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.

Wouldn't sizeof(char)*20 work ?

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

Hmm....then you might want to pass it as a parameter, that's my take on it.

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:

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function(char myarray[])

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This will do:

Code:

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sizeof(array) / sizeof(array[0])

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I'd suggest something like this:

Code:

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#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 */
}

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[EDIT]
Edited to show elements allocated, instead of memory allocated.

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.

Ahh yes, I read size as memory space, prog-bman's method will show you how many elements are allocated.

Quote:

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Originally Posted by prog-bman

This will do:

Code:

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sizeof(array) / sizeof(array[0])

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No it won't. Read the OP.

Quote:

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Now I pass the array to a function:

function(myarray);

How can I find out inside the function the size allocated to the array?

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Code:

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#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
*/

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Arrays degrade to pointers to their first element when passed to functions. No cigar for you.


Quzah.

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.

>>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?

Quote:

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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.

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That is correct.
http://gcc.gnu.org/ml/gcc-bugs/1998-09/msg00834.html :

Quote:

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`sizeof(char) , sizeof(signed char) and sizeof(unsigned char) are 1'

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http://www.ussg.iu.edu/hypermail/lin...01.2/0969.html :

Quote:

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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.

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http://www.parashift.com/c++-faq-lit....html#faq-26.1 :

Quote:

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sizeof(char) is always exactly 1. No exceptions, ever.

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