Hi Phil,
The main reasons for me to prefer static_cast is clarity; I just want to
make the type conversion easily noticeable.
Ralf
On Tue, May 15, 2012 at 3:59 AM, Phil Evans
I can see that problem with pointers but it doesn't seem to apply to double
I am naive about this, I suppose
1) I never learnt C, just C++ 2) I don't use pointers (or very rarely) 3) I read "double(i)" as a constructor of a double from an int ie
class double { public: double(const int& i); };
thus it shouldn't accept eg
thing* instance; d = double(instance);
which should be a compiler error, just as an illegal construction of any other object should be
and indeed double* pd; double d = double(pd);
gives t.cpp:4:23: error: invalid cast from type 'double*' to type 'double'
Phil
On 15 May 2012, at 11:38, Luc Bourhis wrote:
Hi Phil,
Looking at some cctbx C++ code, I note use of syntax like
int i = 1234; double d = static_cast<double>(i);
How is this different or indeed preferable to
double d = double(i);
or
d = i;
?
Short answer: For conversions between non-pointer types, it does not matter but for pointer types, static_cast shall always be used to the exclusion of any other conversion.
The language lawyer:
The construct
double d = double(i)
which is known as functional-style cast is completely equivalent to the so-call cast notation, which comes from C:
double d = (double)i
You may see both in a few places in the cctbx code actually.
The problem with those two constructs is that the compiler will let use convert a pointer to anything into a pointer to anything else. Consider:
#include <iostream>
struct foo { int i,j; foo(int i, int j): i(i), j(j) {} };
struct bar { double d; bar(double d): d(d) {} };
int main() { foo x(1, 2); bar *y = (bar *)&x; std::cout << y->d << std::endl; return 0; }
This prints 4.24399e-314 on my computer but this depends on the compiler. On the contrary,
bar *y = static_cast
(&x) is not legit and the compiler will bug you.
HtH,
Luc
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