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sizeof array problem

Here's the code:


#include <stdafx.h>
#include <iostream>
#include <iomanip>
using namespace std ;

void outputSize( int vals [] )
{
      cout << "In outputSize():  {" << sizeof(vals)/sizeof(vals[0]) << "}" << endl ;
}

int main()
{
      int vals [] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 } ;

      // output array size
      cout << "In main():  {" << sizeof(vals)/sizeof(vals[0]) << "}" << endl ;
      
      // pass array to function, where it's size should be outputted
      outputSize( vals ) ;

      return 0 ;
}


And here's the output:

  In main():  {10}
  In outputSize():  {1}


For some reason, it would seem that 'sizeof(vals)' in the outputSize() function is getting the sizeof a single element within the array, rather than the total amount of memory occupied by the entire array...

- Why is this?
- How can I solve it?


Thanks

btw: Am using VS 2005 Pro now
0
InteractiveMind
Asked:
InteractiveMind
3 Solutions
 
AlexFMCommented:
void outputSize( int vals [] )

The only thing that compiler knows anout vals[] parameter is that it is pointerts int. sizeof(int*) = 4. Compiler doesn't know at compile time what array is passed to outputSize function.
sizeof operator is executed by compiler at compile time, it is replaced by constant and inserted to executable code. So, in the main function, compiler knows that vals is array of 10 integers, and in outputSize function, vals is just pointer to integer.

Solution: add size parameter to the function:

void outputSize( int vals [], int size );
...

outputSize( vals, sizeof(vals)/sizeof(int) ) ;
0
 
grg99Commented:
In C, unlike Pascal, Ada, Java, Delphi, FORTRAN 90, COBOL, SNOBOL, APL, PL/I, PL/M, BASIC+, JavaScript, Simula, REXX, Spitbol, MAtLab, Perl, and a few dozen other languages, what gets passed is just the raw address of the array, and nothing else.  So it's really hard to divine the size of the array, much less the size of each element, when neither bit of info is available.  Well, that's not quite true,  if you defined the receiving array of a specific type, then you can get the sizeof an element, but that's of no use if you want to inquire about the size of arbitrary parameters. Or you can poke through the stabs debug info, if the program was compiled with debug info turned on, AND the variable is a static, global, or local variable, (not a malloc'ed variable), and you can figure out the info.  There's also the RTTi stuff, but that's also a can of worms.

So in general, you can't.

You can FAKE it by using a macro in some cases:

#define CallAndPassSize(Func,Array)       Func( (void* )&Array, (sizeof(Array)/sizeof(Array[0])) )


You use it like this:

 CallAndPassSize(   MyFunct, A )

and receive it like this:

void MyFunct( void * ArbitraryArray, unsigned int ArrayLength ){ .... }



0
 
AxterCommented:
You can get the desired functionallity by making your function a template function.
0
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AxterCommented:
>>You can FAKE it by using a macro in some cases:

I recommed using a template over a macro, because macros enter all namespaces, where-as a template stays within it's own namespace.
0
 
grg99Commented:
OOPS, I goofed, try this to get both pieces of info:

#define CallAndPassSize(Func,Array)       Func( (void* )&Array, (sizeof(Array)/sizeof(Array[0])), (sizeof(Array[0])) )


You use it like this:

 CallAndPassSize(   MyFunct, A )

and receive it like this:

void MyFunct( void * ArbitraryArray, unsigned int ArrayLength, unsigned int ArrayElements ){ .... }


0
 
AxterCommented:
Example code:

template < class T > 
void outputSize( T& vals)
{
      cout << "In outputSize():  {" << sizeof(vals)/sizeof(vals[0]) << "}" << endl ;
}

int main()
{
      int vals [] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 } ;

      // output array size
      cout << "In main():  {" << sizeof(vals)/sizeof(vals[0]) << "}" << endl ;

      // pass array to function, where it's size should be outputted
      outputSize( vals ) ;

      return 0 ;
}
0
 
InteractiveMindAuthor Commented:
Thanks very much all.

Being a Java geek myself, and having limited C++ experience, I had no idea about this -- but you've all done a great job explaining it. It's much clearer now.

:)
0

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