On 09/08/2019 21:55, Maksim Fomin wrote:
‐‐‐‐‐‐‐ Original Message ‐‐‐‐‐‐‐ On Friday, August 9, 2019 8:32 PM, Jonny Grant <jg@xxxxxxxx> wrote:Hi Looks like the bool can't be converted to size_t Seems a shame it can't also convert to size_t What is strange in the example below, is I change it only a single conversion, or as follows, it compiles ok, so there must still be a conversion? Is this an issue worth reporting as a PR? size_t i = a; i += b; The C spec states |true|which expands to the integer constant|1|,|false|which expands to the integer constant|0| #include<cstddef> intmain() { boola = false; boolb = true; //size_t i = a; size_t i = a + b; returni; } #1 with x86-64 gcc (trunk) <source>: In function 'int main()': <source>:9:18: error: conversion to 'size_t' {aka 'long unsigned int'} from 'int' may change the sign of the result [-Werror=sign-conversion] 9 | size_t i = a + b; | ^ cc1plus: some warnings being treated as errors Compiler returned: 1 Please include my email in any replies Thanks, Jonny1. You mention C language but use C++ compiler: #include<cstddef> and 'cc1plus'. 2. You can find answer in C++ language latest draft (http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2019/n4820.pdf) [expr.add], [expr.arith.conv], [conv.prom]. TL;DR '+' promotes bools to signed integer which is not compatible with size_t' {aka 'long unsigned int'}. You can use size_t i = a + (size_t)b.
Fair enough. It is a shame the C++ spec didn't allow them to be treated as signed, or unsigned, they'll certainly never be negative anyway.
Jonny
