Thread: Constant value not known

  1. #1
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    Constant value not known

    Hi,

    When compiling any of these codes, I get the message: "constant value not known" which seems weird to me... at least for the second one I don't understand why.

    Thanks for any help.

    F.
    Code:
    ////////////////////////////////////////////////////////////// V.1
    
    const int size_input_bits = 12382;
    const int size_input_ints = ceil((float)size_input_bits/32);
    int h_input [size_input_ints];
    
    ////////////////////////////////////////////////////////////// V.2
    
    #define size_input_bits 12382;
    ...........
    const int size_input_ints = ceil((float)size_input_bits/32);
    int h_input [size_input_ints];
    Last edited by Salem; 06-12-2007 at 11:05 PM. Reason: Added [code] [/code] tags - learn to use them

  2. #2
    Officially An Architect brewbuck's Avatar
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    You can't call a function (i.e. ceil()) in the initializer for a global variable, much less a const one.

  3. #3
    & the hat of GPL slaying Thantos's Avatar
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    Why not just assign 387 to size_input_ints?

  4. #4
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    Find my comments Inline.

    Code:
    ////////////////////////////////////////////////////////////// V.1
    
    const int size_input_bits = 12382;
    const int size_input_ints = ceil((float)size_input_bits/32);
    int h_input [size_input_ints];
    
    ////////////////////////////////////////////////////////////// V.2
    
    #define size_input_bits 12382;
    U dont need to put ' ; ' when defining a variable.
    ...........
    const int size_input_ints = ceil((float)size_input_bits/32);
    --I think initializers of global variables cannot contain function
    calls in C. You can switch to C++, that introduces the ability to do dynamic initialization of global variables.
    
    int h_input [size_input_ints];
    S_ccess is waiting for u. Go Ahead, put u there.

  5. #5
    and the hat of int overfl Salem's Avatar
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    > #define size_input_bits 12382;
    #defines don't normally have trailing ;

    > int h_input [size_input_ints];
    Standard (C89) C doesn't have variable length arrays of any persuasion, even if you declare the size as an int constant. As far as C is concerned, it is a dynamic run-time expression (which always yields the same value).

    The C and C++ ideas about const are very different.

    Code:
    #define size_input_bits 12382
    ...........
    int h_input [ size_input_bits / 32 ];
    This should work in C.
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.

  6. #6
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    > The C and C++ ideas about const are very different.
    I tested and verified that in C90, using GCC with -W -Wall -ansi -pedantic, declaring the size of a static array as a const int doesn't work, but in C++, it does (of course, C99 doesn't need the const at all). Why is this?

  7. #7
    and the hat of int overfl Salem's Avatar
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    http://david.tribble.com/text/cdiffs...-const-linkage
    In C++, the compiler is free to substitute the known value at compile time, thus achieving what #define does in C.

    In C, const just means "warn me if I try to modify this variable". For example, this is valid C.
    const volatile int clock;
    A read-only variable which is constantly changing.
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.

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