"int" in Ubuntu

Can anyone tell me if, when compiling a c++ program, does g++ (Ubuntu/Linaro 4.4.4-14ubuntu5) 4.4.5 change ints to long ints? if so, how can this be changed? If not, do I just overload operator long? Is it just like overloading operator uint32_t/uint64_t? It seems like a different type of typecasting (no pun intended).

this is causing the errors:

uint128_t.h: In function ‘std::ostream& operator<<(std::ostream&,
uint128_t)’:
uint128_t.h:593: error: conversion from ‘uint128_t’ to ‘long int’ is
ambiguous
uint128_t.h:83: note: candidates are: uint128_t::operator uint64_t()
uint128_t.h:79: note:                 uint128_t::operator uint32_t()
uint128_t.h:75: note:                 uint128_t::oper开发者_如何学运维ator uint16_t()
uint128_t.h:71: note:                 uint128_t::operator uint8_t()
uint128_t.h:67: note:                 uint128_t::operator int()
uint128_t.h:63: note:                 uint128_t::operator char()
uint128_t.h:59: note:                 uint128_t::operator bool()

Try this variation:

out = "0123456789abcdef"[size_t(rhs % div)] + out;

since you've provided conversions to all manner of unsigned types, but not to signed int.

And yes, you can define an implicit conversion to long int just be defining operator long in the same manner as all the other conversions.

Finally, please note that your choice of structure name is reserved by POSIX, and likely to conflict with future versions of the standard library header stdint.h.

No, it doesn't. An int is an int is an int. It's a separate type from long int. As for which operators you overload, it depends entirely on what you want to do. You haven't really described that.

But if you want to define an operator to work on int, then you should overload it for int. If you want it to work on long int, define an overload for that.

What are you trying to do? And why are you passing a uint128_t to the operator, but asking about int and long int?