Why Do Variables at the Same Memory Address Output Different Values in C ?

Variable with Identical Address Outputting Distinct Values

Consider the following code snippet:

const int a1 = 40;
const int* b1 = &a1;
char* c1 = (char *)(b1);
*c1 = 'A';
int *t = (int*)c1;

cout << a1 << " " << *t << endl;
cout << &a1 << " " << t << endl;

Surprisingly, this code outputs:

40 65
0xbfacbe8c 0xbfacbe8c

How is it possible for a variable at the same address (c1 and t) to produce two different values ('A' and 65)?

The explanation lies in the nature of undefined behavior. In this case, the modification of a constant variable (*c1) triggers undefined behavior. According to the C standard, undefined behavior can result in unpredictable outcomes, from program termination to documented behavior (with or without diagnostic messages).

In this specific instance, the undefined behavior allows the compiler to optimize the code in an unexpected way. By modifying the constant variable, the compiler is effectively overriding the original value of a1. However, it is important to note that this behavior is not guaranteed and may vary across different compilers and environments.

Therefore, when working with constant variables and pointers, it is crucial to avoid modifying these variables to ensure predictable program behavior.

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