Determine the real numbers $ a,b, $ such that
$$ [ax+by]+[bx+ay]=(a+b)\cdot [x+y],\quad\forall x,y\in\mathbb{R} , $$
where $ [t] $ is the greatest integer smaller than $ t. $
1. **Substitute \( x = 1 \) and \( y = 0 \):**
   \[
   [a \cdot 1 + b \cdot 0] + [b \cdot 1 + a \cdot 0] = (a + b) \cdot [1 + 0]
   \]
   Simplifying, we get:
   \[
   [a] + [b] = a + b
   \]
   Since \([a]\) and \([b]\) are the greatest integers less than or equal to \(a\) and \(b\) respectively, this implies:
   \[
   a = [a] \quad \text{and} \quad b = [b]
   \]
   Therefore, \(a\) and \(b\) must be integers.

2. **Substitute \( x = \frac{1}{2} \) and \( y = 0 \):**
   \[
   \left[ \frac{a}{2} \right] + \left[ \frac{b}{2} \right] = 0
   \]
   Since \(\left[ \frac{a}{2} \right]\) and \(\left[ \frac{b}{2} \right]\) are integers, the only way their sum can be zero is if both are zero:
   \[
   \left[ \frac{a}{2} \right] = 0 \quad \text{and} \quad \left[ \frac{b}{2} \right] = 0
   \]
   This implies:
   \[
   0 \leq \frac{a}{2} < 1 \quad \text{and} \quad 0 \leq \frac{b}{2} < 1
   \]
   Therefore:
   \[
   0 \leq a < 2 \quad \text{and} \quad 0 \leq b < 2
   \]
   Since \(a\) and \(b\) are integers, the possible values are:
   \[
   a \in \{0, 1\} \quad \text{and} \quad b \in \{0, 1\}
   \]

3. **Case \(a = 0\):**
   \[
   [by] + [bx] = b[x + y]
   \]
   Substitute \( y = -x \):
   \[
   [bx] + [-bx] = 0
   \]
   This implies:
   \[
   [bx] + [-bx] = 0
   \]
   Since \([bx]\) and \([-bx]\) are integers, this is true if and only if \(bx\) is an integer for all \(x\). Therefore, \(b\) must be zero:
   \[
   b = 0
   \]
   Thus, \((a, b) = (0, 0)\) is a solution.

4. **Case \(a = 1\):**
   \[
   [x + by] + [bx + y] = (1 + b)[x + y]
   \]
   Substitute \( x = y = \frac{1}{2} \):
   \[
   \left[ \frac{1}{2} + \frac{b}{2} \right] + \left[ \frac{b}{2} + \frac{1}{2} \right] = (1 + b) \left[ \frac{1}{2} + \frac{1}{2} \right]
   \]
   Simplifying, we get:
   \[
   2 \left[ \frac{1 + b}{2} \right] = 1 + b
   \]
   For this to hold, \(\frac{1 + b}{2}\) must be an integer. Therefore:
   \[
   1 + b = 2k \quad \text{for some integer } k
   \]
   This implies:
   \[
   b = 2k - 1
   \]
   Since \(b\) must be in \(\{0, 1\}\), the only possible value is:
   \[
   b = 1
   \]
   Thus, \((a, b) = (1, 1)\) is a solution.

The final answer is \(\boxed{(1,1), (0,0)}\).