Assume we have these declarations:
int** a;
int b[x][y];
Can I implement a function
foo f(bar c) {}
that lets me
f(a);
f(b);
without needing to overload it?
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You could take a type that has two conversion constructors :p That kind of ruins the point, but it's not overloading that function.Qaz– Qaz2012-12-02 00:04:37 +00:00Commented Dec 2, 2012 at 0:04
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I think if you understand the code a compiler generates for a multi-dimensional array reference, you can answer this yourself. For b[i][j] the compiler generates (b+iy+j) (or it might be (b+jx+i), I never remember). Either way function f() will not know the dimensions of b[][] so it could not properly handle references to c[i][j].brian beuning– brian beuning2012-12-02 00:08:43 +00:00Commented Dec 2, 2012 at 0:08
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1These two types are not compatible, so no.Lightness Races in Orbit– Lightness Races in Orbit2012-12-02 00:17:54 +00:00Commented Dec 2, 2012 at 0:17
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1 Answer
Sure, just use void* :)
And to answer your question, no. A multidimentional array is not the same as a pointer to a pointer. The reason is the indexing scheme. int b [2][2] is a continuous memory block of 4 integers. Indexing into it is equivalent to the following:
b[i][j] == *(b + 2*i + j)
The second dimension is part of the type defintion! The compiler knows that it only needs one dereference due to the memory layout of the array.
Meanwhile, for int** a the indexing is done like this:
a[i][j] == *(*(a+i)+j)
answered Dec 2, 2012 at 0:05
StoryTeller - Unslander Monica
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2 Comments
Lightness Races in Orbit
Do we really need the dangerously misleading first line?
StoryTeller - Unslander Monica
@LightnessRacesinOrbit, is it really misleading? I think it gets the point across that unless you want to give up all type information, you can't do what the OP asked about.