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  1. What will be the output of the program ?

    #include 

 int main()
 {
      char str = "FRESHERGATE";   
      printf("%s\n", str);  
      return 0; 
 }

    2. A.

    Error
    B.

    FresherGATE
    C.

    Base address of str
    D.

    No output

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    Answer : Option A

    Explanation :

    The line char str = "FresherGATE"; generates "Non portable pointer conversion" error.

    To eliminate the error, we have to change the above line to

    char *str = "FresherGATE"; (or) char str[] = "FresherGATE";

    Then it prints "FresherGATE".

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