Ubuntu 中的“int”

Question

Can anyone tell me if, when compiling a c++ program, does g++ (Ubuntu/Linaro 4.4.4-14ubuntu5) 4.4.5 change int s to long int s? 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::operator 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()

Answer 1

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 ?

Answer 2

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.

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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.

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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 .