Question

There i something i dont get, if have the following:

double average (double scores[]){

 double sum = 0;
 int n;
 int nscores = sizeof(scores)/sizeof(double);
 for (n=0 ;n<nscores ;++n){
 sum+=scores [n];
 return sum/nscores;
}

and i send this function an array like this:

  double scores[3]={1,2,3};

why will sizeof(scores) will be 0?

Answer 1

sizeof(scores), inside the function, is equivalent to sizeof(double *): the compiler has no way to tell, at that point, how big the array was.

It won’t be zero, but because sizeof(scores) and sizeof(double) are both integer-type expressions, sizeof(scores)/sizeof(double) is integer division, so if sizeof(scores) < sizeof(double), then sizeof(scores)/sizeof(double) == 0.