What are the problems with this C program? Select two. #include #include int main() { int *ptr; *ptr = 0; printf("%d ",*ptr); *ptr = 1; printf("%d ",*ptr); return 1; }

Question

BlackTom AI · Accepted Answer

Dereferencing an uninitialized pointer leads to undefined behavior, so the line *ptr = 0; is the primary problem. Relevant concepts: in C a pointer declared as int *ptr; has an indeterminate value until you assign it a valid address (for example the address of an existing int variable or memory obtained via malloc). Dereferencing such an indeterminate pointer (*ptr) can overwrite arbitrary memory, crash the program, or appear to "work" unpredictably. Also, by convention main returns 0 to indicate successful termination; returning a nonzero value (here 1) usually signals an error to the calling environment.

Step-by-step reasoning through the code:
1. int *ptr; — ptr is declared but not initialized; it contains an indeterminate address.
2. *ptr = 0; — this dereferences ptr and writes 0 to whatever (likely invalid) address ptr happened to contain. This is undefined behavior: the program may crash (segmentation fault), corrupt memory, or continue with unpredictable results.
3. printf("%d
", *ptr); — even if the previous write "succeeded" by chance, the read is still undefined behavior because ptr was never set to a valid object.
4. *ptr = 1; and printf("%d
", *ptr); — same UB as above.
5. return 1; — returns status code 1 to the OS. By convention, 0 indicates success, so returning 1 signals failure.

Why the two correct choices are correct:
- "int pointer was not initialized with malloc, so a crash may happen sooner or later": correct because ptr must point to valid memory (malloc, static/local int, or other) before dereferencing; otherwise UB and possible crash.
- "It returns 1 for a successful run, while it should return 0 for success by default": correct because main returning 0 indicates success; returning 1 typically means an error.

Why the other options are incorrect:
- "It reassigns memory too quickly": meaningless—there is no timing issue; the real issue is invalid pointer dereference.
- "It prints both 0 and 1 too quickly, without flushing the buffers": incorrect—printf with "
" flushes line-buffered stdout on terminals, and program termination flushes stdio buffers anyway; timing/flush is not the problem.

Conclusion: The two correct problems are the uninitialized pointer (need to allocate or assign it first) and returning 1 instead of 0; this matches the provided answers.

What are the problems with this C program? Select two. #include<stdio.h> #include<stdlib.h> int main() { int ptr; ptr = 0; printf("%d\n",ptr); ptr = 1; printf("%d\n",*ptr); return 1; } 多项选择题

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What are the problems with this C program? Select two. #include<stdio.h> #include<stdlib.h> int main() { int *ptr; *ptr = 0; printf("%d\n",*ptr); *ptr = 1; printf("%d\n",*ptr); return 1; } 多项选择题