C++ const : how come the compiler doesn't give a warning/error
Really simple question about C++ constness.
So I was reading this post, then I tried out this code:
int some_num = 5;
con开发者_如何学编程st int* some_num_ptr = &some_num;
How come the compiler doesn't give an error or at least a warning?
The way I read the statement above, it says:
Create a pointer that points to a constant integer
But some_num is not a constant integer--it's just an int.
The problem is in how you're reading the code. It should actually read
Create a pointer to an integer where the value cannot be modified via the pointer
A const int*
in C++ makes no guarantees that the int
is constant. It is simply a tool to make it harder to modify the original value via the pointer
I agree with Jared Par's answer. Also, check out the C++ FAQ on const correctness.
The const keyword just tells the compiler that you want some stricter checking on your variable. Castring a non const integer to a const integer pointer is valid, and just tells the compiler that it should give an error if you try to change the value of the contents of the const-pointer.
In other words, writing
*some_num_ptr = 6;
should give an error since the pointer points to a const int.
Writing
some_num = 7;
remains valid of course.
int* ptr;
just says you can't change the valued pointed to by ptr
through ptr
. The actual value might still be changed by other means.
(the actual reason the compiler doesn't warn is that casts from T* to const T* are implicit)
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