Why is my variable not increasing after each recursion? - c++

Question

So I'm pretty new to C++ and I'm trying to generate Fibonacci numbers using recursion right? Except when I try to index to the next value I need to increase the index, and the function won't let me increase it.
How I know, is that I've run it in debug mode and stepped over the function - it continuously loops inside the fibGen function with i staying at a constant value of 1 (When I call fibGen in my main function the parameters are (startingSeq, 1, 13), where startingSeq is another vector with values {1,1}
The code builds and compiles fine, but when I run it I get a memory alloc error, which is obv caused by the infinite loop it seems to have.
What have I done wrong? Why doesn't the 'i' increase?
I've tried increasing i by having i++ inside the recursive fibGen call (in the else statement), I've tried having i++ outside the function call, and I've tried what's in there right now, where I have i+=1 and then pass i through.
Also I didn't have the 'return fibSeq' down the bottom originally because it doesn't make sense to have it there, but I put it in because vscode wouldn't compile without it in there, saying that it could never reach the end of fibGen (which makes sense now, and when this problem's fixed I think it can be removed, but at the moment it's just there so that the program will compile)
std::vector<int> fibGen(std::vector<int> fibSeq, int i, int max){
if(fibSeq[i] >= max){
return fibSeq;
}else{
fibSeq.push_back(fibSeq[i] + fibSeq[i-1]);
i+=1;
fibGen(fibSeq, i, max);
}
return fibSeq;
}
The output should be a vector containing the Fibonacci sequence, and I'm getting a mem alloc error (described above)

Answer 1

Actually, your code kind of works. You are just handling the vector of results sub-optimally. With your declaration:
std::vector<int> fibGen(std::vector<int> fibSeq, int i, int max)
you will always pass copies of the vector around (and you are never using the returned value). Instead, what you probably want to do is to work on the same data. To do this, use a reference to a vector (denoted by &). Then, you do not need to return anything:
void fibGen(std::vector<int>& fibSeq, int i, int max) {
if (fibSeq[i] >= max) {
return;
}
else {
fibSeq.push_back(fibSeq[i] + fibSeq[i - 1]);
i += 1;
fibGen(fibSeq, i, max);
}
}
void main(void) {
std::vector<int> fib = { 1, 1 };
fibGen(fib, 1, 34);
for (auto i : fib)
std::cout << i << std::endl;
}

Answer 2

So, I could not reproduce the error, until I realized that in your actual code, you were doing this instead of the code you posted:
std::vector<int> fibGen(std::vector<int> fibSeq, int i, int max){
if(fibSeq[i] >= max){
return fibSeq;
}else{
fibSeq.push_back(fibSeq[i] + fibSeq[i-1]);
i+=1;
fibGen(fibSeq, i, max);
}
}
In Visual Studio 2010, at least, it compiled fine, but threw an error at runtime, which I believe is what you described. So I'm going to assume I reproduced it.
The reason this is throwing an error is because you are invoking C++'s infamous undefined behavior. Undefined behavior in C++ permits anything to happen, like not failing to compile, but throwing a runtime error.
The fix is simple, really. You just need to return a value in all execution paths. Or, simply put, just do this:
std::vector<int> fibGen(std::vector<int> fibSeq, int i, int max){
if(fibSeq[i] >= max){
return fibSeq;
}else{
fibSeq.push_back(fibSeq[i] + fibSeq[i-1]);
i+=1;
return fibGen(fibSeq, i, max);
// Notice we are returning the value of the recursive call to fibGen().
// We can do this because we have already modified the vector the way we need to,
// so just simply returning the value is fine
}
}
Of course, you can take Nico Schertler's suggestion instead too:
void fibGen(std::vector<int>& fibSeq, int i, int max) {
if (fibSeq[i] >= max) {
return;
}
else {
fibSeq.push_back(fibSeq[i] + fibSeq[i - 1]);
i += 1;
fibGen(fibSeq, i, max);
}
}
It should be noted that not returning a value from a void function is not undefined behavior (that I'm aware), but actually how void's intended to work, so this function is fine not returning a value.