Re: function call misinterpreted as a variable

[Thomas Maeder]

sjbrown8@eng.usf.edu writes:

> int main()
> {
>       double xpos;
>       double a;
>       double b;
>       double c;
>       double square;
>       double xneg;
>
>       cout << "This program will compute a quadratic equation\n";
>       cout << "Please enter a number: \n";
>       cin >> a;
>       cout << "Please enter another number: \n";
>       cin >> b;
>       cout << "Please enter a final number: \n";
>       cin >> c;
>
>       square = quad(a,b,c);

Side note:

This is very unsafe code. If one of the input operations fails,
reading its value causes the program to have undefined behavior. A
program should always first check for success of an input operation
before using the value (apparently) read.