pointer arithmetic in C++ using char*

Question

I'm having trouble understanding what the difference between these two code snippets is:

// out is of type char* of size N*D
// N, D are of type int


for (int i=0; i!=N; i++){
    if (i % 1000 == 0){
        std::cout << "i=" << i << std::endl;
    }
    for (int j=0; j!=D; j++) {
        out[i*D + j] = 5;
    }
}

This code runs fine, even for very big data sets (N=100000, D=30000). From what I understand about pointer arithmetic, this should give the same result:

for (int i=0; i!=N; i++){
    if (i % 1000 == 0){
        std::cout << "i=" << i << std::endl;
    }
    char* out2 = &out[i*D];
    for (int j=0; j!=D; j++) {
        out2[j] = 5;
    }
}

However, the latter does not work (it freezes at index 143886 - I think it segfaults, but I'm not 100% sure as I'm not used to developing on windows) for a very big data set and I'm afraid I'm missing something obvious about how pointer arithmetic works. Could it be related to advancing char*?

EDIT: We have now established that the problem was an overflow of the index (i.e. (i*D + j) >= 2^32), so using uint64_t instead of int32_t fixed the problem. What's still unclear to me is why the first above case would run through, while the other one segfaults.

Answer 1

N * D is 3e9; that doesn't fit in a 32 bit int.