char c{ 10 };
int* i = (int*)&c;
*i = 1; // Run-Time Check Failure #2 - Stack around the variable 'c' was corrupted.

But I don't get any error in this case

char* c = new char{ 10 };
int* i = (int*)&c;
*i = 1;
//delete c;

Why is it so?

CodePudding user response：

With

int* i = (int*)&c;

you make i point to the variable c itself, not where c is actually pointing.

Thus *i = 1 will change the value of the pointer variable c not the value of *c.

If you want to get the same (or similar) behavior you should make i point to where c is pointing:

int* i = (int*) c;

As for why it doesn't give you any error, it's because on modern system int is 32 bits wide, while a pointer (like c) will be at least 32 bits wide as well (and 64 bits on a 64-bit system).

Lastly a note about doing C-style casts in C : You should always take it as a sign that you're doing something wrong.