问题
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#include<iostream>
using namespace std;
int main()
{
const int a = 5;
const int *ptr1 = &a;
int *ptr = (int *)ptr1;
*ptr = 10;
cout<<ptr<<" = "<<*ptr<<endl;
cout<<ptr1<<" = "<<*ptr1<<endl;
cout<<&a<<" = "<<a;
return 0;
}
Output:
0x7ffe13455fb4 = 10
0x7ffe13455fb4 = 10
0x7ffe13455fb4 = 5
How is this possible?
回答1:
It is certainly undefined behavior, but I am strong proponent of understanding symptoms of undefined behavior for the benefit of spotting one. The results observed can be explained in following manner:
const int a = 5
defined integer constant. Compiler now assumes that value will never be modified for the duration of the whole function, so when it sees
cout<<&a<<" = "<<a;
it doesn't generate the code to reload the current value of a, instead it just uses the number it was initialized with - it is much faster, than loading from memory.
This is a very common optimization technique - when a certain condition can only happen when the program exhibits undefined behavior, optimizers assume that condition never happens.
回答2:
You shouldn't rely on undefined behaviour. Look what the compiler does with your code, particularly the last part:
cout<<&a<<" = "<<a;
b6: 48 8d 45 ac lea -0x54(%rbp),%rax
ba: 48 89 c2 mov %rax,%rdx
bd: 48 8b 0d 00 00 00 00 mov 0x0(%rip),%rcx # c4 <main+0xc4>
c4: e8 00 00 00 00 callq c9 <main+0xc9>
c9: 48 8d 15 00 00 00 00 lea 0x0(%rip),%rdx # d0 <main+0xd0>
d0: 48 89 c1 mov %rax,%rcx
d3: e8 00 00 00 00 callq d8 <main+0xd8>
d8: ba 05 00 00 00 mov $0x5,%edx <=== direct insert of 5 in the register to display 5
dd: 48 89 c1 mov %rax,%rcx
e0: e8 00 00 00 00 callq e5 <main+0xe5>
return 0;
e5: b8 00 00 00 00 mov $0x0,%eax
ea: 90 nop
eb: 48 83 c4 48 add $0x48,%rsp
ef: 5b pop %rbx
f0: 5d pop %rbp
f1: c3 retq
When the compiler sees a constant expression, it can decide (implementation-dependent) to replace it with the actual value.
In that particular case, g++ did that without even -O1 option!
回答3:
When you invoke undefined behavior anything is possible.
In this case, you are casting the constness away with this line:
int *ptr = (int *)ptr1;
And you're lucky enough that there is an address on the stack to be changed, that explains why the first two prints output a 10.
The third print outputs a 5 because the compiler optimized it by hardcoding a 5 making the assumption that a wouldn't be changed.
来源：https://stackoverflow.com/questions/39106137/change-constant-value