Hi Phil,
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<bar *>(&x)
>
> is not legit and the compiler will bug you.
>
> HtH,
>
> Luc
>
>
>
>
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