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Alessio1989

read double from, inconsistent results with some values

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I was just playing with visual studio 2013 rtm and compile some old demos to see if there were some changes to the standard library (like "max()" and "min()" functions moved to <algorithm>....)

 

And I had a little surprise with a sample code for write and read binary files with the standard library: using the following double values (111.111, 222.222, 333.333, 444.444), when I read back the data I have completely different data (like  -6.27744e+066 -6.27744e+066 -6.27744e+066 -6.27744e+066). It is also strange that the values are the same in debug mode in 32 bit and 64 bit configuration, but they change in release mode in 64 bit and 32 bit configurations. Changing the float precision model ( "precise" 80 bit, strict and fast) doesn't solve the problem.

 

Note also the sample works well if I change the double data to other values... Maybe that is due to some "nice" behaviour of the IEEE standard? ... I don't remember nothing about that in the floating point model, to me that seems just a bug (code or compiler... dunno XD )

 

Here is the code:



#include <fstream>
#include <iostream>

int main()
{
	double a[ 4 ] = { 111.111, 222.222, 333.333, 444.444 };
	double* i = new( double[ 4 ] );

	std::fstream file( "file", std::ios::out | std::ios::binary );

	file.write( ( char* )&a, sizeof( a ) );
	file.close();


	file.open( "file", std::ios::in | std::ios::binary );
	file.read( ( char* )&*i, sizeof ( double )* 4 );
	file.close( );

	for( int j = 0; j < 4; ++j )
	{
		std::cout << i[ j ] << " ";
	}
	std::cout << std::endl;

	return( 0 );
}
Edited by Alessio1989

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yes, that's a copy-typo, but that's not the problem. anyway thanky for make me note that.

Edited by Alessio1989

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    file.write( ( char* )&a, sizeof( a ) );

 

This is incorrect. You want:

 

(char*)a

 

a is already a pointer so taking the address is bobbins.

 

EDIT: I'm surprised (char*)&a even compiles?

Edited by Paradigm Shifter

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    file.write( ( char* )&a, sizeof( a ) );

 

This is incorrect. You want:

 

(char*)a

 

a is already a pointer so taking the address is bobbins.

 

EDIT: I'm surprised (char*)&a even compiles?

 

here the reference is redundant since they are arrays, but that's not the problem. only with that values I got incorrect read-back from the file, with any other value, the read is correct.

Edited by Alessio1989

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    file.write( ( char* )&a, sizeof( a ) );

 

This is incorrect. You want:

 

(char*)a

 

a is already a pointer so taking the address is bobbins.

 

EDIT: I'm surprised (char*)&a even compiles?

No, a is an array, not a pointer, so &a is a pointer to the array. This is the same as just a in this context when the array decays into a pointer to its first element, but &a is not wrong.

Edited by Brother Bob

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Alessio, you still have the third incorrect parameter in your call to open(). You need to OR the last two parameters there as well.

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Well that's silly. a, &a and &a[0] are all the same? &a is not like the others and should really be a compiler error, if C was a sensible language (which it isn't, of course). a and &a[0] are of course the same. &a should be of type char** (since it is the address of a decayed array).

 

I guess I don't know the correct answer since I'd never use the address of an array. Taking the address of an array suggests to me it could be changed!?

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They all have the same value, but they all have different types.

  1. a is of type double[4].
  2. &a is of type double (*)[4].
  3. &a[0] is of type double *.

And no, &a should not have the type double ** since a is not a pointer. See point 2, a is of type double[4], so &a is a pointer to a double[4], thus a double (*)[4].

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-6.27744e+066

That looks a lot like accidental/uninitialised memory reinterpreted as a double - the upper bytes are 0xCDCDCDCE...

 

Are you sure you are running the exact same program you have reproduced here? Apart from the missing bitwise-OR already mentioned, there isn't anything particularly wrong with your code snippet, and a compiler bug is beyond unlikely.

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