```python
def retry_delays(attempts, base, cap):
    if isinstance(attempts, bool) or not isinstance(attempts, int):
        raise TypeError("attempts must be a non-negative integer")
    if attempts < 0:
        raise ValueError("attempts must be non-negative")
    if base < 0:
        raise ValueError("base must be non-negative")
    if cap < 0:
        raise ValueError("cap must be non-negative")
    
    result = []
    for i in range(attempts):
        # Compute base * 2**i without causing overflow by checking if 2**i would exceed cap/base
        # If base == 0, then base * 2**i == 0, so min(cap, 0) = 0
        if base == 0:
            result.append(0.0)
            continue
        
        # Check if 2**i * base > cap, i.e., 2**i > cap / base
        # To avoid computing huge 2**i, we can check if i is large enough that 2**i * base > cap
        # Since cap and base are non-negative, if cap == 0, then min(cap, ...) = 0
        if cap == 0:
            result.append(0.0)
            continue
        
        # We need min(cap, base * 2**i)
        # If base * 2**i >= cap, then result is cap
        # Otherwise, result is base * 2**i
        
        # To avoid overflow, check if 2**i * base > cap
        # This is equivalent to 2**i > cap / base
        # We can compute this safely by checking if i is large enough
        
        # If base * 2**i would overflow, we know it's > cap (since cap is finite), so result = cap
        # But we need to be careful: if base * 2**i <= cap, we compute it normally
        
        # Check if 2**i * base > cap
        # If i is large, 2**i can be huge. We can check: if i > some threshold, then 2**i * base > cap
        # The threshold is when 2**i > cap / base, i.e., i > log2(cap / base)
        
        # To avoid computing huge powers, we can check if i is large enough that 2**i * base > cap
        # If cap / base is small, then for large i, 2**i * base > cap
        
        # Let's compute cap / base safely
        # If base > 0 and cap > 0, cap / base is a float
        # We need to check if 2**i > cap / base
        
        # If i is large enough that 2**i > cap / base, then min(cap, base * 2**i) = cap
        # Otherwise, compute base * 2**i
        
        # To avoid overflow, we can check if i is large enough that 2**i * base > cap
        # We can do this by checking if i > log2(cap / base)
        
        # But computing log2 might have precision issues. Instead, we can iteratively check.
        
        # Alternative approach: compute 2**i as an integer, but if it's too large, we know the result is cap
        # Since cap is a float, and base is a float, base * 2**i could overflow if 2**i is huge
        
        # Let's check if 2**i * base > cap
        # If base * 2**i > cap, result = cap
        # Otherwise, result = base * 2**i
        
        # To avoid overflow, we can check if i is large enough that 2**i * base > cap
        # We can do this by checking if i > some threshold
        
        # Let's compute the threshold: i such that 2**i * base > cap
        # i > log2(cap / base)
        
        # If cap / base is very small, log2(cap / base) is negative, so for i >= 0, 2**i * base > cap
        # If cap / base is large, log2(cap / base) is large
        
        # We can compute log2(cap / base) safely
        import math
        if base > 0 and cap > 0:
            threshold = math.log2(cap / base)
            if i > threshold:
                result.append(cap)
            else:
                result.append(base * (2 ** i))
        else:
            # base == 0 or cap == 0