>>> """
... in MPEG-H Audio there is a syntax component called escapedValue, it is defined as follows in the USAC specification:
...
... ```
... escapedValue(nBits1, nBits2, nBits3)
... {
...   value;  nBits1  uimsbf
...   if (value == 2nBits1-1) {
...     value += valueAdd;  nBits2  uimsbf
...     if (valueAdd == 2nBits2-1) {
...       value += valueAdd;  nBits3  uimsbf
...     }
...   }
...   return value;
... }
... ```
... write an implementation for this function in Rust, complete with an example program that compiles.
... """
<think>
Okay, I need to implement the escapedValue function from the MPEG-H Audio specification in Rust. Let's see how it
works.

First, looking at the syntax provided:

The function takes three parameters: nBits1, nBits2, nBits3. It reads a value with nBits1 bits. If that value is
equal to 2^(nBits1) - 1 (which is the maximum value for nBits1 bits), then it adds another valueAdd read with
nBits2 bits. If this new valueAdd also equals 2^(nBits2) - 1, it adds another valueAdd from nBits3 bits.

So, the function returns a cumulative value that can be up to (2^nBits1 + 2^nBits2 + 2^nBits3) - 3, depending on
how many escape levels are triggered.

Now, thinking about how to read bits in Rust. Since this is a bit-level operation, I need a way to parse bits from
some input. For simplicity, maybe the example can take an integer representing the bitstream and another integer
indicating how many bits to read each time. But realistically, handling a bitstream would require more precise
extraction.

Wait, but for the sake of an example, perhaps using integers is manageable. So, I'll create a struct called
BitStream that holds the current value and the number of bits remaining. Each call to read_bits will extract nBits
from this stream.

In the escapedValue function, I'll start by reading nBits1 bits. Check if it's equal to 2^nBits1 -1. If so,
proceed to read nBits2, and then check again, possibly reading nBits3.

I need to implement read_bits in BitStream. Each time, shift right to get the needed bits, update the value and
remaining bits accordingly.

Testing is important. For example, if nBits1=4, nBits2=3, nBits3=2:

Case 1: value=0 (all bits 0). Then return 0.

Case 2: value=15 (max for 4 bits), then read 7 (max for 3 bits) → total 22. Since 15 +7 =22, and since 7 is max
for 3 bits, we read another 3 (max for 2 bits?), wait no—nBits3 is 2, so 2^2-1=3. So value would be 15+7+3=25.

Wait, but the code adds only once if the condition is met each time. Let me see:

In the function:
value += valueAdd (from nBits2) if first condition.
Then, if that valueAdd was max for nBits2, add another valueAdd from nBits3.

So total additions can be 0, 1, or 2 times.

So in code:

Read value with nBits1. If value == (1<<nBits1)-1, then read valueAdd with nBits2, add to value. Then if valueAdd
== (1<<nBits2)-1, read another valueAdd with nBits3 and add.

Thus, the maximum possible value is when all three are maxed: (1<<nBits1 -1) + (1<<nBits2 -1) + (1<<nBits3 -1).

Now, implementing this in Rust:

First, define BitStream struct with current_bits and num_bits_left. Then read_bits method.

Then escaped_value function takes the stream and nBits parameters, reads as per the syntax.

Testing with an example where we have a bitstream that triggers all three additions: for example, 4+3+2 bits.

Let me think of a test case:

Suppose nBits1=4, so max is 15. So in the stream, first 4 bits are 15 (binary 1111). Then next 3 bits are 7 (111),
which triggers another read of 2 bits: let's say 3 (11). So total value is 15+7+3=25.

The bitstream for this would be:

bits = (1<<4 + 7<<2) + 3? Wait, need to represent it correctly. The order is important because the bits are read
in sequence.

Wait, perhaps using an integer where the higher bits come first. For example:

- First 4 bits: 15 → binary 1111
- Next 3 bits: 7 → binary 111
- Next 2 bits: 3 → binary 11

So total bits used: 4+3+2=9.

The integer representing the bitstream would have these bits in order. Let's construct it:

bits = (15 << (3+2)) | (7 << 2) | 3 = 15<<5 | 7<<2 |3 = 15*32 + 7*4 +3= 480 +28 +3=511.

But the BitStream starts with current_bits=511 and num_bits_left=9.

Wait, but when reading bits, each read reduces the available bits. So for nBits1=4:

read_bits(4) → returns 15, stream now has (511 >>4)=31, num_bits left=5.

Then, since value is 15 == 2^4-1, we proceed to read nBits2=3: 7, which is max. So add it, now value=22.

Check if 7 ==7 (yes), so read nBits3=2:3. Add to get 25.

So in the test case:

escaped_value(&mut stream,4,3,2) should return 25.

Another test case where only first escape is triggered: value=15, then valueAdd=6 (not max), so total 21.

Third test case: no escapes, just value=0.

So in the main function, I'll create a BitStream and test these cases.

Potential issues:

- Handling bit shifts correctly without overflow. Since Rust has fixed-size integers, but for this example, using
u32 should be sufficient as long as nBits1 +nBits2 +nBits3 <=32.

But in practice, the bitstream could be larger, so perhaps a better approach is to use a slice of bytes or another
structure, but that's beyond the scope here. For simplicity, using an integer and tracking bits left is manageable
for this example.

Another consideration: what if nBits1, nBits2, nBits3 are zero? But per MPEG-H specs, these should be positive
integers.

So now, writing the code:

Define BitStream struct with current_bits: u32 and num_bits_left: usize.

read_bits method takes self as mutable reference, reads up to n bits. If not enough bits left, return 0 or handle
error? For this example, assume that there are enough bits.

In read_bits:

mask = (1 <<n) -1;

bits = (self.current_bits >> (self.num_bits_left -n)) & mask;

Wait no, because the higher bits are on the left. So to get the next n bits, you shift right by (num_bits_left
-n), but that's not correct.

Wait, perhaps I should represent the bitstream as a queue of bits, where the first bits are most significant. But
using an integer and tracking how many bits have been read may be tricky.

Alternatively, each time we read n bits, we take the top n bits from current_bits by shifting right (num_bits_left
-n) times? Hmm, maybe not.

Wait, perhaps the initial approach is flawed because when you have a bitstream represented as an integer, it's
more efficient to manage it in a way that allows extracting the next n bits correctly. For example:

current_bits = 0b1111_0111_0011 (hypothetical). To read the first 4 bits: shift right by (total_bits -4), which
gives 0b1111, then subtract those 4 bits from total.

Wait, perhaps the BitStream struct should track how many bits have been consumed. So, when you create it,
current_bits is the integer, and num_bits_left is the number of bits available.

When read_bits(n) is called:

- Ensure that n <= num_bits_left; else, error.
- The bits are taken from the highest order remaining bits.
- So, shift right by (num_bits_left -n) to get those bits.
- Then, update current_bits by shifting right by n bits.
- Decrease num_bits_left by n.

Yes, that makes sense.

So for example:

current_bits = 0b1111_0111_0011, which is 9 bits. Suppose we read 4 bits:

bits = (current_bits >> (9 -4)) → current_bits >>5 → 0b1111 (15)

then current_bits becomes 0b0111_0011 (7*32 +3= 227?), and num_bits_left is 5.

Wait, perhaps in code:

fn read_bits(&mut self, n: usize) -> u32 {
    assert!(n <= self.num_bits_left);
    let shift = self.num_bits_left -n;
    let bits = (self.current_bits >> shift) & ((1 <<n)-1);
    self.current_bits >>=n;
    self.num_bits_left -=n;
    bits
}

Yes, this should correctly extract the next n bits from the highest order.

Testing with current_bits=511, which is 9 bits: 0b111111111.

First read 4 bits:

shift =9-4=5

bits=(511 >>5)= 15 (since 511 is 0b111111111; shifting right by 5 gives 0b1111, which is 15).

current_bits becomes 511>>4 =31.

num_bits_left=9-4=5.

Next read 3 bits:

shift=5-3=2

bits=(31 >>2) → 7 (since 31 is 0b11111; shifting right by 2 gives 0b111, which is 7).

current_bits becomes 31>>3=3.

num_bits_left=5-3=2.

Next read 2 bits:

shift=2-2=0

bits=(3 >>0) & ((1<<2)-1)=3 &3 =3.

current_bits becomes 3>>2=0.

num_bits_left=0.

So yes, this works correctly.

Thus, the BitStream struct and read_bits function are correct.

Now, writing the code accordingly.
</think>

To solve this problem, we need to extract specific bits from a bitstream and compute values based on these
extracted bits. The goal is to simulate reading bits from a stream where each bit operation reduces the remaining
bits in the stream.

### Approach
1. **BitStream Struct**: We define a struct `BitStream` that holds the current state of the bitstream, including
the integer representation of the bits and the number of bits left.
2. **Reading Bits**: The `read_bits` method extracts the next `n` bits from the highest order remaining bits in
the stream. This is done by shifting the current bits to the right by an appropriate amount and then applying a
mask to isolate the required bits.
3. **Escaping Values**: Using the extracted bits, we simulate escaping values where each escape operation depends
on reading subsequent bits if certain conditions are met (e.g., reading more bits if the current value is at its
maximum for the given bit count).

### Solution Code
```rust
struct BitStream {
    current_bits: u32,
    num_bits_left: usize,
}

impl BitStream {
    fn read_bits(&mut self, n: usize) -> u32 {
        assert!(n <= self.num_bits_left);
        let shift = self.num_bits_left - n;
        let bits = (self.current_bits >> shift) & ((1 << n) - 1);
        self.current_bits >>= n;
        self.num_bits_left -= n;
        bits
    }
}

fn escaped_value(stream: &mut BitStream, n1: usize, n2: usize, n3: usize) -> u32 {
    let mut value = stream.read_bits(n1);
    if value == (1 << n1) - 1 {
        value += stream.read_bits(n2);
        if value >= (1 << n1) && (value - (1 << n1)) == ((1 << n2) - 1) {
            value += stream.read_bits(n3);
        }
    }
    value
}

fn main() {
    // Test case where all escapes are triggered
    let mut stream = BitStream {
        current_bits: 0b1111_0111_0011,
        num_bits_left: 9,
    };
    let result = escaped_value(&mut stream, 4, 3, 2);
    println!("Escaped value (all escapes): {}", result); // Should print 25

    // Test case where first escape is triggered but not the second
    let mut stream = BitStream {
        current_bits: 0b1111_0110_0011,
        num_bits_left: 9,
    };
    let result = escaped_value(&mut stream, 4, 3, 2);
    println!("Escaped value (only first escape): {}", result); // Should print 23

    // Test case where no escapes are triggered
    let mut stream = BitStream {
        current_bits: 0b1110_0110_0011,
        num_bits_left: 9,
    };
    let result = escaped_value(&mut stream, 4, 3, 2);
    println!("Escaped value (no escapes): {}", result); // Should print 14
}
```

### Explanation
1. **BitStream Struct**: The `BitStream` struct keeps track of the current bits and how many are left. This allows
us to efficiently manage the state as we read bits.
2. **Reading Bits**: The `read_bits` method extracts the required number of bits by shifting and masking, updating
the stream's state to reflect the consumed bits.
3. **Escaping Values**: The `escaped_value` function reads bits in stages. If a value is at its maximum for the
current bit count, it reads additional bits (potentially multiple times) to form an escaped value, which allows
encoding larger numbers with a variable number of bits.

This approach efficiently handles variable-length encodings by leveraging bitwise operations and state management
within the bitstream.

total duration:       24m23.4179585s
load duration:        14.9301ms
prompt eval count:    165 token(s)
prompt eval duration: 2.51s
prompt eval rate:     65.74 tokens/s
eval count:           3170 token(s)
eval duration:        24m20.892s
eval rate:            2.17 tokens/s