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Home/ Questions/Q 9109159
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Editorial Team
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Editorial Team
Asked: June 17, 20262026-06-17T02:58:35+00:00 2026-06-17T02:58:35+00:00

On a project I’ve been working on I’ve made use of boost’s signals2. This

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On a project I’ve been working on I’ve made use of boost’s signals2. This worked fine untill I started making use of virtual classes. Now slots don’t get called anymore. The following is a stripped down version of my project.

Class Glue creates instances of classes VSig and Process and connects the signal in VSig to the slot in Process.

IVSig.h:

class IVSig {
public:
    boost::signals2::signal<void (void)> sigTest;
    virtual void update() = 0;
};

VSig.h:

class VSig : public IVSig {
public:
    boost::signals2::signal<void (void)> sigTest;
    void update();
};

VSig.cpp:

void VSig::update()
{
    printf("updating!\n");
    sigTest();
}

Process.h:

class Process {
public:
    void process();
};

Process.cpp:

void Process::process()
{
    printf("hi!\n");
}

Glue.h:

class Glue {
public:
    Glue();
    void callUpdate();

private:
    IVSig *sig;
    Process *proc;
};

Glue.cpp:

Glue::Glue()
{
    sig = new VSig();
    proc = new Process();

    sig->sigTest.connect(boost::bind(&Process::process, proc));

    callUpdate();
}

void Glue::callUpdate()
{
    sig->update();
}

This should print ‘updating!’ followed by ‘hi!’ but only outputs ‘updating!’.

Any help or suggestions are greatly appreciated!

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-17T02:58:37+00:00Added an answer on June 17, 2026 at 2:58 am

    The problem is because you are redeclaring sigTest in VSig class, and when you use the sig->sigTest you are accessing the sigTest member of the IVSig class ( the declared type ), but when the update is executed you are calling the virtual update method, that will use it’s local sigTest variable. Not sure if I was clear enough.

    So, the rule of thumb is: never redefine a variable with the same name inside an inherited class.

    Hope that helps!

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