Write the source code

[jhooper3581]

Write a C++ source code that will read an input of radius of a circle and outputs the area of the circle.

[jhooper3581]

My solution.

code:

#include <iostream> #include <cmath> using namespace std; int main() {   // Mathematically it's obvious that r >= 0, where r is the radius.   double r;   cout << "Enter a radius where it is greater than or equal to 0: ";   cin >> r;   const double PI(3.1416);   double area(PI * pow(r, 2));   cout << "The area of the circle with radius " << r << " is " << area << "." << endl;     system("PAUSE");    return 0; }

If any other interesting solutions are available, please post them.

[saltpro15]

Nice. My only recommendation is to use a longer value of Pi, as really, really big numbers could give you precision errors. I like to use the first 9 digits.

[Insectoid]

Is ^ not a function in C++? I notice you use pow(x, 2) which I assume means x^2. How hard IS it to put that into a language? (Or is it already reserved for a different command?)

[saltpro15]

in C++ "^" is used as XOR.

and why the heck would you not just use x * x ?

[A.J]

You could also output the area to a given number of decimal places like so:

c++:

#include <iostream> #include <cmath> using namespace std; double r, area; int DecimalPlaces; int main() {     cout << "Enter a radius where it is greater than or equal to 0: " << endl;     cin >> r;     cout << "To how many decimal places do you want the area to be given to you?" << endl;     cin >> DecimalPlaces;     area = M_PI * (r*r);     cout << "The area of the circle with radius " << r << " is ";     cout.setf (ios_base::fixed, ios_base::floatfield);     cout.precision (DecimalPlaces);     cout << area << "." << endl;     system ("PAUSE");     return 0; }

(Note: M_PI is a built-in constant that can be accessed by importing cmath/math.h)

[saltpro15]

Just because I'm waiting for my pancakes to finish cooking, here's another version

c++:

#include<cstdio> #include<cmath> double r, area; main() {     printf("Enter a radius greater than or equal to 0\n");     scanf("%f", &r); // scanf needs a pointer when inputting variables, hence &r instead of r     area= M_PI * (r*r);     printf("%.5f\n", area); // the .5 will print to 5 decimal places }

[DtY]

insectoid @ Mon Aug 31, 2009 8:20 am wrote:

Is ^ not a function in C++? I notice you use pow(x, 2) which I assume means x^2. How hard IS it to put that into a language? (Or is it already reserved for a different command?)

I believe the reason that there is no power operator in C/++ is because there's no good way to implement it (that works with decimals and all that), and giving an operator will make people dependant on it (like using x**2 instead of x*x), instead it's a lot more work to #include <math.h> and then link it with the math library (if you want to do decimal powers)

[CodeMonkey2000]

insectoid @ Mon Aug 31, 2009 8:20 am wrote:

Is ^ not a function in C++? I notice you use pow(x, 2) which I assume means x^2. How hard IS it to put that into a language? (Or is it already reserved for a different command?)

As saltpro15 already said, ^ is the XOR operation. The function pow is part of the cmath library, which is a C library. You could theoretically create your own number class, and overload the ^ operator for powers. Like so:

c++:

#include <iostream> #include <cmath> class INT {         public:                 //does all standard operations like a good integer should                 INT();                 INT(int value) {val=value;}                 INT operator*(INT other) const {return INT(val*other.val);}                 INT operator/(INT other) const {return INT(val/other.val);}                 INT operator+(const INT &other) const {return INT(val+other.val);}                 INT operator-(const INT &other) const {return INT(val-other.val);}                 INT operator^(const INT &other) const {return INT((int)pow((double)val,(double)other.val));}                 INT operator-() const {return INT(val*-1);}                         const INT &operator*=(const INT &other){val*=other.val; return *this;}                 const INT &operator/=(const INT &other){val/=other.val; return *this;}                 const INT &operator+=(const INT &other){val+=other.val; return *this;}                 const INT &operator-=(const INT &other){val-=other.val; return *this;}                 const INT &operator=(const INT &other){val=other.val; return *this;}                                 //some code shamelessly stolen from wtd                 friend std::ostream& operator<<(std::ostream& out, const INT& other);         private:                 int val; }; std::ostream& operator<<(std::ostream& out, const INT& other) {     return out <<other.val; } int main() {         INT a=2;         INT b=3;         std::cout<<"a="<<a<<std::endl                     <<"b="<<b<<std::endl                     <<"a+b="<<(a+b)<<std::endl                     <<"b-a="<<(b-a)<<std::endl                     <<"a^b="<<(a^b)<<std::endl;         return 0; }

[md]

saltpro15 @ 2009-08-31, 8:57 am wrote:

Just because I'm waiting for my pancakes to finish cooking, here's another version

c++:

#include<cstdio> #include<cmath> double r, area; main() {     printf("Enter a radius greater than or equal to 0\n");     scanf("%f", &r); // scanf needs a pointer when inputting variables, hence &r instead of r     area= M_PI * (r*r);     printf("%.5f\n", area); // the .5 will print to 5 decimal places }

That's C; using C++ links to C headers.

Also... I am moving this out of C++ Help, since obviously it has nothing to do with help

[saltpro15]

[md]

Nein. "math.h" is the C math library - infact the exact same as cmath in most cases (sometimes some things are undefinded).

However, using the C math library is really the correct thing to do in this case - if you use it as "cmath".

[jhooper3581]

How do you guys use that type of code typing posting in this forum? (Colored texts, et cetra).

[DtY]

md @ Mon Aug 31, 2009 7:17 pm wrote:

Nein. "math.h" is the C math library - infact the exact same as cmath in most cases (sometimes some things are undefinded).

However, using the C math library is really the correct thing to do in this case - if you use it as "cmath".

Does c++ have it's own math library?

[bbi5291]

I guess there really was no reason to introduce a new math library in C++. You can't add templates, since the math functions are coded in assembly. (I suppose I/O does get a new library because there's much more room for flexibility)