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What will be the output of the following code?
#include <iostream>
using namespace std;
using namespace std;
int main() {
int a = 5;
auto lambda = [&a]() { return a; };
a = 10;
cout << lambda();
return 0;
}
int a = 5;
auto lambda = [&a]() { return a; };
a = 10;
cout << lambda();
return 0;
}
- a)5
- b)10
- c)Compiler Error
- d)Undefined Behavior
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
What will be the output of the following code?#include <iostream>...
Explanation:
Lambda Function Capture:
- In the given code, a lambda function is created which captures the variable 'a' by reference using the lambda capture syntax '[&a]'.
- This means that the lambda function will always refer to the current value of 'a' when it is called.
Changing Value of 'a':
- Initially, the value of 'a' is set to 5.
- The lambda function is created before changing the value of 'a' to 10.
- Even though the value of 'a' is changed to 10 after creating the lambda function, the lambda function still captures the reference to the original value of 'a' (which was 5).
Output:
- When the lambda function is called using 'lambda()', it returns the value of 'a' that was captured when the lambda function was created.
- So, the output of the code will be 5, as the lambda function returns the original value of 'a' (which is 5) and not the updated value (which is 10).
Therefore, the correct output of the code is:
a) 5
Lambda Function Capture:
- In the given code, a lambda function is created which captures the variable 'a' by reference using the lambda capture syntax '[&a]'.
- This means that the lambda function will always refer to the current value of 'a' when it is called.
Changing Value of 'a':
- Initially, the value of 'a' is set to 5.
- The lambda function is created before changing the value of 'a' to 10.
- Even though the value of 'a' is changed to 10 after creating the lambda function, the lambda function still captures the reference to the original value of 'a' (which was 5).
Output:
- When the lambda function is called using 'lambda()', it returns the value of 'a' that was captured when the lambda function was created.
- So, the output of the code will be 5, as the lambda function returns the original value of 'a' (which is 5) and not the updated value (which is 10).
Therefore, the correct output of the code is:
a) 5
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Community Answer
What will be the output of the following code?#include <iostream>...
The lambda expression captures the variable a by reference. So, when a is modified to 10, the lambda still refers to the updated value.
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