Let $ f, g: \mathbb {R} \to \mathbb {R} $ function such that $ f (x + g (y)) = - x + y + 1 $ for each pair of real numbers $ x $ e $ y $. What is the value of $ g (x + f (y) $?
To solve the problem, we need to find the value of \( g(x + f(y)) \) given the functional equation \( f(x + g(y)) = -x + y + 1 \) for all \( x, y \in \mathbb{R} \).

1. **Prove that \( f \) and \( g \) are injective:**

   - Suppose \( g(a) = g(b) \). Then, substituting \( x = 0 \) and \( y = a \) and \( y = b \) in the original equation, we get:
     \[
     f(g(a)) = -0 + a + 1 = a + 1
     \]
     \[
     f(g(b)) = -0 + b + 1 = b + 1
     \]
     Since \( g(a) = g(b) \), we have \( f(g(a)) = f(g(b)) \), which implies \( a + 1 = b + 1 \). Therefore, \( a = b \), proving that \( g \) is injective.

   - Now, suppose \( f(a) = f(b) \). For arbitrary \( y \), we have:
     \[
     f(a) = -a + g(y) + y + 1
     \]
     \[
     f(b) = -b + g(y) + y + 1
     \]
     Since \( f(a) = f(b) \), we get:
     \[
     -a + g(y) + y + 1 = -b + g(y) + y + 1
     \]
     This implies \( -a = -b \), so \( a = b \). Therefore, \( f \) is injective.

2. **Prove that \( f \) and \( g \) are surjective:**

   - To show that \( f \) is surjective, note that for any \( y \in \mathbb{R} \), we can vary \( x \) to cover all real numbers. Thus, \( f \) is surjective.

   - Since \( f \) is bijective (both injective and surjective), it is invertible. From the equation \( f(g(y)) = y + 1 \), we can solve for \( g(y) \):
     \[
     g(y) = f^{-1}(y + 1)
     \]
     Varying \( y \) gives the surjectivity of \( g \).

3. **Find the explicit forms of \( f \) and \( g \):**

   - Since \( f \) is bijective, it has an inverse. Using the original equation \( f(x + g(y)) = -x + y + 1 \), set \( y = g^{-1}(0) \):
     \[
     f(x) = -x + g^{-1}(0) + 1
     \]
     Let \( c = g^{-1}(0) + 1 \), then:
     \[
     f(x) = -x + c
     \]

   - Substitute \( f(x) = -x + c \) back into the equation \( f(g(y)) = y + 1 \):
     \[
     -g(y) + c = y + 1
     \]
     Solving for \( g(y) \):
     \[
     g(y) = -y + c - 1
     \]
     Let \( d = c - 1 \), then:
     \[
     g(y) = -y + d
     \]

4. **Calculate \( g(x + f(y)) \):**

   - Using the forms \( f(y) = -y + c \) and \( g(y) = -y + d \):
     \[
     g(x + f(y)) = g(x - y + c)
     \]
     Substitute \( g(z) = -z + d \) with \( z = x - y + c \):
     \[
     g(x - y + c) = -(x - y + c) + d
     \]
     Simplify:
     \[
     g(x + f(y)) = -x + y - c + d
     \]
     Since \( c = d + 1 \), we have:
     \[
     g(x + f(y)) = -x + y - (d + 1) + d = -x + y - 1
     \]

Conclusion:
\[
\boxed{-x + y - 1}
\]