Question

Int ++a=100; .println (++a); what will be the output of the above fraction of code??​

A. 100

B. 101

C. 102

Answer 1

The corrected code would output '101' after fixing the provided syntax error, as the variable 'a' is incremented before being printed.Correct option is b).

Step-by-step explanation:

The question is asking about the output of a code snippet in which a variable 'a' is pre-incremented twice. The code, as provided, contains a syntax error because 'int ++a=100;' is not a valid statement in any programming language. Assuming the intention was to declare an integer variable 'a' and set it to 100 before the print statement, we will correct the syntax to a valid form and then discuss the output. A corrected version of the code might look like this:

int a = 100;
System.out.println(++a);

In this case, the output would be '101' because the pre-increment operator (++a) increases 'a' by 1 before it is used in the println statement.

The code int ++a=100; .println (++a); contains a syntax error. The correct syntax for incrementing a variable and printing it in Java would be:

int a = 100;

++a;

System.out.println(a);

With this correct syntax, the output of the code would be 101, which corresponds to option B.