I'm trying to understand the memory implications of this realloc loop in C. My professor claims that the memory overhead is constant, but I have some doubts. Here's the code I was given:
```c
int *f(int n) {
int i, *a = NULL;
for(i = 0; i < n; i++) {
a = realloc(a, (i + 1) * sizeof(int));
a[i] = i;
}
return a;
}
```
Can someone break down what this means for memory usage and if the professor's statement holds up?
2 Answers
It sounds like your professor might be mixing some concepts. Memory overhead isn't truly constant in this loop. While the function may only ever use `n` integers worth of space at the end, the realloc process can repeatedly request memory that's larger than necessary and could allocate more than you need, especially as `n` grows. There's a risk here too—if `realloc()` fails and returns NULL, it could lead to a segmentation fault since your pointer will still reference old data. It would be a lot safer and more efficient to allocate memory for the full array initially using `calloc(n, sizeof(int))` before the loop begins.
When we talk about memory overhead, it's important to clarify what we mean. The overall memory usage for this function is O(n), which means it increases with the size of `n`. However, the memory allocations might not be constant—especially since calling `realloc()` can lead to varying overhead depending on several factors. Instead of reallocating every single integer increment, it's often a better practice to double the size when reallocating to improve efficiency. This approach reduces the total number of reallocations needed which can make a difference, especially in larger datasets.
Good point! I’ve also read that the specifics of how `realloc()` works can really vary by implementation, affecting performance in ways that could surprise you.