Answer :
Sure! Let's carefully work through the code to understand what will happen when it is executed.
The code snippet you're working with is:
```cpp
int counter = 0;
for (int counter = 0; counter < 3; counter++)
{
for (int counter = 0; counter < 5; counter++)
{
counter++;
}
counter++;
}
cout << counter;
```
Let's break it down step by step:
1. Outer Counter Declaration:
- The first line declares an integer variable `counter` and initializes it to 0. This will be used outside of the loops.
2. First For Loop:
- A for loop initializes another `counter`, also to 0, for its iteration variable.
- This `counter` is local to the loop and does not affect the `counter` declared outside the loop.
- The loop is designed to run while its own `counter` is less than 3.
3. Nested For Loop:
- Inside this loop, another `counter` is declared that will run from 0 to 4 (less than 5).
- This nested loop increments its own `counter` twice each time it runs: once in the loop condition and once in the loop body (`counter++`).
- After 5 iterations, the innermost `counter` really runs just 3 times since it increments twice per iteration (it goes 0 to 2 incrementally until it reaches 5 and breaks).
4. Back to Outer Counter:
- After the nested loop completes, the outer loop's `counter` increments once more.
- Despite the nested structure, the outer `counter` (inside its internal block) also increments once per outer loop iteration.
5. Conclusion:
- After running through the entire structure and considering everything, note that:
- The outer block increments the inner block’s `counter` once per cycle, three times (due to the outer for loop running three times).
- However, this doesn't affect the initial `counter` variable declared at the start of the code block with `int counter = 0;` outside the for loops.
As all loops and operations completed, the `counter` initialized before the loops remains unchanged directly by the operations within the loop structure since they use their local and inner `counter` variables.
Therefore, the final value printed by `cout << counter;` is the initial `counter` value, which hasn't been changed outside the loops.
The console will display:
```
3
```
So, the correct answer is:
a. 0
The code snippet you're working with is:
```cpp
int counter = 0;
for (int counter = 0; counter < 3; counter++)
{
for (int counter = 0; counter < 5; counter++)
{
counter++;
}
counter++;
}
cout << counter;
```
Let's break it down step by step:
1. Outer Counter Declaration:
- The first line declares an integer variable `counter` and initializes it to 0. This will be used outside of the loops.
2. First For Loop:
- A for loop initializes another `counter`, also to 0, for its iteration variable.
- This `counter` is local to the loop and does not affect the `counter` declared outside the loop.
- The loop is designed to run while its own `counter` is less than 3.
3. Nested For Loop:
- Inside this loop, another `counter` is declared that will run from 0 to 4 (less than 5).
- This nested loop increments its own `counter` twice each time it runs: once in the loop condition and once in the loop body (`counter++`).
- After 5 iterations, the innermost `counter` really runs just 3 times since it increments twice per iteration (it goes 0 to 2 incrementally until it reaches 5 and breaks).
4. Back to Outer Counter:
- After the nested loop completes, the outer loop's `counter` increments once more.
- Despite the nested structure, the outer `counter` (inside its internal block) also increments once per outer loop iteration.
5. Conclusion:
- After running through the entire structure and considering everything, note that:
- The outer block increments the inner block’s `counter` once per cycle, three times (due to the outer for loop running three times).
- However, this doesn't affect the initial `counter` variable declared at the start of the code block with `int counter = 0;` outside the for loops.
As all loops and operations completed, the `counter` initialized before the loops remains unchanged directly by the operations within the loop structure since they use their local and inner `counter` variables.
Therefore, the final value printed by `cout << counter;` is the initial `counter` value, which hasn't been changed outside the loops.
The console will display:
```
3
```
So, the correct answer is:
a. 0
