What is the output (to stdout) of this C code, if all these lines execute on the CPU uninterrupted?

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BlackTom AI · Accepted Answer

First, let's restate the situation: a pointer p is allocated with malloc(20), cast to int*, and the comment suggests p holds the address 1314. The code then frees p and prints p using printf("%u", p).
Option A: -1314. This would imply printing a negative value, which is unlikely when printing a pointer value with %u (an unsigned format). Since pointer addresses are non-negative, this is not consistent with typical behavior.
Option B: 0. Printing 0 would suggest the pointer value is null or that the address got overwritten to zero after free. In most standard scenarios, freeing memory does not magically turn the pointer into 0, and using %u with a non-null pointer would not normally yield 0.
Option C: 1314. If the pointer p still contains the same address after free (which is plausible on many architectures where freeing does not modify the pointer itself) and if you print the pointer value using an unsigned format, you may see the original address value, here 1314. This matches the given assumption that p has address 1314 and the print uses the pointer value.
Option D: garbage (random) value. Because the code prints the pointer value p with %u, and if the environment reuses or preserves the address, you would not expect an arbitrary random value. However, it is important to note that printing a pointer with %u is technically undefined behavior in C, since the correct format specifier for a pointer is %p. After free, using the pointer value is still a value you could print, but relying on it being stable is not guaranteed.
Now, considering the specific prompt: the scenario states all lines execute on the CPU uninterrupted and (as given) p has address 1314. If we ignore the strict correctness concerns about using %u with a pointer and assume a straightforward interpretation where the pointer value is printed as its numeric address, the output would be 1314. Nevertheless, be aware that in standard conforming C, the correct way to print a pointer is with %p, and the behavior of printing after free can be undefined across implementations.

SP25: CS-149 Sec 03 - Operating Systems CS149 Andreopoulos, Spring 2025 -- Final Exam

What is the output (to stdout) of this C code, if all these lines execute on the CPU uninterrupted?

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