C++ Boost: what's the cause of this warning?

2019-01-11 09:02发布

问题:

I have a simple C++ with Boost like this:

#include <boost/algorithm/string.hpp>

int main()
{
  std::string latlonStr = "hello,ergr()()rg(rg)";
  boost::find_format_all(latlonStr,boost::token_finder(boost::is_any_of("(,)")),boost::const_formatter(" "));

This works fine; it replaces every occurrence of ( ) , with a " "

However, I get this warning when compiling:

I'm using MSVC 2008, Boost 1.37.0.

1>Compiling...
1>mainTest.cpp
1>c:\work\minescout-feat-000\extlib\boost\algorithm\string\detail\classification.hpp(102) : warning C4996: 'std::copy': Function call with parameters that may be unsafe - this call relies on the caller to check that the passed values are correct. To disable this warning, use -D_SCL_SECURE_NO_WARNINGS. See documentation on how to use Visual C++ 'Checked Iterators'
1>        c:\program files (x86)\microsoft visual studio 9.0\vc\include\xutility(2576) : see declaration of 'std::copy'
1>        c:\work\minescout-feat-000\extlib\boost\algorithm\string\classification.hpp(206) : see reference to function template instantiation 'boost::algorithm::detail::is_any_ofF<CharT>::is_any_ofF<boost::iterator_range<IteratorT>>(const RangeT &)' being compiled
1>        with
1>        [
1>            CharT=char,
1>            IteratorT=const char *,
1>            RangeT=boost::iterator_range<const char *>
1>        ]
1>        c:\work\minescout-feat-000\minescouttest\maintest.cpp(257) : see reference to function template instantiation 'boost::algorithm::detail::is_any_ofF<CharT> boost::algorithm::is_any_of<const char[4]>(RangeT (&))' being compiled
1>        with
1>        [
1>            CharT=char,
1>            RangeT=const char [4]
1>        ]

I could certainly disable the warning using

-D_SCL_SECURE_NO_WARNINGS

but I'm a bit reluctant to do that before I find out what's wrong, or more importantly if my code is incorrect.

回答1:

It is nothing to worry about. In the last few releases of MSVC, they've gone into full security-paranoia mode. std::copy issues this warning when it is used with raw pointers, because when used incorrectly, it can result in buffer overflows.

Their iterator implementation performs bounds checking to ensure this doesn't happen, at a significant performance cost.

So feel free to ignore the warning. It doesn't mean there's anything wrong with your code. It is just saying that if there is something wrong with your code, then bad things will happen. Which is an odd thing to issue warnings about. ;)



回答2:

You can also disable this warning in specific headers:

#if defined(_MSC_VER) && _MSC_VER >= 1400 
#pragma warning(push) 
#pragma warning(disable:4996) 
#endif 

/* your code */ 

#if defined(_MSC_VER) && _MSC_VER >= 1400 
#pragma warning(pop) 
#endif 


回答3:

If you feel safe about disabling this error:

  • Go to the properties of your C++ project
  • Expand the "C/C++"
  • Highlight "Command Line"
  • Under "Additional Options" append the following to any text that might be in that box

" -D_SCL_SECURE_NO_WARNINGS"



回答4:

The warning comes from Visual Studio's non-standard "safe" library checks introduced starting with MSVC 8.0. Microsoft has identified "potentially dangerous" APIs and has injected warnings discouraging their use. While it is technically possible to call std::copy in an unsafe way, 1) receiving this warning does not mean you are doing so, and 2) using std::copy as you normally should is not dangerous.



回答5:

  • Go to the properties of your C++ project

  • Expand the "C/C++"

  • Advanced: Disable Specific Warnings: 4996



回答6:

Alternatively, if you use C++11 and don't want to turn off warnings, you have the painful option of replacing

boost::is_any_of(L"(,)")

with the following lambda expression

[](wchar_t &c) { for (auto candidate : { L'(', L',', L')' }) { if (c == candidate) return true; }; return false; }

You can also possibly pack that into a macro