Assign a BigInteger to a normal integer

Question

For example, I can do this :

uint64_t a = 100;
BigInteger b = a;

But I can't do this :

BigInteger a = 100;
uint64_t b = a;

Is there any way? Thanks.

http://codereview.stackexchange.com/questions/142588/converting-bits-to-decimal-integer-string-is-so-slow — , Sep 29 '16 at 08:49
@xersi _"Is there any way?"_ You can provide a conversion operator overload for example. — πάντα ῥεῖ, Sep 29 '16 at 08:59
@πάνταῥεῖ , it's possible to overload ```operator=``` for ```uint64_t``` type ? — isnullxbh, Sep 29 '16 at 09:01
@xersi See [here](http://stackoverflow.com/questions/4421706/operator-overloading/16615725#16615725) please. — πάντα ῥεῖ, Sep 29 '16 at 09:02
@isnullxbh A conversion operator overload seems to be the more natural way.. In the end the result might be equivalent. — πάντα ῥεῖ, Sep 29 '16 at 09:05

score 1 · Answer 1 · answered Sep 29 '16 at 09:05

1

Problem solved!

operator uint64_t() {return get_integer();}

Thanks for your help.

answered Sep 29 '16 at 09:05

score 1 · Accepted Answer · answered Sep 29 '16 at 09:05

1

The simplest solution is to make BigInteger implicitly convertible to uint64_t. This can be achieved with a conversion operator:

BigInteger::operator uint64_t();

answered Sep 29 '16 at 09:05

eerorika

2 Answers2