#include int main(){ int i = 3; printf("%d", (++i)++); return 0;}What is the output of the above program?a)3b)4c)5d)Compile-time errorCorrect answer is option 'D'. Can you explain this answer?

Question

#include <stdio.h>int main(){  int i = 3;  printf("%d", (++i)++);  return 0;}What is the output of the above program?a)3b)4c)5d)Compile-time errorCorrect answer is option 'D'. Can you explain this answer?

Answer

Explanation:

- The code provided will result in a compile-time error.

- In C language, the expression `++i` increments the value of `i` by 1 and returns the updated value of `i`.

- However, the expression `(++i)++` is attempting to increment a value that is not an lvalue (i.e., it cannot be modified).

- In other words, the result of the `++i` expression is not a variable or a memory location that can be modified.

- Therefore, trying to apply the `++` operator to the result of `++i` is invalid and will result in a compile-time error.

- This is because the `++` operator can only be applied to variables or lvalues, not to the result of an expression.

- As a result, the code will not compile and a compile-time error will be generated.

Answer

In C, prefix and postfix operators need l-value to perform operation and return r-value. The expression (++i)++ when executed increments the value of variable i(i is a l-value) and returns r-value. The compiler generates the error(l-value required) when it tries to post-incremeny the value of a r-value.

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