sizeof(array) returns differnt values for same array?

Code:

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#include <iostream.h>
#include <stdlib.h>
#include <time.h>


void printArray(int array[]) {
        cout<<"size of array: "<<sizeof(array)<<endl;
        for (int i=0;i<sizeof(array);i++)
                cout<<array[i]<<" ";
        cout<<endl;
}


int main () {
        int array[10];
        srand ( time(NULL) );

        for (int i=0;i<sizeof(array);i++)
                array[i] = rand() % 100;
        cout<<"size of array: "<<sizeof(array)<<endl;
        printArray(array);
        return 0;
}

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prints:
size of array: 40
size of array: 4
7 89 51 6 (or some other random numbers, key thing is that there are only 4 displayed)

why is the array differnt sizes for the function and main body?
also what is a mathematical way of using sizeof() to get the length of the array (I want to get 10 from it).

>>why is the array differnt sizes for the function and main body?
That's because arrays are passed to functions as pointers, in main sizeof returns the size of the array, but in the other function sizeof returns the size of a pointer to int, in your case 4 bytes :-)

>>also what is a mathematical way of using sizeof() to get the length of the array (I want to get 10 from it).

Code:

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int len = sizeof(array) / sizeof(array[0]);

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that makes sense.

would i have to use:

Code:

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

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in the function to get 10 as the length?

Your code is a c++ code, you're in a C board and not C++.
I'll answer you about the way to get the sizeof the array like you told, you want to receive 10.

You have an array of 10 elements and they're as Integer (4 bits), so just do:

Code:

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(sizeof(array)/sizeof(int))

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Of course if it's an array of float change to sizeof(float) and so on..

>>in the function to get 10 as the length?
You can't use sizeof on an array that was passed as a function parameter because it's just a pointer, sizeof won't work like you want. You have to pass the size with it :-)

Code:

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#include <stdio.h>

void print_array(int array[], int size)
{
  int i;

  for (i = 0; i < size; i++)
  {
    printf("%d\n", array[i]);
  }
}

int main(void)
{
  int a[] = {1,2,3,4,5};

  print_array(a, sizeof(a) / sizeof(a[0]));

  return 0;
}

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pardon my mistake, thought i was in C++

must have been a wrong click

thnx cela