A function $g$ is such that for all integer $n$:
$$g(n)=\begin{cases}
1\hspace{0.5cm} \textrm{if}\hspace{0.1cm}  n\geq 1 & \\
0 \hspace{0.5cm} \textrm{if}\hspace{0.1cm} n\leq 0 &
\end{cases}$$

A function $f$ is such that for all integers $n\geq 0$ and $m\geq 0$:
$$f(0,m)=0 \hspace{0.5cm} \textrm{and}$$
$$f(n+1,m)=\Bigl(1-g(m)+g(m)\cdot g(m-1-f(n,m))\Bigr)\cdot\Bigl(1+f(n,m)\Bigr)$$

Find all the possible functions $f(m,n)$ that satisfies the above for all integers $n\geq0$ and $m\geq 0$
To find all possible functions \( f(n, m) \) that satisfy the given conditions, we will proceed step-by-step.

1. **Understanding the function \( g(n) \):**
   \[
   g(n) = \begin{cases}
   1 & \text{if } n \geq 1 \\
   0 & \text{if } n \leq 0
   \end{cases}
   \]
   This function \( g(n) \) acts as an indicator function that is 1 for positive integers and 0 for non-positive integers.

2. **Initial condition for \( f(n, m) \):**
   \[
   f(0, m) = 0
   \]
   This is given directly in the problem statement.

3. **Recursive definition of \( f(n+1, m) \):**
   \[
   f(n+1, m) = \left(1 - g(m) + g(m) \cdot g(m-1-f(n, m))\right) \cdot \left(1 + f(n, m)\right)
   \]
   We need to analyze this expression for different values of \( m \).

4. **Case \( m = 0 \):**
   \[
   g(0) = 0 \implies f(n+1, 0) = (1 - 0 + 0 \cdot g(-1 - f(n, 0))) \cdot (1 + f(n, 0)) = 1 + f(n, 0)
   \]
   Since \( f(0, 0) = 0 \), by induction:
   \[
   f(n, 0) = n
   \]

5. **Case \( m \geq 1 \):**
   \[
   g(m) = 1 \implies f(n+1, m) = \left(1 - 1 + 1 \cdot g(m-1-f(n, m))\right) \cdot \left(1 + f(n, m)\right) = g(m-1-f(n, m)) \cdot (1 + f(n, m))
   \]
   We need to consider the value of \( g(m-1-f(n, m)) \):
   - If \( f(n, m) \leq m-1 \), then \( m-1-f(n, m) \geq 0 \) and \( g(m-1-f(n, m)) = 1 \):
     \[
     f(n+1, m) = 1 \cdot (1 + f(n, m)) = 1 + f(n, m)
     \]
   - If \( f(n, m) = m-1 \), then \( f(n+1, m) = 1 \cdot (1 + (m-1)) = m \):
     \[
     f(n+1, m) = 0 \quad \text{(since } f(n+1, m) \text{ is taken modulo } m\text{)}
     \]

6. **Induction hypothesis:**
   Assume \( f(n, m) = n \mod m \) for some \( n \geq 0 \) and \( m \geq 1 \).

7. **Inductive step:**
   \[
   f(n+1, m) = \begin{cases}
   1 + f(n, m) & \text{if } f(n, m) < m-1 \\
   0 & \text{if } f(n, m) = m-1
   \end{cases}
   \]
   This matches the behavior of \( n \mod m \):
   - If \( f(n, m) < m-1 \), then \( f(n+1, m) = (n+1) \mod m \).
   - If \( f(n, m) = m-1 \), then \( f(n+1, m) = 0 \).

Thus, by induction, we conclude that:
\[
f(n, m) = n \mod m
\]

The final answer is \( \boxed{ f(n, m) = n \mod m } \).