[solved]Can anyone explain why the output is "Try Again"

Question

Check the code in this post --
https://www.sololearn.com/post/281725/?ref=app

Answer 1

If macro was not defined, then ! operator with nonexisting macro name is evaluated to true. it is not an error for #if directive (not to be confused with if statement for which it will be an error).

Answer 2

Akshay Jain Do not confuse preprocessor directives with conditional if statement. Revise the course again if needed or read on it on the internet. Also see the example that I have shared.

Answer 3

#if is preprocessor directive which uses to make conditional compilation.
If condition in #if directive evaluates to true then the subsequent code (till #else) will be included in the compilation process. If it evaluates to false then code followed after #else directive (till #endif) will be included in the compilation process.

Condition of #if directive is !SOLO. It is evaluated to true because there is no any difinition of SOLO (! means not). So, after prepocessing stage the compiler will get follow:

int main(){
printf("Try again");
return 0;
}

As result "Try again" will be printed.

Answer 4

SOLO is not defined, so #if !SOLO will return true and it prints 'Try Again'.

If I define SOLO then it will result in false and the else part is executed.
https://code.sololearn.com/chC8zcRItFd9/?ref=app

Answer 5

Avinesh As SOLO is not defined so it should show error...how it can return True???
If SOLO is defined to be 0 then only it should show "Try Again".

Answer 6

Got It...thanks for answering andriy kan Avinesh