size of array - why function gives size ONE only
the user-defined function size_of_array(numA) is meant to give the size of numA which is 5 but it gives 1...wonder where went wrong?
#include <iostream>
using namespace std;
int size_of_array(int numA[]);
int main()
{
int numA[]={0,1,2,3,4};
int i=0;
int total=0;
for(i=0; i<size_of_array(numA); i++)
{
total+=numA[i];
}
float average=total/i;
cout << "average is "<<average;
return 0;
}
int size_of_array(int numA[])
{
int num_of_elements=sizeof(numA)/sizeof(int);
return num_of_elements;
}
#include <iostream>
using namespace std;
int size_of_array(int numA[]);
int main()
{
int numA[]={0,1,2,3,4};
int i=0;
int total=0;
for(i=0; i<size_of_array(numA); i++)
{
total+=numA[i];
}
float average=total/i;
cout << "average is "<<average;
return 0;
}
int size_of_array(int numA[])
{
int num_of_elements=sizeof(numA)/sizeof(int);
return num_of_elements;
}
because anything divided by itself is 1 and I think that is whats happening. its pointing at the first data type in numA and dividing it by the size of int which would equal 1. I may be wrong though
i dont know if you know about pointers yet
http://www.cs.stanford.edu/cslibrary/PointerFunCppBig.avi
Ok. In memory when you pass an array into a function it points to the first datatype. THen when you increment it goes to the next position etc. Its a little confusing. But that is just how it works
http://www.cs.stanford.edu/cslibrary/PointerFunCppBig.avi
Ok. In memory when you pass an array into a function it points to the first datatype. THen when you increment it goes to the next position etc. Its a little confusing. But that is just how it works
The compiler has no idea how many elements are present in the array when it is compiling the size_of_array() function.
It is likely taking the array pointer and giving you the size of that which just happens to be the same size as an int.
It is likely taking the array pointer and giving you the size of that which just happens to be the same size as an int.
Topic archived. No new replies allowed.