Hi,
I have 8 char stored in a char array.
char char_buff[65536];
char_buff[x]=68;
char_buff[x+1]=59;
...
char_buff[x+7]=-56;
I would like to combine those 8 char into one double type variable. I thought about using bit manipulation and temporarily store all the char into a long int and then somehow convert that to a double.
Can anyone please point me in the right direction?
-56 is not a valid ascii character code
http://www.asciitable.com/
look up strtod for double conversion
Last edited by jimtuv; 06-14-2010 at 08:56 AM.
well for some reason when I store the binary numbers in my char array and do a "quickwatch" of the array index I get -27 which, according to the Visual Studio, corresponds to char å.
it was just an example, extracted from exactly what I'm seeing.
I don't think my question clear, so let me clarify it:
I am storing binary code into the char array from a double in the first place. So I really don't care about character representation.
I am NOT doing this: 45.6777 and store it as characters then convert it to double.
What I am doing is this:
1- I have a double say -.1378
2- All 8 bytes of this double are stored in the char array, each byte is stored into an index
3- What I need is to read those 8 bytes from the array back into a double
Last edited by doia; 06-14-2010 at 09:06 AM.
Are you certain you can allocate that much memory on the stack?
1. Get rid of gets(). Never ever ever use it again. Replace it with fgets() and use that instead.
2. Get rid of void main and replace it with int main(void) and return 0 at the end of the function.
3. Get rid of conio.h and other antiquated DOS crap headers.
4. Don't cast the return value of malloc, even if you always always always make sure that stdlib.h is included.
hasn't given me any issues so far.Originally Posted by claudiu
Probably union ?
Code:union foo { char x[8]; double real; };
Can you please be more specific?
The values are stored in a char array. Are you suggesting I transfer the 8 bytes from the char array into the union and then just call the value for double? Isn't there a chance my double will be corrupted?
converting byte array to double - c - Stack Overflow
Not sure it'll help.
How aboutYou only want to convert 8 bytes, right? People are worried about an array of size 65536 bytes.Code:union foo { unsigned char x[65536]; unsigned long long y[8192]; double z[8192]; };
I'm not certain you really want the floating point representation.
as you move bytes of double to char array - move them back in the same way
All problems in computer science can be solved by another level of indirection,
except for the problem of too many layers of indirection.
– David J. Wheeler
Thanks nonoob and Bayint.
Not really, I need to convert a lot more than just 8 bytes, I am going to have a large amount of values stored into this array. The values stored are going to be in different formats too (double, uint, int, float, etc.), I have a map for this array telling me where each value starts and ends and what data type it should be converted to. I just wanted to get the general idea how to do it.
The array values comes from a buffer that is made available to my program from an independent process, so I don't know how they were stored there. I wouldn't ask the question if I already had the answer.as you move bytes of double to char array - move them back in the same way
Last edited by doia; 06-14-2010 at 04:01 PM.
You already have the answer. Use a union.Quzah.Code:union foo { char c[sizeof(double)]; double d; } bar; char buf[ BUFSIZ ]; char *p = buf; ... for( this; that; theotherthing ) { memcpy( bar.c, p, sizeof( double ) ); p += sizeof( double ); printf( "Let's pretend a double can be represented correctly! %f\n", bar.d ); }
Hope is the first step on the road to disappointment.
The easiest is probably to use memcpy(), or make a function that takes a void pointer as argument, then just assign the void pointer to a double pointer, dereference and return. Or do a cast directly, more messy but it works as well.
okay I used the union from quzah's code and it works!
Thanks to everyone
Isa
