问题
I would like to understand what is going on here, more precisely why I do NOT receive a segmentation fault while writing on a memory location, which according to my understanding is not allocated.
Let's say I want to define a 2D array of int (testptr). One dimension (4) is allocated statically (as an "array"), the second dimension (2) dynamically (as a "pointer").
// First dimension 4 rows static
int *testptr[4];
for (i=0; i<4; i++)
testptr[i] = calloc(2, sizeof(int));
// Second dimension 2 columns "dynamically" (in this example it is really just a constant)
Now I write to some locations:
testptr[0][0] = 5;
testptr[1][0] = 6;
testptr[2][1] = 7;
testptr[3][1] = 7;
All the above I expect to work fine, as they are within a 4x2 "array".
Now I write to a location which should not be allocated:
testptr[2][3] = 8;
And to make sure I write to many of them:
for (i=0; i<1000; i++)
testptr[3][i] = i;
In none of these I get a segmentation fault nor other errors.
- Is it correct to say that we are writing on unallocated memory?
- Is it just luck that we are not receiving an error, as those unallocated memory locations are not reserved by other variables/processes?
- Can we assume that doing this will cause problems (segfaults) in other points of the programs?
Thanks for your answers.
回答1:
Your program invokes undefined behavior. It will give either expected or unexpected results. There is no guarantee that accessing arrays out of bounds will produce a segmentation fault or it will gonna crash your hard disk!
来源：https://stackoverflow.com/questions/27850221/why-dont-i-get-sigsegv