passing size of an array to a function

...,you must pass to the function the array size. (Note:Another practise is to pass a开发者_如何学C pointer to the beginning of the array and a pointer to the location just beyond the end of the array.The difference of the two pointers is the length of the array and resulting code is simpler.)

I couldn't write the codes for what it's saying in the note. Can anyone help me?

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The problem is that if you cannot pass an array to a function. When you do so, the name of the array "decays" to a pointer to its first element. That means you no longer know the size of that array, as you can only pass around a pointer to an element in that array. So you have to pass it a size, so the function receiving the "array" knows how long it is.

e.g.

//the below declaration is even the same as e.g.
//void foo(unsigned char array[4], size_t length)
//the array notation doesn't really apply to a function argument list
//it'll mean a pointer anyhow.
void foo(unsigned char *array, size_t length)
{
    size_t i;
    for(i = 0; i < length ; i++) {
       printf("%02X ", array[i]);
    }
}

Or in the case where you pass in a pointer one element past the end of the array:

void bar(unsigned char *array, unsigned char *end)
{
     while(array != end) {
       printf("%02X ", *array);
       array++;
    }
}

And you call it like

unsigned char data[] = {1,2,3,4};
foo(data,sizeof data); //same as foo(&data[0], siseof data);
bar(data, data + 4); //same as bar(&data[0], &data[4]);

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Something like this:

int accumulate( int *first, int *last )
{
  int result = 0;

  for( ; first != last; ++first ) {
    result += *first;
  }

  return result;
}

int main()
{
  int arr[] = {1,2,3,4,5};
  int sum = accumulate( arr, arr + 5 );

  return 0;
}

The code above sums all the elements in the range [first, last); where first in included in the range and last is excluded. So you need to pass a pointer to the beginning of the range you want to sum and another to one location past the last element you want added to the sum.

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Let's assume we have this array of integers:

#define CAPACITY 42
int foo[CAPACITY];

And these two functions that process it:

void func1(int* begin, int* end) {
    // Loop over the array
    while(begin < end) {
        int val = *begin++; // process element
    }
}

void func2(int* begin, int size) {
    // Loop over the array
    for(int i = 0; i < size; ++i) {
        int val = begin[i]; // process element
    }
}

The first one would be called like func1(foo, foo + CAPACITY): passing in a pointer to the start of the array, and a pointer to just beyond the last element.

The second would be called like func2(foo, CAPACITY): passing in a pointer to the start of the array and the array size.

Due to how pointer arithmetic works, both versions are equivalent.

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void f(int* arr, int len) { ... }

void f(int* start, int* end) { ... }

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If you only want to pass the array size you could do pass the following to the function :

function(sizeof(array)/sizeof(array[0]));

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 int A[] = {1,2,3,4,5,6,7,8,9,0};
 printf("Size of the Array %d \n", sizeof(A)/sizeof(int));//size of the array


you can use this value to function 

void getSize(int A[], size_t size){
//do somthing
}

void function()
{
//from where you call this 
 int A[] = {1,2,3,4,5,6,7,8,9,0};
 getSize(A, sizeof(A)/sizeof(int));
}