[RESOLVED] setbase not setting the base for cout

[MattInSD73]

I'm having trouble understanding why this code doesn't work. I am copying almost verbatim from msdn help for the setw iomanip library function. It should set the base for output, but it isn't.

#include <iostream>
#include <iomanip>

using namespace std;

    int paws;
    int base = 10;

    const double      d1 = 16.0;
    const double      d2 = 256.0;
    const double      d3 = 1024.0;
    const double      d4 = 4096.0;
    const double      d5 = 65536.0;


void printBases(double input)
{
    cout << "input: " << input << ", oct:" << setbase(8) << input << ", hex:" << setbase(16) << input << "\n";
}

void DisplayLongs( )
{
   cout << setbase(10);
   cout << endl << "setbase(" << base << ")" << endl;
   cout << setbase(base);
   cout << "l1 = " << d1 << endl;
   cout << "l2 = " << d2 << endl;
   cout << "l3 = " << d3 << endl;
   cout << "l4 = " << d4 << endl;
   cout << "l5 = " << d5 << endl;
}

int main()
{


    printBases(d1);
    printBases(d2);
    printBases(d3);
    printBases(d4);
    printBases(d5);

    base = 16;
    DisplayLongs();
    
    base = 8;
    DisplayLongs();
    
    base = 10;
    DisplayLongs();

    return (EXIT_SUCCESS);
}

output:

C:\My Programs\pretest3\Debug>pretest3
input: 16, oct:16, hex:16
input: 256, oct:256, hex:256
input: 1024, oct:1024, hex:1024
input: 4096, oct:4096, hex:4096
input: 65536, oct:65536, hex:65536

setbase(16)
l1 = 16
l2 = 256
l3 = 1024
l4 = 4096
l5 = 65536

setbase(8)
l1 = 16
l2 = 256
l3 = 1024
l4 = 4096
l5 = 65536

setbase(10)
l1 = 16
l2 = 256
l3 = 1024
l4 = 4096
l5 = 65536

C:\My Programs\pretest3\Debug>

[Paul McKenzie]

I'm having trouble understanding why this code doesn't work. I am copying almost verbatim from msdn help for the setw iomanip library function. It should set the base for output, but it isn't.

Change those doubles to longs. What are your results?

Similarly, what if you changed one of the double values to 16.5 instead of 16.0? What would you have expected your output to be?

Regards,

Paul McKenzie

[MattInSD73]

I assumed that the doubles would get converted to the proper base, but apparently cout and setbase don't play that well enough together as to accept a non-integral type. Luckily, I don't need the fractional values, so I can truncate with a cast to long.

Thanks for the help!

void printBases(double input)
{
    cout << setbase(10) << "input: " << (long) input << ", oct:" << setbase(8) << (long) input << ", hex:" << setbase(16) << (long) input << "\n";
}