Problem 26 Consider the following C++ program. void a(int &x, int y) { x = y; y = x + 1; } int b(int &x, int y) { y = x + 1; x = y; return y; } void c(int x[], int y) { if (y1) return; cout << x[y- 1] - y; c(x, y - 1); } int main() { } 3 (a) What is the ouput at line (a)? Answer: 22 (b) What is the ouput at line (b)? Answer: int x [6] = {1, 2, 3, 4, 5, 6}; int y [3] = {5, 4, 3}; cout << y[x [1]] % x[y[1]] << endl; a(x [0], x[1]); 4 cout << x[0] << x[1] << endl; cout <

C++ Programming: From Problem Analysis to Program Design
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Chapter15: Recursion
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Please explain the answers and how they were gotten step by step. The programming language is C + +.

### Problem 26
Consider the following C++ program.

```cpp
void a(int &x, int y) {
    x = y;
    y = x + 1;
}

int b(int &x, int y) {
    y = x + 1;
    x = y;
    return y;
}

void c(int x[], int y) {
    if (y < 1) return;
    cout << x[y - 1] - y;
    c(x, y - 1);
}

int main() {
    int x[6] = {1, 2, 3, 4, 5, 6};
    int y[3] = {5, 4, 3};
    cout << y[x[1]] << x[y[1]] << endl;       // line (a)
    a(x[0], x[1]);
    cout << x[0] << x[1] << endl;             // line (b)
    cout << b(2), x[3] << endl;               // line (c)
    cout << x[2] << x[3] << endl;             // line (d)
    c(y, 3); cout << endl;                    // line (e)
}
```

#### (a) What is the output at line (a)?
**Answer:**  
3

#### (b) What is the output at line (b)?
**Answer:**  
22

#### (c) What is the output at line (c)?
**Answer:**  
4

#### (d) What is the output at line (d)?
**Answer:**  
44

#### (e) What is the output at line (e)?
**Answer:**  
024 

### Explanation:

- **Line (a)**: `cout << y[x[1]] << x[y[1]] << endl;`
  - `x[1]` is 2, so `y[x[1]]` is `y[2]` which is 3.
  - `y[1]` is 4, so `x[y[1]]` is `x[4]` which is 5.
  - Therefore the output is `3`.

- **Line (b)**: `a(x[0], x[1]);`
  - `x[0]` becomes equal to `x[1]` which
Transcribed Image Text:### Problem 26 Consider the following C++ program. ```cpp void a(int &x, int y) { x = y; y = x + 1; } int b(int &x, int y) { y = x + 1; x = y; return y; } void c(int x[], int y) { if (y < 1) return; cout << x[y - 1] - y; c(x, y - 1); } int main() { int x[6] = {1, 2, 3, 4, 5, 6}; int y[3] = {5, 4, 3}; cout << y[x[1]] << x[y[1]] << endl; // line (a) a(x[0], x[1]); cout << x[0] << x[1] << endl; // line (b) cout << b(2), x[3] << endl; // line (c) cout << x[2] << x[3] << endl; // line (d) c(y, 3); cout << endl; // line (e) } ``` #### (a) What is the output at line (a)? **Answer:** 3 #### (b) What is the output at line (b)? **Answer:** 22 #### (c) What is the output at line (c)? **Answer:** 4 #### (d) What is the output at line (d)? **Answer:** 44 #### (e) What is the output at line (e)? **Answer:** 024 ### Explanation: - **Line (a)**: `cout << y[x[1]] << x[y[1]] << endl;` - `x[1]` is 2, so `y[x[1]]` is `y[2]` which is 3. - `y[1]` is 4, so `x[y[1]]` is `x[4]` which is 5. - Therefore the output is `3`. - **Line (b)**: `a(x[0], x[1]);` - `x[0]` becomes equal to `x[1]` which
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