C Question
Is accessing members through offsetof well defined?
When doing pointer arithmetic with
offsetofConsider the following example:
#include <stddef.h>
#include <stdio.h>
typedef struct {
const char* a;
const char* b;
} A;
int main() {
A test[3] = {
{.a = "Hello", .b = "there."},
{.a = "How are", .b = "you?"},
{.a = "I\'m", .b = "fine."}};
for (size_t i = 0; i < 3; ++i) {
char* ptr = (char*) &test[i];
ptr += offsetof(A, b);
printf("%s\n", *(char**)ptr);
}
}
This should print "there.", "you?" and "fine." on three consecutive lines, which it currently does with both clang and gcc, as you can verify yourself on wandbox. However, I am unsure whether any of these pointer casts and arithmetic violate some rule which would cause the behavior to become undefined.
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Answer Source
As far as I can tell, it is well-defined behavior. But only because you access the data through a char type. If you had used some other pointer type to access the struct, it would have been a "strict aliasing violation".
Strictly speaking, it is not well-defined to access an array out-of-bounds, but it is well-defined to use a character type pointer to grab any byte out of a struct. By using offsetof you guarantee that this byte is not a padding byte (which could have meant that you would get an indeterminate value).
Note however, that casting away the const qualifier does result in poorly-defined behavior.
Similarly, the cast (char**)ptr is an invalid pointer conversion - this alone is undefined behavior.
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