How to find the size of an int

Question

I have  int list      1  2  3     How to I get the size of list   I know that for a char array  we can use strlen array  to find the size  or check with   0  at the end of the array      I tried sizeof array    sizeof array 0   as some answers said  but it only works in main  For example   int size int arr1         return sizeof arr1    sizeof arr1 0       int main         int list      1  2  3        int size1   sizeof list    sizeof list 0       ok     int size2   size list 1      no        size1 and size2 are not the same     Why

User · Answer

You have to use sizeof   function   Code Snippet    include lt bits stdc   h gt  using namespace std   int main           ios  sync with stdio false          int arr     5  3  6  7          int size   sizeof arr    sizeof arr 0          cout lt  lt size lt  lt endl         return 0

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You could use boost  size  which is basically defined this way   template  lt typename T  std  size t N gt  std  size t size T const   amp   N         return N      Note that if you want to use the size as a constant expression  you ll either have to use the sizeof a   sizeof a 0  idiom or wait for the next version of the C   standard

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This method work when you are using a class  In this example you will receive a array  so the only method that worked for me was these one   template  lt typename T  size t n  size t m gt     Matrix amp  operator   T   amp a  n  m             int arows   n      int acols   m       p   new double  arows        for  register int r   0  r  lt  arows  r                  p r    new double acols             for  register int c   0  c  lt  acols  c                          p r  c    a r  c     A rows  columns                 https   www geeksforgeeks org how-to-print-size-of-an-array-in-a-function-in-c

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If you want to know how much numbers the array have  you want to know the array length  The function sizeof var  in C gives you the bytes in the computer memory  So if you know the memory the int occupy you can do like this  int arraylength int array          return sizeof array    sizeof int      Size of the Array divided by the int size

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You can make a template function  and pass the array by reference to achieve this   Here is my code snippet  template  lt typename TypeOfData gt    void PrintArray TypeOfData  amp arrayOfType    int main           char charArray      my name is        int intArray       1 2 3 4 5 6         double doubleArray       1 1 2 2 3 3          PrintArray charArray        PrintArray intArray        PrintArray doubleArray        template  lt typename TypeOfData gt   void PrintArray TypeOfData  amp arrayOfType          int elementsCount   sizeof arrayOfType    sizeof arrayOfType 0          for  int i   0  i  lt  elementsCount  i                    cout  lt  lt   Value in elements at position    lt  lt  i   1  lt  lt    is    lt  lt  arrayOfType i   lt  lt  endl

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Try this   sizeof list    sizeof list 0      Because this question is tagged C    it is always recommended to use std  vector in C   rather than using conventional C-style arrays     An array-type is implicitly converted into a pointer-type when you pass it to a function  Have a look at  this   In order to correctly print the sizeof an array inside any function  pass the array by reference to that function  but you need to know the size of that array in advance    You would do it like so for the general case  template lt typename T int N gt     template argument deduction int size T   amp arr1  N     Passing the array by reference       return sizeof arr1  sizeof arr1 0      Correctly returns the size of  list        or    return N    Correctly returns the size too  cool trick  -

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You can t do that for a dynamically allocated array  or a pointer   For static arrays  you can use sizeof array  to get the whole array size in bytes and divide it by the size of each element    define COUNTOF x   sizeof x  sizeof  x     To get the size of a dynamic array  you have to keep track of it manually and pass it around with it  or terminate it with a sentinel value  like   0  in null terminated strings    Update  I realized that your question is tagged C   and not C  You should definitely consider using std  vector instead of arrays in C   if you want to pass things around   std  vector lt int gt  v  v push back 1   v push back 2   std  cout  lt  lt  v size    lt  lt  std  endl     prints 2

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Assuming you merely want to know the size of an array whose type you know  int  but whose size  obviously  you don t know  it is suitable to verify whether the array is empty  otherwise you will end up with a division by zero  causing a Float point exception    int array size int array          if sizeof array     0            return 0            return sizeof array  sizeof array 0

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The  standard  C way to do this is  sizeof list    sizeof list 0

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Since you ve marked this as C    it s worth mentioning that there is a somewhat better way than the C-style macro   template  lt class T  size t N gt  size t countof const T  amp array N     return N      This has the advantage that if you accidentally try to pass something other than an array to it  the code simply won t compile  whereas passing a pointer to the C macro will compile but produce a bad result  The disadvantage is that this doesn t give you a compile-time constant  so you can t do something like this   int a 20    char x countof a      In C  11 or newer  you can add constexpr to get a compile-time constant   template  lt class T  size t N gt  constexpr size t countof const T  amp array N     return N      If you really want to support the same on older compilers  there is a way  originally invented by Ivan Johnson  AFAIK    define COUNTOF x                                                   0   sizeof  reinterpret cast lt const   Bad arg to COUNTOF  gt  x           0   sizeof    Bad arg to COUNTOF  check type  x    amp  x                 sizeof x    sizeof  x  0                                          class Bad arg to COUNTOF   public     class Is pointer     class Is array          template lt typename T gt     static Is pointer check type const T   const T  const       static Is array check type const void   const void         This uses sizeof x  sizeof x 0   to compute the size  just like the C macro does  so it gives a compile-time constant  The difference is that it first uses some template magic to cause a compile error if what you ve passed isn t the name of an array  It does that by overloading check type to return an incomplete type for a pointer  but a complete type for an array  Then  the really tricky part  it doesn t actually call that function at all -- it just takes the size of the type the function would return  which is zero for the overload that returns the complete type  but not allowed  forcing a compile error  for the incomplete type   IMO  that s a pretty cool example of template meta programming -- though in all honesty  the result is kind of pointless  You really only need that size as a compile time constant if you re using arrays  which you should normally avoid in any case  Using std  vector  it s fine to supply the size at run-time  and resize the vector when if needed

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int arr1      8  15  3  7   int n   sizeof arr1  sizeof arr1 0      So basically sizeof arr1  is giving the size of the object being pointed to  each element maybe occupying multiple bits so dividing by the number of bits per element  sizeof arr1 0   gives you the actual number of elements you re looking for  i e  4 in my example

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when u pass any array to some function  u are just passing it s starting address  so for it to work u have to pass it size also for it to work properly  it s the same reason why we pass argc with argv   in command line arguement

User · Answer

Besides Carl s answer  the  standard  C   way is not to use a C int array  but rather something like a C   STL std  vector lt int gt  list which you can query for list size

[c++] How to find the size of an int[]?

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