Let $$f : R \to R$$ and $$g : R \to R$$ be defined as $$f(x) = \begin{cases} x+a, & x < 0 \\ |x-1|, & x \geq 0 \end{cases}$$ and $$g(x) = \begin{cases} x+1, & x < 0 \\ (x-1)^2 + b, & x \geq 0 \end{cases}$$, where $$a, b$$ are non-negative real numbers. If $$g \circ f(x)$$ is continuous for all $$x \in R$$, then $$a + b$$ is equal to ________.

We need $$g(f(x))$$ to be continuous for all $$x \in R$$. First, note that $$f(x) = x + a$$ for $$x < 0$$ and $$f(x) = |x-1|$$ for $$x \geq 0$$.

For $$x < 0$$: $$f(x) = x + a$$. If $$a \geq 0$$, then as $$x \to 0^-$$, $$f(x) \to a \geq 0$$. For $$x$$ sufficiently negative, $$f(x) < 0$$. The transition happens at $$x = -a$$.

For $$x < -a$$: $$f(x) = x + a < 0$$, so $$g(f(x)) = f(x) + 1 = x + a + 1$$. For $$-a \leq x < 0$$: $$f(x) = x + a \geq 0$$, so $$g(f(x)) = (f(x) - 1)^2 + b = (x+a-1)^2 + b$$. For $$x \geq 0$$: $$f(x) = |x-1| \geq 0$$, so $$g(f(x)) = (|x-1|-1)^2 + b$$.

Checking continuity at $$x = -a$$: From the left, $$g(f(-a^-)) = -a + a + 1 = 1$$. From the right, $$g(f(-a^+)) = (-a + a - 1)^2 + b = 1 + b$$. For continuity: $$1 = 1 + b$$, so $$b = 0$$.

Checking continuity at $$x = 0$$: From the left (with $$-a \leq x < 0$$), $$g(f(0^-)) = (0 + a - 1)^2 + b = (a-1)^2$$. From the right, $$g(f(0^+)) = (|0-1|-1)^2 + b = (1-1)^2 + 0 = 0$$. For continuity: $$(a-1)^2 = 0$$, so $$a = 1$$.

Therefore $$a + b = 1 + 0 = 1$$.

The answer is $$1$$.