Array inside an 2dimensional array

First of all I would like to apologize if my question has been answered in the past...


I want to create a 30 columns *2 rows array (A[30][2]) and store at every element of it another array.

Practically, I want to store 8 bit binary addresses in a 2dimensional array, however I want these addresses to be stored as arrays and not as integers as I want to be able to read them byte by byte.

Therefore

Code:

int A[30][2];

as declaration is not what I want.

What I want is to store

Code:

B[8]={0,0,0,0,1,1,1,1}, C[8]=..., etc at every element of A[30][2].

If somebody could help me I would be grateful!!!

Thank you very much in advance.

Maybe a 3 dimensional array? Or have I just missed your point?

Originally Posted by mike_g

Maybe a 3 dimensional array? Or have I just missed your point?

Yes, you 're right, I didn' t explain well what I want...

If I declare an

Code:

int A[30][2][8];

array I will have to fill it up like this:

Code:

A[0][0][0] = 0;
A[0][0][1] = 0;
A[0][0][2] = 0;
A[0][0][3] = 0;
A[0][0][4] = 1;
A[0][0][5] = 1;
A[0][0][6] = 1;
A[0][0][7] = 1;

which represents the first byte 11110000 (from MSB to LSB).

I will have to repeat this 30*2 = 60 times.

Could you please tell how I could put in every element of the 30*2 array directly a full byte as an array?
How can I just declare

Code:

B[8] = {0,0,0,0,1,1,1,1};

and then just put it inside A[0][0]?

Thank you again for your time...

Code:

unsigned char array[2][30];
array[0][0] = 0xF0;

?

Use a union and a struct with bitfields:

Code:

union
{
    unsigned char thebyte;
    struct
    {
        unsigned int bit1:1;
        unsigned int bit2:1;
        unsigned int bit3:1;
        unsigned int bit4:1;
        unsigned int bit5:1;
        unsigned int bit6:1;
        unsigned int bit7:1;
        unsigned int bit8:1;
    }bits;
}byte;

int main(void)
{
    byte.thebyte = 0;

    byte.bits.bit8 = 1;  // equivalent to 1000 0000 (msb to lsb order)


    printf("Byte is: %d\n", byte.thebyte);

    return 0;
}

Originally Posted by IceDane

Use a union and a struct with bitfields:

The approach has portability issues. I prefer plain old bytes.

Originally Posted by Dave_Sinkula

Code:

unsigned char array[2][30];
array[0][0] = 0xF0;

?

This would be just fine if I just wanted to be able to write a full 8 bit binary at once and read it at once. However, I mentioned that I' ll have to be able to read the data bit by bit as I will send them through a serial port to a external hardware device bit by bit . So it is very helpful for me that each bit is an array element.

p.s. I' m sorry for the "stupid" questions, however I' m much more familiar with hardware programming than with software programming.

Maybe you could use bitwise operators to filter out the bits in the byte you want?

Untested, but maybe something like:

Code:

int GetBit(unsigned char byte, unsigned char bit)
{
    bit = 1 << bit;
    return (byte & bit);
}

int main()
{
    unsigned char byte = 0xFF;
    int state = GetBit(byte, 4);
}

Originally Posted by aserf

This would be just fine if I just wanted to be able to write a full 8 bit binary at once and read it at once. However, I mentioned that I' ll have to be able to read the data bit by bit as I will send them through a serial port to a external hardware device bit by bit . So it is very helpful for me that each bit is an array element.

This serial port doesn't use a byte-sized shift register like every other CPU under the sun?

Originally Posted by mike_g

Maybe you could use bitwise operators to filter out the bits in the byte you want?

Untested, but maybe something like:

Code:

int GetBit(unsigned char byte, unsigned char bit)
{
    bit = 1 << bit;
    return (byte & bit);
}

int main()
{
    unsigned char byte = 0xFF;
    int state = GetBit(byte, 4);
}

This seems practical. I will try it and give you the feedback in a while.

Thank you for your time and patience...

Originally Posted by Dave_Sinkula

This serial port doesn't use a byte-sized shift register like every other CPU under the sun?

I wish it did... it would have made my work much easier.

It' s a Nios II Stratix II board (for those who know) of which I can control the I/O pins. Therefore I have to create a program that will work as hardware does.
Specifically, I use the one pin as a clock (practically by constantly changing its output from 0 to 1) and in the same time I use the other pin to send bit by bit the data synchronized with the clock. That' s why I can' t send the whole byte at once, as I'm not using an existing hardware.

@Dave_Sinkula

I used this:

Code:

#include "stdafx.h"
#include <stdio.h>
#include <windows.h>
#include "math.h"


int GetBit(unsigned char byte, unsigned char bit)
{
    bit = 1 << bit;
    return (byte & bit);
}

int _tmain(int argc, _TCHAR* argv[])
{
	unsigned char byte = 0x2B;
    int i;
	
	for (i=7;i>-1;i--)
	{
		int state = GetBit(byte, i);
        
		if (state == 0)
		{
			printf("%d",state);
		}
		else
		{
			state = state/(pow(2,i));
			printf("%d",state);
		}
	}

	for(;;)
	{
	}
}

to test it and it' s just what I wanted! Now I can use a 2 dimension array with unsigned char elements.
I just had to add

Code:

state = state/(pow(2,i));

because without this state had as a value the power of 2 of the location of the bit.

You have been really helpful

If you want the get bit function to return 0 or 1 you could use:

Code:

int GetBit(unsigned char byte, unsigned char bit)
{
    bit = 1 << bit;
    if(byte & bit) return 1;
    return 0;
}

It would save using an expensive pow function

Originally Posted by mike_g

If you want the get bit function to return 0 or 1 you could use:

Code:

int GetBit(unsigned char byte, unsigned char bit)
{
    bit = 1 << bit;
    if(byte & bit) return 1;
    return 0;
}

It would save using an expensive pow function

True!!!

I made the change and it works just fine!

Thank you all! You have been really helpful!!!

If you want the get bit function to return 0 or 1 you could use:

or maybe just

Code:

return (byte>>bit) & 1;