost of language syntax complexity  verbosity  and difficulty architecting code  and which I dislike and have ranted about many times before  ex  see  quot My Thoughts on C   quot  here    For additional tests and experimentation  see also the link just below  References See links above  Also   My code experimentation online  https   onlinegdb com B1RsrBDFD

User · Answer

Passing a multidimensional array as argument to a function  Passing an one dim array as argument is more or less trivial  Let s take a look on more interesting case of passing a 2 dim array  In C you can t use a pointer to pointer construct  int      instead of 2 dim array  Let s make an example   void assignZeros int  arr  5   const int rows        for  int i   0  i  lt  rows  i              for  int j   0  j  lt  5  j                      arr   i    j    0                 or equivalent assignment             arr i  j    0                    Here I have specified a function that takes as first argument a pointer to an array of 5 integers   I can pass as argument any 2 dim array that has 5 columns   int arr1 1  5  int arr1 2  5      int arr1 20  5        You may come to an idea to define a more general function that can accept any 2 dim array and change the function signature as follows   void assignZeros int    arr  const int rows  const int cols        for  int i   0  i  lt  rows  i              for  int j   0  j  lt  cols  j                      arr   i    j    0                      This code would compile but you will get a runtime error when trying to assign the values in the same way as in the first function  So in C a multidimensional arrays are not the same as pointers to pointers     to pointers  An int  arr  5  is a pointer to array of 5 elements   an int  arr  6   is a pointer to array of 6 elements  and they are a pointers to different types   Well  how to define functions arguments for higher dimensions  Simple  we just follow the pattern  Here is the same function adjusted to take an array of 3 dimensions   void assignZeros2 int  arr  4  5   const int dim1  const int dim2  const int dim3        for  int i   0  i  lt  dim1  i              for  int j   0  j  lt  dim2  j                  for  int k   0  k  lt  dim3  k                            arr   i    j    k    0                     or equivalent assignment                 arr i  j  k    0                                    How you would expect  it can take as argument any 3 dim arrays that have in the second dimensions 4 elements and in the third dimension 5 elements  Anything like this would be OK   arr 1  4  5  arr 2  4  5      arr 10  4  5        But we have to specify all dimensions sizes up to the first one

User · Answer

Arrays are always passed by reference if you use a   or  a  int  printSquares int a    int size  int e             for int i   0  i  lt  size  i              e i    i   i            return e     int  printSquares int  a  int size  int e          for int i   0  i  lt  size  i              e i    i   i            return e

User · Answer

You are passing the value of the memory location of the first member of the array   Therefore when you start modifying the array inside the function  you are modifying the original array   Remember that a 1  is   a 1

User · Answer

In C  except for a few special cases  an array reference always  decays  to a pointer to the first element of the array  Therefore  it isn t possible to pass an array  by value   An array in a function call will be passed to the function as a pointer  which is analogous to passing the array by reference   EDIT  There are three such special cases where an array does not decay to a pointer to it s first element    sizeof a is not the same as sizeof   amp a 0     amp a is not the same as  amp   amp a 0    and not quite the same as  amp a 0    char b      foo  is not the same as char b      amp   foo

User · Answer

If you want to pass a single-dimension array as an argument in a function  you would have to declare a formal parameter in one of following three ways and all three declaration methods produce similar results because each tells the compiler that an integer pointer is going to be received   int func int arr                               int func int arr SIZE                              int func int  arr                              So  you are modifying the original values   Thanks

[c] Passing an array as an argument to a function in C

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