Thread: not an lvalue

  1. #1
    Registered User
    Join Date
    Aug 2004

    not an lvalue

    k obviously i am a noob, but i have observed the basic courtasies for posting. i searched for it in posts and did not find an answer, anyways...

    im trying to write a code to output (part of) the fib sequence, but i keep running into the same problem.
    #include <iostream.h>
    int fib1[50]; 
    int a=0;  
    int b=1;
    int c=2;
    int main()
     fib1[a]=1; //sets fib1[whatever # a is] to start out at 1
     for(int x=0;x<50;x++) //set for 50 repetitions
      fib1[a]+fib1[b]=fib1[c];   //value in fib1[a]+value in fib1[b]=value in fib1[b]. but for somereason my compiler is saying its not an lvalue
      a++; //increments a
      cout<<fib1[a]<<", ";   //outputs #s
    return 0;
    i have tried putting the values from the array into int variables, but i had the same problem. same thing happended when i used pointers.

  2. #2
    Hardware Engineer
    Join Date
    Sep 2001

    l-value means left-value...

    The value on the left side of the = sign, is the "unknown".

    x = 5; // This is OK
    5 = X; // This is NOT OK

    fib1[c] = fib1[a] + fib1[b]; // Try This

    In C++ (and in programming in general), the equal-sign does not indicate an equation. It is the assignment operator. It assigns the value of the statement on the right to the variable on the left.

  3. #3
    Crazy Fool Perspective's Avatar
    Join Date
    Jan 2003
    >> fib1[a]+fib1[b]=fib1[c];

    you have to put the thing being assigned on the left (lvalue)

    fib1[c] = fib1[a]+fib1[b];

    edit: d'oh, beaten by doug

  4. #4
    Registered User
    Join Date
    Aug 2004
    ahhh, i see, i should have know that. thanks so much for the help

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