Array of Function Pointers as Parameter

Question

After creating an array of function pointers void(*array[d1])(int);

How can I pass it as a parameter of a function?

callFunctions(int array[],int size){
    for(int i=0;i<size;i++){
        *array[i](0);
    }
}

Compiler keeps saying my parameter type is wrong

interesting, maybe I'm just too novice to see the difference — R.Velazquez
– R.Velazquez, Commented Jan 31, 2017 at 1:44
IMO, It is clearer, in particular to pass/return argument: void foo(const std::array<void(*)(int), d1>& a) against void foo(void(*(&a)[d1])(int)). using/typedef might also be used. — Jarod42
– Jarod42, Commented Jan 31, 2017 at 9:44

P0W · Accepted Answer · 2017-01-30 05:29:46Z

3

Call the functions using the function pointer array like following:

 (*array[i]) ( 0 );

Update you callFunctions as:

/* void */ callFunctions( void (*fptr[ ])(int), int size )
{
   // (*fptr[ i ])( 0 );
}

answered Jan 30, 2017 at 5:22

P0W

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3 Comments

Thanks for that, but still doesn't like how I am passing the array in the function

@R.Velazquez Sorry didn't noticed. I updated the post.

array[i] ( 0 ); is also a legal call but dereference notation helps to figure out how to write the pointer