How can I get rid of C4100 warning when using typeid in a template method?

Question

The following C++ code uses typeid to print out the runtime class of the parameter:

#include <iostream>

class Foo
{
};

class Bar: public Foo
{
};

template <class O> void printTypeName(O& object)
{
    std::cout << typeid(object).name();
}

int main(void)
{
    Bar x;
    printTypeName(x);
}

Since Foo is not polymorphic, VS C++ doesn't use the object to determine type information and raises

C4100 warning ("unreferenced formal parameter").

Is there any way to get rid of the warning, while preserving the possibility to print out the object type with a simple method call? I would prefer not to have to disable the warning.

I think that the warning is a good and easy way to get a diagnostic if you apply `typeid` to a non-polymorphic type. Try adding a virtual destructor to `Foo`. It may well be possible that MSVC omits to warn then. — Johannes Schaub - litb, Sep 17 '11 at 14:35
@Johannes What is wrong with applying `typeid` to a non-polymorphic type? — quant_dev, Sep 17 '11 at 14:37
@quant for non-polymorphic classes `typeid` will not give you a dynamic type. It just yields the static (compile time) type. It will not touch the parameter at runtime (it's like "sizeof" then). — Johannes Schaub - litb, Sep 17 '11 at 16:17

score 3 · Answer 1 · answered Sep 16 '11 at 13:25

3

You can use

#pragma warning(disable : 4100)
//.. stuff
#pragma warning(default : 4100)

to turn the warning off and then on again when you're done.

answered Sep 16 '11 at 13:25

Puppy

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I know ;-) is there any less brutal way? – quant_dev Sep 16 '11 at 13:26
@sharptooth: That question gives absolutely no reason whatsoever for the posted answers. – Puppy Sep 16 '11 at 13:40
@quant_dev: You gave absolutely no such requirements in the question. Define "brutal" and edit the question with that definition in and we'll talk. You can't complain about my answer not fulfilling your requirements when those requirements are utterly subjective and were never mentioned in the first place. – Puppy Sep 16 '11 at 13:41
@quant_dev: What? I stated some facts. – Puppy Sep 16 '11 at 17:11
1

Never mind, you won't get it anyway. Bots are notoriously poor when it comes to nuances of human language. – quant_dev Sep 16 '11 at 19:20

score 1 · Accepted Answer · edited Sep 17 '11 at 14:32

1

There's an UNREFERENCED_PARAMETER macro you can use for that.

==== Edited by the OP: one can also use

(void) object;

and avoid using the macro (credits to David Rodriguez for his comment about it).

edited Sep 17 '11 at 14:32

quant_dev

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answered Sep 16 '11 at 13:28

Mat

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2

Alternatively the standard approach would be: `(void)object;` which does not depend on a macro being defined somewhere. – David Rodríguez - dribeas Sep 16 '11 at 13:35
@quant_dev: That should be a way to suppress a potential "statement has no effect" warning. – UncleBens Sep 16 '11 at 15:13
2

To be honest, it looks more and more like playing some stupid game. – quant_dev Sep 17 '11 at 14:33

score -1 · Answer 3 · answered Sep 16 '11 at 15:03

-1

This works for me without any errors:

template <typename T>
void prn(const T&){
    std::cout << typeid(T).name() << std::endl;
}

answered Sep 16 '11 at 15:03

j_kubik

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How can I get rid of C4100 warning when using typeid in a template method?

3 Answers3