```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):
        # Avoid computing 2**i as a huge int that can't convert to float
        # Instead, compute base * 2**i carefully
        # If base == 0, then base * 2**i == 0 for all i
        if base == 0:
            value = 0.0
        else:
            # Compute base * 2**i without creating huge ints
            # We can use logarithms or iterative multiplication
            # But simpler: if 2**i is too large, the result will exceed cap
            # So we can check if base * 2**i > cap without computing the full value
            # Use: if i is large enough that base * 2**i > cap, then value = cap
            # We need to determine when base * 2**i > cap
            
            # If base > 0 and cap > 0, find the smallest i where base * 2**i > cap
            # base * 2**i > cap => 2**i > cap / base => i > log2(cap / base)
            # But we need to handle edge cases
            
            # Simpler approach: iteratively compute, but stop early if value >= cap
            # Actually, let's just compute iteratively to avoid huge ints
            pass
        
        # Let me redo this properly
        # We want min(cap, base * 2**i)
        # If base == 0, result is 0
        # If cap == 0, result is 0
        # Otherwise, compute base * 2**i but avoid huge int conversion
        
        # Use iterative approach: start with base, multiply by 2 each step
        # But we need to handle the case where the value exceeds cap
        
        # Let's use a different approach: compute the value step by step
        # Start with current = base
        # For each i, value = min(cap, current)
        # Then current *= 2 for next iteration
        
        # But we need to be careful: if current already >= cap, all subsequent values are cap
        
        # Let me restructure the loop
        pass
    
    # Redo the function properly
    result = []
    current = base
    for i in range(attempts):
        value = min(cap, current)
        result.append(value)
        current *= 2
        # If current already exceeds cap, all future values will be cap
        # But we still need to append cap for remaining iterations
        # Actually, we can break early and fill with cap
        if current > cap and cap > 0:
            # All remaining values are cap
            result.extend([cap] * (attempts - i - 1))
            break
    return result
```