I am sure that the following code should not compile. But, in g++, it does compile! See it compile at http://codepad.org/MR7Dsvlz .
The code:
#include <iostream> using namespace std; int main() { int x = 32 ; // note: if x is, instead, a const int, the code still compiles, // but the output is "32". const int * ptr1 = & x ; *((int *)ptr1) = 64 ; // questionable cast cout << x ; // result: "64" } Is g++ in error by compiling this?
No. According to §5.4.4 of the C++ standard, the casts that can be performed by a C-style cast are:
— a const_cast (5.2.11), — a static_cast (5.2.9), — a static_cast followed by a const_cast, — a reinterpret_cast (5.2.10), or — a reinterpret_cast followed by a const_cast This is widely known as "casting away const-ness", and the compiler would be non-conformant to that part of the standard if it did not compile that code.
As ildjarn points out, modifying a const object via casting away constness is undefined behaviour. This program does not exhibit undefined behaviour because, although an object that was pointed to by the pointer-to-const, the object itself is not const (thanks R.Martinho and eharvest for correcting my bad reading).
const. The object being modified is the int x = 32 ;.No. g++ is not in error by compiling your code. the cast you have done is valid.
(int *)ptr1 is a c cast. the equivalent in c++ is const_cast<int*>(ptr1). the second style is clearer to read.
but, the need to do this cast (to modify a const variable) shows a problem in the design.
The line *((int *)ptr1) = 64 is equivalent to *(const_cast<int*>(ptr1)) = 64 The const_cast is the first cast that is performed when you use the cast notation.
const_cast<int*>(ptr1)- although the C cast will work too, as you've just seen.