/* bigInt.c
 *
 * Demonstrates use of passing an array to a function.
 *
 * Also shows why sizeof() is not a reliable way to discover array size. It
 *   *only* works correctly inside the function it's declared in. When passed
 *   to a function, the *address* of teh begining of the array is passed and so
 *   sizeof() returns the sizeof an address. Here's the compiler output
 *   (I ignored the compile warnings to run it)
 *
 * wightman% gcc bigInt.c
 * bigInt.c:32:49: warning: sizeof on array function parameter will return size of
 * 'int *' instead of 'int []' [-Wsizeof-array-argument]
 * printf("bigInt: sizeof(a): %lu bytes\n", sizeof(arr));
 *                                                    ^
 * bigInt.c:29:16: note: declared here
 * int bigInt(int arr[], int iNarr){
 *                ^
 * 1 warning generated.
 */
#include <stdio.h>
#include <stdlib.h>
#define N 5

int bigInt(int arr[], int iNarr);

int main()
{
	int i;
	int iErr;
	int a[5];

	// sizeof() return a long unsigned int (%lu)
	printf("main: sizeof(a): %lu bytes\n", sizeof(a));

  printf("enter some integers,followed by ^D on blank line: ");
	i = 0;
  iErr = 1;
	while( iErr != 0 && i < N ){
		iErr = scanf("%d",&a[i]);
		i++;
	}

	printf("Biggest value: %d\n", bigInt(a,N));

	return EXIT_SUCCESS;
}

int bigInt(int arr[], int iNarr){
	int i;
	int big = arr[0];
	printf("bigInt: sizeof(a): %lu bytes\n", sizeof(arr));
	for(i=1; i<iNarr; i++){
		if(arr[i] > big)
		  big = arr[i];
	}
	return big;
}