On Solaris 64-bit:
sizeof(long) = 8
sizeof(double) = 8 !!!
So i figure i can't store a long with the same precision in a double.
But, when i create a long and assign its largest value to it, then
assign this to a double, then subtract the double from the long, I get
0.
I see from the include files the exponent is 11 bits. So the Mantissa
must be 53 (or less) bits. So how can the double store my long to the
same precision??
I've posted the code and output below.
Any help appreciated.
Nishant
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#include <stdlib.h>
#include <iostream.h>
#include <stdio.h>
int main()
{
cout << sizeof(long) << endl;
cout << sizeof(double) << endl;
cout << sizeof(float) << endl;
long l = 0x7FFFFFFFFFFFFFFF;
//long l = 9223372036854775807LL;
cout << "l = " << l << endl;
// want to confirm l is indeed the largest long
long nl = l + 1;
cout << "nl = " << nl << endl;
double d = l;
cout << "d = " << d << endl;
double r = d / l;
cout << "r = " << r << endl;
double rr = l - d;
cout << "rr = " << rr << endl;
}
--------
output:
8
8
4
l = 9223372036854775807
nl = -9223372036854775808
d = 9.22337e+18
r = 1
rr = 0
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