w****g
发帖数: 206 | 1 运行一下 程序：
int main() {
const int i = 0;
int * j;
cout << "&j=" << &j << endl;
cout << "&i=" << &i << endl;
cout << "j=" << j << endl;
j = const_cast(&i);
*j = 2;
cout << "i=" << i << endl;
cout << "*j=" << *j << endl;
cout << "j=" << j << endl;
cout << "&i=" <<&i <
}
得到结果：
&j=0xbf93a658
&i=0xbf93a65c
j=0
i=0
*j=2
j=0xbf93a65c
&i=0xbf93a65c
这就是说：j 和&i是一样的，但是为什么i和*j不一样呢？ |
s*****n
发帖数: 5488 | 2 looks like j is int pointer. and const-cast just removes the const of the
pointer.
【在 w****g 的大作中提到】
: 运行一下 程序：
: int main() {
: const int i = 0;
: int * j;
: cout << "&j=" << &j << endl;
: cout << "&i=" << &i << endl;
: cout << "j=" << j << endl;
: j = const_cast(&i);
: *j = 2;
: cout << "i=" << i << endl;
|
M********5
发帖数: 715 | 3 This is undefined behavior of const_cast. So that's the reason why you need
to take care when you use those cast features.
You can google const_cast undefined behavior. |
r**u
发帖数: 1567 | 4 This doesn't explain his question. Since i & *j points to the same address,
why i = 0 but *j = 2?
【在 s*****n 的大作中提到】
: looks like j is int pointer. and const-cast just removes the const of the
: pointer.
|
x***y
发帖数: 633 | 5 change the definition of "const int i = 0" to
"volatile const int = 0", then i will be 2. The reason is due to
optimization, that i will be read from register, not from memory.
【在 w****g 的大作中提到】
: 运行一下 程序：
: int main() {
: const int i = 0;
: int * j;
: cout << "&j=" << &j << endl;
: cout << "&i=" << &i << endl;
: cout << "j=" << j << endl;
: j = const_cast(&i);
: *j = 2;
: cout << "i=" << i << endl;
|
h**k
发帖数: 3368 | 6 volatile关键字会强迫从内存中读取，而不会只返回寄存器中的值。
我觉得是compiler直接优化，把输出i的值的语句中的i直接替换成0，因为i声明成
const变量。
【在 x***y 的大作中提到】
: change the definition of "const int i = 0" to
: "volatile const int = 0", then i will be 2. The reason is due to
: optimization, that i will be read from register, not from memory.
|
d*********i
发帖数: 628 | |
f****4
发帖数: 1359 | |
