Find all functions $f: \mathbb{R}\to{R}$ such that for any $a<b$, $f([a,b])$ is an interval of length $b-a$
1. **Given Condition**: We need to find all functions \( f: \mathbb{R} \to \mathbb{R} \) such that for any \( a < b \), \( f([a,b]) \) is an interval of length \( b-a \).

2. **Consider Three Points**: Let \( a < b < c \). Note that \( f([a,c]) = f([a,b]) \cup f([b,c]) \). Since \( f([a,b]) \) and \( f([b,c]) \) are intervals, we have:
   \[
   |f([a,b]) \cup f([b,c])| \le |f([a,b])| + |f([b,c])|
   \]
   with equality if and only if \( f([a,b]) \) and \( f([b,c]) \) intersect in exactly one point or not at all.

3. **Length of Intervals**: Given the condition, we know:
   \[
   |f([a,b])| = b-a \quad \text{and} \quad |f([b,c])| = c-b
   \]
   Therefore:
   \[
   |f([a,c])| = c-a
   \]

4. **Equality Condition**: Since \( |f([a,c])| = |f([a,b])| + |f([b,c])| \), it implies that \( f([a,b]) \) and \( f([b,c]) \) are disjoint except possibly at a single point. This common point must be \( f(b) \).

5. **Monotonicity**: Suppose \( f(a) \le f(b) \). If \( f(c) < f(b) \), then \( f([a,b]) \) and \( f([b,c]) \) both contain the interval \( [\max\{f(a), f(c)\}, f(b)] \), which contradicts the condition that their intersection is a single point. Hence, \( f(c) \ge f(b) \). This implies that \( f \) is non-decreasing on \( [b, +\infty) \). By similar logic, \( f \) is non-decreasing on all of \( \mathbb{R} \).

6. **Function Form**: For all \( 0 < y \), we have:
   \[
   |f([0,y])| = f(y) - f(0) = y
   \]
   Thus:
   \[
   f(y) = f(0) + y
   \]
   Similarly, this holds for all \( y \in \mathbb{R} \).

7. **Non-Increasing Case**: If \( f(a) \ge f(b) \), we get that \( f \) is non-increasing. By the same logic, we have:
   \[
   f(y) = -y + f(0)
   \]

8. **Conclusion**: The function \( f \) must be either of the form \( f(x) = x + c \) or \( f(x) = -x + c \) for some constant \( c \).

The final answer is \( \boxed{ f(x) = x + c } \) or \( f(x) = -x + c \).