q: abt static-cast function

[technologist]

Below code - simply experimenting with it to learn more abt static cast functions. THe purpose of the code was to show type conversion. What I don't understand is why in static_cast<float>(j)/k; fx you only need one integer from the two integers(j,k) to make a conversion to float to work. Plz see quotation below.

#include<iostream>

using namespace std;

int main()
{

	int j(10); int k(6);
	float m = j/k;
	float d = static_cast<float>(j)/k;
	cout<<"m = "<<m<<endl;
	cout<<"d = "<<d<<endl;

		return 0;
}

In this example, m = j/v; produces an answer of type int because both j and v are integers. Conversely, d = static_cast<float>(j)/v; produces an answer of type float. The static_cast operator converts variable j to a type float. This allows the compiler to generate a division with an answer of type float. All static_cast operators resolve at compile time and do not remove any const or volatile modifiers.

[cire]

See: http://stackoverflow.com/a/5563131

[technologist]

Thanks that got most of the problem. Remaining question : how do I static_cast a variable then static cast that variable back to what its intial type was? For example int-->float-->int. The ide forces me to use a new variable(s):

int j = 41;
	  int v = 4;
	  float m = j/v;
	  float d = static_cast<float>(j)/v;
	  cout << "m = " << m << endl;
	  cout << "d = " << d << endl;

	  float e = static_cast<int>(j)/v;
	  cout<<"e = "<<e<<endl;
	  float x = static_cast<float>(j)/v;
	  cout<<"x = "<<x<<endl;

so I want to change float d back to integer instead of using a new variable like x or e .. 'll start losing track of thing otherwise

[LB]

If you want to start and end with integer, why do you use float at all? You would get the same precision either way.

[JLBorges]

> how do I static_cast a variable then static cast that variable back to what its intial type was?
> For example int-->float-->int.

>> If you want to start and end with integer, why do you use float at all?
>> You would get the same precision either way.

This is an interesting exercise.
Those unfamiliar with floating point may be surprised by the results.

For instance:

#include <iostream>
#include <iomanip>
#include <limits>

int main()
{
    std::cout << std::fixed << std::setprecision(2) << std::showpos ;

    int i = std::numeric_limits<int>::max() ;
    std::cout << "int: " << i << "   int=>float: " << float(i)
              <<  "   int=>float=>int: " << int( float(i) ) << '\n'
              << "int: " << i << "  int=>double: " << double(i)
              << "  int=>double=>int: "  << int( double(i) ) << "\n\n" ;

    i = std::numeric_limits<int>::min() ;
    std::cout << "int: " << i << "   int=>float: " << float(i)
              <<  "   int=>float=>int: " << int( float(i) ) << '\n'
              << "int: " << i << "  int=>double: " << double(i)
              << "  int=>double=>int: "  << int( double(i) ) << "\n\n" ;

    long long ll = std::numeric_limits<long long>::max() ;
    std::cout << "long long: " << ll << "   long long=>float: " << float(ll)
              <<  "    long long=>float=>long long: " << (long long)( float(ll) ) << '\n'
              << "long long: " << ll << "  long long=>double: " << double(ll)
              << "   long long=>double=>long long: "  << (long long)( double(ll) ) << "\n\n" ;

    ll = std::numeric_limits<long long>::min() ;
    std::cout << "long long: " << ll << "   long long=>float: " << float(ll)
              <<  "    long long=>float=>long long: " << (long long)( float(ll) ) << '\n'
              << "long long: " << ll << "  long long=>double: " << double(ll)
              << "   long long=>double=>long long: "  << (long long)( double(ll) ) << "\n\n" ;
}

With defaulted floating point related compiler options:

Linux stacked-crooked 3.2.0-74-virtual #109-Ubuntu SMP Tue Dec 9 17:04:48 UTC 2014 x86_64 x86_64 x86_64 GNU/Linux

clang version 3.6.0 (tags/RELEASE_360/final 235480)

int: [phone]   int=>float: +2147483648.00   int=>float=>int: +4201984
int: [phone]  int=>double: +2147483647.00  int=>double=>int: [phone]

int: -2147483648   int=>float: -2147483648.00   int=>float=>int: -2147483648
int: -2147483648  int=>double: -2147483648.00  int=>double=>int: -2147483648

long long: +9223372036854775807   long long=>float: +9223372036854775808.00    long long=>float=>long long: +4201984
long long: +9223372036854775807  long long=>double: +9223372036854775808.00   long long=>double=>long long: +4201984

long long: -9223372036854775808   long long=>float: -9223372036854775808.00    long long=>float=>long long: -9223372036854775808
long long: -9223372036854775808  long long=>double: -9223372036854775808.00   long long=>double=>long long: -9223372036854775808



g++ (GCC) 5.2.0

int: [phone]   int=>float: +2147483648.00   int=>float=>int: [phone]
int: [phone]  int=>double: +2147483647.00  int=>double=>int: [phone]

int: -2147483648   int=>float: -2147483648.00   int=>float=>int: -2147483648
int: -2147483648  int=>double: -2147483648.00  int=>double=>int: -2147483648

long long: +9223372036854775807   long long=>float: +9223372036854775808.00    long long=>float=>long long: +9223372036854775807
long long: +9223372036854775807  long long=>double: +9223372036854775808.00   long long=>double=>long long: +9223372036854775807

long long: -9223372036854775808   long long=>float: -9223372036854775808.00    long long=>float=>long long: -9223372036854775808
long long: -9223372036854775808  long long=>double: -9223372036854775808.00   long long=>double=>long long: -9223372036854775808

http://coliru.stacked-crooked.com/a/cb465a10a5bbaee5

Microsoft Windows [Version 6.2.9200]
msc++: 1800

int: [phone]   int=>float: +2147483648.00   int=>float=>int: -2147483648
int: [phone]  int=>double: +2147483647.00  int=>double=>int: [phone]

int: -2147483648   int=>float: -2147483648.00   int=>float=>int: -2147483648
int: -2147483648  int=>double: -2147483648.00  int=>double=>int: -2147483648

long long: +9223372036854775807   long long=>float: +9223372036854775800.00    long long=>float=>long long: -9223372036854775808
long long: +9223372036854775807  long long=>double: +9223372036854775800.00   long long=>double=>long long: -9223372036854775808

long long: -9223372036854775808   long long=>float: -9223372036854775800.00    long long=>float=>long long: -9223372036854775808
long long: -9223372036854775808  long long=>double: -9223372036854775800.00   long long=>double=>long long: -9223372036854775808

http://rextester.com/CGD9093

[technologist]

Hi LB & JLBorges, yes I agree its not a practical exercise.

Why not go:
float e = static_cast<int>(j)/v;
cout<<"e = "<<e<<endl;
float e = static_cast<float>(j)/v;

But it is a good thought experiment for me to learn more abt this. My desire to use an int-->float-->int again, but using the same assignment variable. I want to recycle the same assigment var going from int to float to int., if its possible(e.g.int-float below)

  int j = 41;
	  int v = 4;

          float e = static_cast<int>(j)/v;
	  cout<<"e = "<<e<<endl;
	  float x = static_cast<float>(j)/v;

       if I do this:
         float e = static_cast<int>(j)/v;
	  cout<<"e = "<<e<<endl;
	  float e = static_cast<float>(j)/v

error redeclaration

etc.

[JLBorges]

> error redeclaration

Don't redeclare; reuse the same variable.

int v = 41 ; // declare and initialise v
std::cout << v << '\n' ;

float f = static_cast<float>(v) ; // declare and initialise f
std::cout << f << '\n' ;

v = static_cast<int>(f) ; // assign to v 
std::cout << v << '\n' ;

f = static_cast<float>(v) ; // assign to f
std::cout << f << '\n' ;

v = static_cast<int>(f) ; // assign to v 
std::cout << v << '\n' ;

f = static_cast<float>(v) ; // assign to f
std::cout << f << '\n' ;

// etc.