What will be output of the following c code?
#include<stdio.h>
int main()
{
volatile int a=11;
printf("%d",a);
return 0;
}
a)
11
b)
Garbage
c)
-2
d)
Cannot Predict
Correct answer is option 'D'. Can you explain this answer?

Question

What will be output of the following c code?

#include<stdio.h>
int main()
{
volatile int a=11;
printf("%d",a);
return 0;
}

a)
11
b)
Garbage
c)
-2
d)
Cannot Predict

Correct answer is option 'D'. Can you explain this answer?

Answer

- **Explanation:**

- **Declaration of a Variable:**
- In the given code, a variable 'a' is declared as a volatile integer with an initial value of 11.

- **Printing the Value:**
- The value of variable 'a' is then printed using the printf() function with the format specifier %d.

- **Volatile Keyword:**
- The volatile keyword is used to indicate that a variable may be changed by external factors outside the program's control.
- In this case, since 'a' is declared as volatile, the value of 'a' could potentially change during program execution due to external factors.

- **Output Prediction:**
- Since the value of 'a' is printed immediately after its declaration and initialization, it is expected to output 11 in most cases.
- However, due to the volatile nature of the variable, the value of 'a' could potentially change before it is printed, leading to unpredictable output.

- **Conclusion:**
- The output of the code cannot be accurately predicted due to the volatile nature of the variable 'a'. It may print 11 or some other unpredictable value depending on external factors.

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