Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
287 views
in Technique[技术] by (71.8m points)

c++ - Using concurrently 2 versions of boost

I'm using RHEL 5.3, which is shipped with gcc 4.1.2 and boost 1.33. There're some features I want, that are missing in the boost 1.33. Therefore the thought was to upgrade to fresh boost release 1.43.

  1. Is it possible to use concurrently some header-only library(s) from boost 1.43 and the rest from 1.33? For example I want to use unorded_map, which is missing in boost 1.33.

  2. Is it possible to use concurrently binary boost libraries from different releases?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

NO -- never do this!

It is impossible, you'll likely to get accidental crashes.

The only way to get it done correctly is using namespace renaming: i.e. create alternative boost version placed in different namespace.

Latest version of BCP provides this option. So you will use something like boost_1_43 instead of boost. But it will be quite transparent for you. But you should still be aware of that you can't use two versions of boost in same cpp file.

Also take a look on this discussion: Creating Library with backward compatible ABI that uses Boost

The liked script renames namespace, defines and includes so you can actually include two versions of boost like

#include <boost/foo.hpp>
#include <myboost/bar.hpp>

boost::foo f;
myboost::bar b;

Boost BCP does not allow this.

But still you should be careful as some libraries export extern "C" symbols without boost prefix, boost::thread and boost::regex's C API (regexec, regcomp)

Edit

As example of such issue create following files:

a.cpp:

template<typename Foo>
Foo add(Foo a, Foo b)
{
        return a+b;
}


int foo(int x,int y)
{
        return add(x,y);
}

b.cpp:

template<typename Foo>
Foo add(Foo a, Foo b)
{
        return a-b;
}


int bar(int x,int y)
{
        return add(x,y);
}

test.cpp:

#include <iostream>

int foo(int,int);
int bar(int,int);

int main()
{
        std::cout<< foo(10,20) <<" " <<bar(10,20) << std::endl;
}

Compile them:

g++ a.cpp b.cpp test.cpp

You would expect:

30 -10

But you'll get

30 30

or

-10 -10

Depending on linking order.

So using two boost versions you may accidentally use symbols from other boost and crash same as in this program symbol int add<int>(int,int) is resolved to same symbol even if it is placed in different compilation units.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...