https://gcc.gnu.org/bugzilla/show_bug.cgi?id=70318
Bug ID: 70318
Summary: `std::sqrt<int>(int)` Produces -Wfloat-conversion
Warning, Erroneous After C++11.
Product: gcc
Version: 5.2.1
Status: UNCONFIRMED
Severity: normal
Priority: P3
Component: c++
Assignee: unassigned at gcc dot gnu.org
Reporter: ian at geometrian dot com
Target Milestone: ---
Created attachment 38032
--> https://gcc.gnu.org/bugzilla/attachment.cgi?id=38032&action=edit
Sample given in text, as a file.
As required by the C++ standard, after C++11, the `std::sqrt` function in
`<cmath>` is required to have "A set of overloads or a function template
accepting an argument of any integral type."
Consider the following command-line tool for computing integer-square root:
//Compile like so: g++ isqrt.cpp -std=c++11 -Wfloat-conversion -o isqrt
#include <cmath>
#include <iostream>
int main(int argc, char* argv[]) {
if (argc!=2) {
std::cout << "Usage: \"./isqrt [integer]\"" << std::endl;
return -1;
} else {
int arg = atoi(argv[1]);
int isqrt = std::sqrt<int>(arg);
std::cout << isqrt << std::endl;
return isqrt;
}
}
When compiling as the comment says, a warning is produced:
isqrt.cpp: In function ‘int main(int, char**)’:
isqrt.cpp:10:33: warning: conversion to ‘int’ from
‘__gnu_cxx::__enable_if<true, double>::__type {aka double}’ may alter its value
[-Wfloat-conversion]
int isqrt = std::sqrt<int>(arg);
^
Internally, the argument is indeed converted to `double`, before doing the
square root with `double`s (at it should be). However, since the overload is
required to exist separately and explicitly (and it evidently doesn't), I think
this warning is erroneous.
For your convenience, my sample is also attached as a file.
