From this reference, In C it seems the following behavior is undefined.
int my_array[100][50];
int *p = my_array[0];
p[50]; // UB
Is there a reference in C++03 or C++11 which confirms this?
2 Answers
Yes in the description of the + operator. You may not dereference that pointer in C because it is a past the end pointer for the first subarray. In C++ this currently is legal because the pointer points to a valid integer (the points to relation is defined somewhere in clause 3). However in both standards adding more than 50 yields undefined behavior.
A DR was recently sent to the c++ committee about the rule that dereferencing such "valid out of thin air" pointers may be dereferenced, so i would not rely on that.
answered Dec 8, 2011 at 9:33
Johannes Schaub - litb
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2 Comments
Šimon Tóth
Damn, I don't have C89 standard here, but C99 talks about a pointer that points to an array in a case of the multidimensional array, I would personally interpret this as define behaviour since the pointer doesn't point outside of the original array, only outside of the sub-array.
M.M
To clarify, when you say "Yes", you mean "Yes the behaviour is undefined"; (the question title is the negative of the question in the actual post!). And the standard section that says you can't add more than 50 is
[expr.add]#5.int my_array[3][3] = {{1,2,3},{4,5,6},{7,8,9}};
for(i=0;i<3*3;i++)
{
printf("%d,",*(*my_array+i));
}
output is 1,2,3,4,5,6,7,8,9,
I think you can do it like that.
p[50]involves undefined behaviour?gcc -Wall -Wextraand get no warnings about any undefined behavior; changep[50]top[50] = 1;to quiet the "no effect" warning and make sure it knows it should generate the code. (Add in someprintf(3)calls if you want to make sure it doesn't optimize away the line.)