pointer to a const int and modify its value via pointer

Asked Jan 28 '17 at 19:18

Active Jan 28 '17 at 19:22

Viewed 33 times

I am learning pointers and got confused when I tried to modify the value of a const int through its address. I am not sure if it has any practical meaning. I just want to learn how the pointer works for const int. Here is the code:

const int a = 10;
*((int*)&a) = 20;
cout << a << endl;
cout << *(&a) << endl;

It prints 10 and 20. But I thought it should print 20 and 20, because in the second line, I casted &a and modified its value. So the value of a should change, because the bit pattern at that address changed.

I also tried the following code:

int a = 10;
*((int*)&a) = 20;
cout << a << endl;
cout << *(&a) << endl;

It prints 20 and 20 as I expected. I just don't understand how does const works in this case. I really appreciate it if anyone could give any suggestions.

edited Jan 28 '17 at 19:22

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asked Jan 28 '17 at 19:18

user3204065

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`const int a = 10; *((int*)&a) = 20;` is undefined behavior in C. The rest is moot. – chux - Reinstate Monica Jan 28 '17 at 19:20
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`cout << a << endl;` is not valid C. Perhaps you are coding in another language like C++? – chux - Reinstate Monica Jan 28 '17 at 19:22
thanks for your reply! could you please give more details on this undefined behavior? And why it works as expected if const is removed as in the second scenario? Thanks! – user3204065 Jan 28 '17 at 19:23
It prints the same result if I change cout << a << endl; to printf("%d\n", a); – user3204065 Jan 28 '17 at 19:25
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@user3204065 Certainly the [dupe](http://stackoverflow.com/questions/2508605/modifying-a-const-through-a-non-const-pointer) explains it. – chux - Reinstate Monica Jan 28 '17 at 19:26

pointer to a const int and modify its value via pointer

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