C++ variable type

Question

I have a function which takes two variables:

call((void**) &var, float);
call((void**) &var, int);
call((void**) &var, string);

Now I want to create a function to call that special function:

function start_call(/*what to put here?*/)
{
    call((void**) &var1, float);
    call((void**) &var2, int);
    call((void**) &var3, string);
     //code
}

So, what to put in the "what to put here?" place, in order for the function to accep any variable?

I tried my berst to explain it...

@JoachimPileborg by using a language other than c++, for one — Taylor Brandstetter, Dec 06 '13 at 22:29
Have you looked at this: http://stackoverflow.com/questions/2635715/call-a-void-as-a-function-without-declaring-a-function-pointer or this:http://stackoverflow.com/questions/1485983/calling-c-class-methods-via-a-function-pointer — FuzzyBunnySlippers, Dec 06 '13 at 22:32
Can you show how you declare `call`, and exactly what you'd like `start_call` to look like? — Alan Stokes, Dec 06 '13 at 22:36
I don't know how to declare call. start_call must take one variable. That variable should be any type. Then, on the function I want to find out the variable type. — SkyRipper, Dec 06 '13 at 22:39
@SkyRipper It might be better for you to describe the problem you're trying to solve, instead of asking us how to implement what you think might be a solution. — Praetorian, Dec 06 '13 at 22:42
Here is a function used in nVidia CUDA: cudaMalloc( (void**) &d_A, N * sizeof(float)); I want to call that function from another function. — SkyRipper, Dec 06 '13 at 22:43

score 1 · Accepted Answer · answered Dec 06 '13 at 22:52

1

You could use a template:

template <class T> start_call(T in);

answered Dec 06 '13 at 22:52

chbaker0

1,758
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No problem :) templates are a very powerful feature of C++. – chbaker0 Dec 06 '13 at 23:13

C++ variable type

1 Answers1