finding size of empty char array

This is a discussion on finding size of empty char array within the C Programming forums, part of the General Programming Boards category; Lets say I create a char array: char myarrya[20]; Now I pass the array to a function: function(myarray); How can ...

  1. #1
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    finding size of empty char array

    Lets say I create a char array:
    char myarrya[20];

    Now I pass the array to a function:

    function(myarray);

    How can I find out inside the function the size allocated to the array? Assuming no string is placed in array yet. Strlen() and sizeof() only give the size of a string placed in the array.

  2. #2
    Fear the Reaper...
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    Wouldn't sizeof(char)*20 work ?
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  3. #3
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    The size of the array entered into the function will be undetermined until run time:
    myarray[n]
    function(myarray)

    The function needs to know the value of n

  4. #4
    Fear the Reaper...
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    Hmm....then you might want to pass it as a parameter, that's my take on it.
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  5. #5
    Registered User whiteflags's Avatar
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    You're sure that you need to know the size of an empty character array? I'm sure that character arrays are capable of being passed without a parameter for the size. You just need to change the function a bit.
    Code:
    function(char myarray[])

  6. #6
    Sweet
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    This will do:
    Code:
    sizeof(array) / sizeof(array[0])
    Woop?

  7. #7
    Awesomefaceradcore bivhitscar's Avatar
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    I'd suggest something like this:

    Code:
    #include <stdio.h>
    #include <stdlib.h>
    
    void test( char arr[], int size );
    
    int main(int argc, char *argv[])
    {
        
        char array[20];
        
        test(array, sizeof(array) / sizeof(array[0]) );
        
        system("PAUSE");	
        return 0;
    }
    
    void test( char arr[], int size )
    {
         /* code */
    }

    [EDIT]
    Edited to show elements allocated, instead of memory allocated.
    Last edited by bivhitscar; 05-28-2006 at 09:38 PM.
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  8. #8
    Sweet
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    That code will only work if the size of a char is only one which I am not sure that is guranteed to be, I think so though. Anyways if you want to use it for more than chars you will have to use my method.
    Woop?

  9. #9
    Awesomefaceradcore bivhitscar's Avatar
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    Ahh yes, I read size as memory space, prog-bman's method will show you how many elements are allocated.
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  10. #10
    ATH0 quzah's Avatar
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    Quote Originally Posted by prog-bman
    This will do:
    Code:
    sizeof(array) / sizeof(array[0])
    No it won't. Read the OP.
    Now I pass the array to a function:

    function(myarray);

    How can I find out inside the function the size allocated to the array?
    Code:
    #include<stdio.h>
    
    void foo( int array[] )
    {
        printf( "sizeof( array ) is %d\n", sizeof( array ) );
        printf( "sizeof( array[ 0 ] ) is %d\n", sizeof( array[ 0 ] ) );
        printf( "sizeof( array ) / sizeof( array[ 0 ] ) is %d\n",
            sizeof( array ) / sizeof( array[ 0 ] ) );
    }
    
    void bar( char array[] )
    {
        printf( "sizeof( array ) is %d\n", sizeof( array ) );
        printf( "sizeof( array[ 0 ] ) is %d\n", sizeof( array[ 0 ] ) );
        printf( "sizeof( array ) / sizeof( array[ 0 ] ) is %d\n",
            sizeof( array ) / sizeof( array[ 0 ] ) );
    }
    
    int main( void )
    {
        int array[ 10 ];
        char barray[ 10 ];
    
        foo( array );
        bar( barray );
    
        return 0;
    }
    
    /*
    sizeof( array ) is 4
    sizeof( array[ 0 ] ) is 4
    sizeof( array ) / sizeof( array[ 0 ] ) is 1
    sizeof( array ) is 4
    sizeof( array[ 0 ] ) is 1
    sizeof( array ) / sizeof( array[ 0 ] ) is 4
    */
    Arrays degrade to pointers to their first element when passed to functions. No cigar for you.


    Quzah.
    Last edited by quzah; 05-28-2006 at 11:24 PM. Reason: Added bar.
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  11. #11
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    I don't want to pass the size of array as arguement because I want the function to do error checking on the passed array, to be certain that it doesn't overwrite array boundries. The size of array n is undetermined before compile time and needs to be known.

    function(char * p)

    function(char p[])

    The problem with sizeof here is that it probably measures the size of the pointer. At any rate sizeof(p) and strlen always return 4, regardless of the actual size of the array.

    The size of the array must be stored somewhere, and a function recieving the array as an arguement ought to be able to access it.

  12. #12
    The Richness... Richie T's Avatar
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    >>The size of the array must be stored somewhere

    No it doesn't - arrays do not have built in bounds checking

    >>The size of array n is undetermined before compile time

    So you are using a dynamic memory allocation i guess - how
    does that know/calculate the size of the array?

    Just pass it as a parameter. I suppose your function could read
    the array until the string termination character - but does your
    program guarantee that the char array passed is an actual
    string?
    No No's:
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  13. #13
    Frequently Quite Prolix dwks's Avatar
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    That code will only work if the size of a char is only one which I am not sure that is guranteed to be, I think so though. Anyways if you want to use it for more than chars you will have to use my method.
    That is correct.
    http://gcc.gnu.org/ml/gcc-bugs/1998-09/msg00834.html :
    `sizeof(char) , sizeof(signed char) and sizeof(unsigned char) are 1'
    http://www.ussg.iu.edu/hypermail/lin...01.2/0969.html :
    A7.4.8 Sizeof Operator
    The sizeof operator yields the number of bytes required to store an
    object of the type of its operand. [...] When sizeof is applied to a
    char, the result is 1. [...] The operator may not be applied to [...]
    a bit-field.
    http://www.parashift.com/c++-faq-lit....html#faq-26.1 :
    sizeof(char) is always exactly 1. No exceptions, ever.
    dwk

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