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I can treat array as if it were a 1-based array ...
C Interview Questions and Answers
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I can treat array as if it were a 1-based array ...
Q: Here's a neat trick: if I write
int realarray[10];
int *array = &realarray[-1];
I can treat array as if it were a 1-based array.
A: Although this technique is attractive (and was used in old editions of the book Numerical Recipes in C), it is not strictly conforming to the C Standard. Pointer arithmetic is defined only as long as the pointer points within the same allocated block of memory, or to the imaginary ``terminating'' element one past it; otherwise, the behavior is undefined, even if the pointer is not dereferenced. The code above computes a pointer to memory before the beginning of realarray and could fail if, while subtracting the offset, an illegal address were generated (perhaps because the address tried to ``wrap around'' past the beginning of some memory segment).
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