Length of array in function argument

Question

This is well known code to compute array length in C   sizeof array  sizeof type    But I can t seem to find out the length of the array passed as an argument to a function    include  lt stdio h gt   int length const char  array        return sizeof array  sizeof char       int main       const char  friends        John    Jack    Jim       printf   d  d   sizeof friends  sizeof char    length friends       3 1     I assume that array is copied by value to the function argument as constant pointer and reference to it should solve this  but this declaration is not valid   int length const char   amp  array     I find passing the array length as second argument to be redundant information  but why is the standard declaration of main like this   int main int argc  char   argv     Please explain if it is possible to find out the array length in function argument  and if so  why is there the redundancy in main

User · Answer

First  a better usage to compute number of elements when the actual array declaration is in scope is   sizeof array   sizeof array 0    This way you don t repeat the type name  which of course could change in the declaration and make you end up with an incorrect length computation  This is a typical case of don t repeat yourself   Second  as a minor point  please note that sizeof is not a function  so the expression above doesn t need any parenthesis around the argument to sizeof   Third  C doesn t have references so your usage of  amp  in a declaration won t work   I agree that the proper C solution is to pass the length  using the size t type  as a separate argument  and use sizeof at the place the call is being made if the argument is a  real  array   Note that often you work with memory returned by e g  malloc    and in those cases you never have a  true  array to compute the size off of  so designing the function to use an element count is more flexible

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Best example is here  thanks  define SIZE 10  void size int arr SIZE         printf  size of array is  d n  sizeof arr       int main         int arr SIZE       size arr       return 0

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The array decays to a pointer when passed   Section 6 4 of the C FAQ covers this very well and provides the K amp R references etc     That aside  imagine it were possible for the function to know the size of the memory allocated in a pointer   You could call the function two or more times  each time with different input arrays that were potentially different lengths  the length would therefore have to be passed in as a secret hidden variable somehow   And then consider if you passed in an offset into another array  or an array allocated on the heap  malloc and all being library functions - something the compiler links to  rather than sees and reasons about the body of    Its getting difficult to imagine how this might work without some behind-the-scenes slice objects and such right     Symbian did have a AllocSize   function that returned the size of an allocation with malloc    this only worked for the literal pointer returned by the malloc  and you d get gobbledygook or a crash if you asked it to know the size of an invalid pointer or a pointer offset from one   You don t want to believe its not possible  but it genuinely isn t   The only way to know the length of something passed into a function is to track the length yourself and pass it in yourself as a separate explicit parameter

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sizeof only works to find the length of the array if you apply it to the original array   int a 5     real array  NOT a pointer sizeof a           However  by the time the array decays into a pointer  sizeof will give the size of the pointer and not of the array   int a 5   int   p   a  sizeof p           As you have already smartly pointed out main receives the length of the array as an argument  argc   Yes  this is out of necessity and is not redundant   Well  it is kind of reduntant since argv is conveniently terminated by a null pointer but I digress   There is some reasoning as to why this would take place  How could we make things so that a C array also knows its length   A first idea would be not having arrays decaying into pointers when they are passed to a function and continuing to keep the array length in the type system  The bad thing about this is that you would need to have a separate function for every possible array length and doing so is not a good idea   Pascal did this and some people think this is one of the reasons it  lost  to C   A second idea is storing the array length next to the array  just like any modern programming language does   a - gt   5   0 0 0 0 0    But then you are just creating an invisible struct behind the scenes and the C philosophy does not approve of this kind of overhead  That said  creating such a struct yourself is often a good idea for some sorts of problems   struct       size t length      int   elements        Another thing you can think about is how strings in C are null terminated instead of storing a length  as in Pascal   To store a length without worrying about limits need a whopping four bytes  an unimaginably expensive amount  at least back then   One could wonder if arrays could be also null terminated like that but then how would you allow the array to store a null

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As stated by  Will  the decay happens during the parameter passing  One way to get around it is to pass the number of elements  To add onto this  you may find the  countof   macro useful - it does the equivalent of what you ve done

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length of an array type int  with sizeof  sizeof array  sizeof int

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Regarding int main     According to the Standard  argv points to a NULL-terminated array  of pointers to null-terminated strings    5 1 2 2 1 1    That is  argv    char      argv 0        argv argc - 1   0      Hence  size calculation is performed by a function which is a trivial modification of strlen     argc is only there to make argv length calculation O 1    The count-until-NULL method will NOT work for generic array input  You will need to manually specify size as a second argument

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int arsize int st1          int i   0      for  i    st1 i   amp   1  lt  lt  30    i         return i      This works for me

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This is a old question  and the OP seems to mix C   and C in his intends examples  In C  when you pass a array to a function  it s decayed to pointer  So  there is no way to pass the array size except by using a second argument in your function that stores the array size   void func int A        should be instead  void func int   A  const size t elemCountInA    They are very few cases  where you don t need this  like when you re using multidimensional arrays   void func int A 3  whatever here      That s almost as if read  int  A 3     Using the array notation in a function signature is still useful  for the developer  as it might be an help to tell how many elements your functions expects  For example   void vec add float out 3   float in0 3   float in1 3     is easier to understand than this one  although  nothing prevent accessing the 4th element in the function in both functions    void vec add float   out  float   in0  float   in1    If you were to use C    then you can actually capture the array size and get what you expect   template  lt size t N gt  void vec add float   amp out  N   float   amp in0  N   float   amp in1  N         for  size t i   0  i  lt  N  i             out i    in0 i    in1 i       In that case  the compiler will ensure that you re not adding a 4D vector with a 2D vector  which is not possible in C without passing the dimension of each dimension as arguments of the function   There will be as many instance of the vec add function as the number of dimensions used for your vectors

[c] Length of array in function argument

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