sizeof() only works as parameter, not as variable in my function?

[platypus1130]

The following is my code to average the doubles in an array:

#include <iostream>
using namespace std;

long double average(long double n[],long int num_of_terms){
    long double sum=0;
	long int i;
    for(i=0; i<num_of_terms; i++)sum+=n[i];
    return sum/num_of_terms;
}
int main(){
	long double n[]={1,2,3,4,5,6,7};
    cout<<average(n,sizeof(n)/16)<<endl;
}

apparently, I discovered that the sizeof returned 16x the number of elements in the array,

which is y my num_of_terms parameter is sizeof(n)/16

however, if i make sizeof(n)/16 a variable in the average function, it returns

nan every time? What the hell?? pls help...

[helios]

Try sizeof(n)/sizeof(*n).

however, if i make sizeof(n)/16 a variable in the average function, it returns

How do you mean?

[platypus1130]

long double average(long double n[]){
long int num_of_terms=sizeof(n)/16;//no longer a parameter
long double sum=0;
long int i;
for(i=0; i<num_of_terms; i++)sum+=n[i];
return sum/num_of_terms;
}
int main(){
long double n[]={1,2,3,4,5,6,7};
cout<<average(n)<<endl;
}

this returns nan
why?? and what to do...

[guestgulkan]

why are you using the value of 16 directly?

If you want to find out the size of a type or object - then use sizeof(type) or sizeof(object)

so in this case:
long int num_of_terms = sizeof(n) / sizeof(long double)

#include <iostream>
using namespace std;

long double average(long double n[],long int num_of_terms)
{
    long double sum=0;
    long int i;
    for(i=0; i<num_of_terms; i++)sum+=n[i];
    return sum/num_of_terms;
}

int main()
{
   long double n[]={1,2,3,4,5,6,7};
   cout<<average(n,sizeof(n)/sizeof(long double) )<<endl;
}

[platypus1130]

thats great, but it doesnt explain why the value of num_of_terms is always 0 when sizeof(n)/16 is a variable rather than a parameter

[guestgulkan]

** your second method which I have shown below will not work**

long double average(long double n[])
{
    long int num_of_terms=sizeof(n)/16;//problem here - see comment below
    long double sum=0;
    long int i;
    for(i=0; i<num_of_terms; i++)sum+=n[i];
    return sum/num_of_terms;
}


int main()
{
    long double n[]={1,2,3,4,5,6,7};
    cout<<average(n)<<endl;
}

It will not work because the parameter being passed to the function is
only a pointer to the array not the array itself - so sizeof(n) will only report
4 bytes or so (the size of a pointer).
So long int num_of_terms=sizeof(n)/16; will be 4 /16 = 0

so return sum/num_of_terms; will a whatever sum happens to be DIVIDED BY 0. In this case it is 0.0/0
Divide by zero is a fatal error - it either crashes the program or is this case
the system catches it and reports NAN

[platypus1130]

that makes sense... so how would I be able to find the size of the array and not the pointer in the code you copied above???

[helios]

It's impossible to know the size of an array, unless it's been created in the stack, and then only from the function that created it:

void f2(int *);

void f(){
	int v1[4];
	//Ok. The size of v1 is known because it was created in the stack and we're
	//inside the function that created it.
	int size1=sizeof(v1)/sizeof(*v1);
	int *v2=new int[4];
	//Wrong. v2 was created in the heap, not in the stack.
	int size2=sizeof(v2)/sizeof(*v2);
	f2(v1);
}

void f2(int *array){
	//Wrong. This function did not create v1, so it doesn't know how big it is.
	//Furthermore, at this point it's impossible to know the size of array.
	int size1=sizeof(array)/sizeof(*array);
}

The only way for other functions to know the size of an array is by passing a size parameter (e.g. void f(int *array,int size);).