another simple question

[cracksevi]

im still a newbie tryin to learn the basics...
now...
the prob is double & static_cast functions :

int main()
{
int total, // sum of grades
gradeCounter, // number of grades entered
grade; // one grade
double average; // number with decimal point for average

// initialization phase
total = 0;
gradeCounter = 0;

// processing phase
cout << "Enter grade, -1 to end: ";
cin >> grade;

while ( grade != -1 ) {
total = total + grade;
gradeCounter = gradeCounter + 1;
cout << "Enter grade, -1 to end: ";
cin >> grade;
}

// termination phase
if ( gradeCounter != 0 ) {
average = static_cast< double >( total ) / gradeCounter;
cout << "Class average is " << setprecision( 2 )
<< setiosflags( ios::fixed | ios::showpoint )
<< average << endl;




why -> total <- is in parantheses @ average = static_cast< double >( total ) / gradeCounter;

and is there any connection between double and static_cast??
cuz when i tried a proggie with double...
for example the prog was 11/2
and it showed the result as 5 or 6 (i dont remember well)
and when i tried with static_cast as extra , it showed result as 5.5
an explanation can be so nice friends....



and sorry for asking dumbish questions...

[joe_bruin]

what's going on here is the order of type promotions is getting shifted around.

/* #1 */
average =  total / gradeCounter;
/* #2 */
average = static_cast< double >( total ) / gradeCounter;

in line #1, what you are doing is taking one int (total), and dividing it by another int (gradeCounter). that result is then cast to a double and stored into average. so, in integer math, 11 / 2 = 5. this is then cast to a double, and average is 5.0.

in line #2, the first thing that happens is that "total" is promoted to a double. now, you can't divide a double by an int, so "gradeCounter" is promoted to a double too. 11.0 / 2.0 = 5.5. the result is already double, so no promotion is necessary.

now, perhaps someone better at c++ than i can explain to us the value of static_cast over a plain cast.

[Valmont]

"Plain Cast"
I assume you main "tradittional cast" like (doube)anInt/anInt;

The C++ Language, section 6.2.7., B. Stroustrup
This Cstyle cast is far more dangerous than the named conversion
operators because the notation is harder to spot in a large program and the kind of conversion intended by the programmer is not explicit.

[joe_bruin]

thanks.

[cracksevi]

aha now it begins to make some sense...
thanx for the info bros!
cheers !