Undeclared identifier uppur, please help me!

This is a discussion on Undeclared identifier uppur, please help me! within the C Programming forums, part of the General Programming Boards category; Hello, Do you know what is wrong with this? Code: /* print fahr-celc table for fahr = 0, 10, 20, ...

  1. #1
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    Question Undeclared identifier uppur, please help me!

    Hello,

    Do you know what is wrong with this?
    Code:
    /* print fahr-celc table for fahr = 0, 10, 20, ..., 100 */
    
    
    int main (void)
    {
    int celc, fahr;
    int lower, upper, step;
    
    
    lower = 0;        /* lower limit of temperature scale */
    upper = 100;    /* upper limit */
    step = 10;        /* step size */
    
    
    fahr = lower;
    while (fahr <= uppur);    
        {
        celc = 5*(fahr-32)/9;
        printf("%d\t%d\n", fahr, celc);
        fahr = fahr + step;
        }
    }
    When I press execute, pelles c says:
    Building main.obj.
    D:\Documenten\Mijn_documenten\c programma's van mij\derde project\main.c(37): error #2048: Undeclared identifier 'uppur'.
    *** Error code: 1 ***
    Done.

    I don't know what is wrong with this.

    Please help me.

  2. #2
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    Check your spelling. Where do you define a variable with the name uppur?

    Jim

  3. #3
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    Check your spelling:

    Code:
    int lower, upper, step;
    
    // ...
    
    while (fahr <= uppur);
    When you fix this and see that the program appears to do nothing, think about whether or not you really want that semi-colon at the end of your "while()".

  4. #4
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    Also the following will probably not produce your desired results:
    Code:
    celc = 5*(fahr-32)/9;
    You are using integer math, there are no fractions in integer math. So something like 1/2 will yield zero, with the fractional part discarded.

    Jim

  5. #5
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    Quote Originally Posted by jimblumberg View Post
    Also the following will probably not produce your desired results:
    Code:
    celc = 5*(fahr-32)/9;
    You are using integer math, there are no fractions in integer math. So something like 1/2 will yield zero, with the fractional part discarded.

    Jim
    The other issue is that C uses BODMAS. So divisions happen before multiplications, and the expression becomes 5*((fahr-32)/9). fahr is rounded down to the lowest multiple of 5. If you write (5*(fahr-32))/9 instead, you will get what is probably an acceptably accurate integer result.
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  6. #6
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    Quote Originally Posted by Malcolm McLean
    The other issue is that C uses BODMAS. So divisions happen before multiplications
    No, C has precedence rules akin to BODMAS, but not quite the same. In particular, the / and * operators have the same precedence. Furthermore, precedence determines grouping, not necessarily the order of evaluation.

    Quote Originally Posted by Malcolm McLean
    the expression becomes 5*((fahr-32)/9).
    No, the grammar is such that * and / are left associative. Hence the expression 5*(fahr-32)/9 is equivalent to (5*(fahr-32))/9 rather than 5*((fahr-32)/9).
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
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