int *A [10];
A[2][3]=15;
This statement gives no compile time error, yet when I run the program it gives a runtime error. Why?
Edit: I specifically want to know why there is no compile time error?
-
2Why should this be a runtime error? Where's the logic to do the kind of checking that would create an error?Kerrek SB– Kerrek SB2017-01-03 09:43:29 +00:00Commented Jan 3, 2017 at 9:43
Add a comment
|
3 Answers
This statement gives no compile time error in c
Because there is no syntactical error or constraint violation that compiler should be judging. C does not do any bound-checking for arrays (or pointer arithmetics).
You'll be perfectly allowed to write a code which makes use of invalid memory (example: dereference an invalid memory location) but in case, the compiler has produced a binary for such code, running the binary would invoke undefined behaviour.
yet when I run the program, it gives a runtime error
In your code,
int *A [10];
A is an array of 10 int *s, and they are not initialized explicitly. Looking from the snippet, it appears A is not in global scope, i.e., not having static storage, so the contents of each of those pointers are indeterminate.
So, later in process of writing A[2][3]=15;, you're trying to access A[2] (a pointer), which points to an invalid memory. This invokes undefined behavior.
7 Comments
StoryTeller - Unslander Monica
A[2] is a valid element of the array A... The wording in your last sentence can be better
Sourav Ghosh
@StoryTeller right, but it does not point to anything valid. Please suggest or feel free to edit in, no worries. :)
Sourav Ghosh
@StoryTeller Thanks, I've also added a little more. :)
cdcdcd
Sourav, to be pedantic, A[2] is likely, if uninitialised to be pointing to invalid memory, but it may not. The values for each pointer may well, depending what was on the stack at "instantiation" (during run time), may well have (and just by fluke) have a value that is within the process' memory space. This of-course is much more dangerous as you will end up pulling the ground from beneath the process elsewhere.
Sourav Ghosh
@cdcdcd it may, it may not, it's not guaranteed, the one thing guaranteed is
indeterminate and attempt to use indeterminate "value" leads to UB, right? |
Because accessing the uninitialized pointer A[2] invokes undefined behavior, which means anything is allowed to happen as far as the C standard is concerned.
why there is no compile time error
Because the standard doesn't require a diagnostic (such as a compile error) to be issued in this case.
1 Comment
StoryTeller - Unslander Monica
I'd argue that better wording would be "uninitialized pointer at A[2]". But it's just me.
It's for the same reason that
int a, d = 0;
a = 1 / d;
compiles cleanly. To detect the invalid operation the compiler would need to run the program.
