Thread: Truth tables and binary

Hi everyone, its been a while since i last posted, i started teaching myself C a while ago and i'm trying to get back to it. Been really busy at work with fun languages like Ocean!

Anyway, i like to learn by having a project and working towards it. So i thought i would try to write a program which could display a truth table given a boolean expression.

My first stumbling block has been dynamic arrays. I searched through the forums and found references to malloc(). I'm not asking these to be explained but i thought now would be a good time to ask if anyone has a better method of how to go about this? I don't know how well C deals with binary.

This is what i have so far:

Code:

#include <stdio.h>

int main()
{
//Initialise Variables
int x, i;
int max_value;
int inputs[x];
int bcd_dec;
int count;
int bit;
int reg;

x = 4; //width of array
max_value = 1;
for (i = 0; i < 2; i++)
	{
	max_value *= x;	//Given 'x' binary bits, max_value is all bits set to 1 +1
	}

printf("max_value = %d\n", max_value);
for (i = max_value - 1; i >= 0; i--)	//for each value
{	
count = max_value / 2; //MSB
printf("i = %d\n", i);
bit = 0;
reg = i;
	while(count > 0)
	{
	printf("count = %d\n", count);	//Count corresponds to value of each bit.
	printf("reg = %d\n", reg);
	bit = x - bit;
	printf("bit = %d\n", bit);
		if (reg >= count)	//if value is greater than or equal to 
					//corresponding bit, set it
		{
		printf("reg>count");
		inputs[bit] = 1;
		reg = reg - count;
		}
		else			//or not
		{
		printf("reg<count");
		inputs[bit] = 0;
		}

	printf("[%d]", inputs[bit]);	
		bit = bit++; 		//move to next bit
		count = count / 2;	//and corresponding value
		
	}
printf("\n");
}
}

of course this doesn't work, cause i'm stupid and didn't know you couldn't do that with arrays. This is only the part which will walk through the binary representation of 0 to 15 (for example)

Sorry for the long post. Your help would be appreciated.

Also, does anyone have any ideas for projects? Maybe someone wants to set me a not to difficult task?
Kai.

Thanks for your reply,
Tha part where i set x is because i want the array (i.e the number of inputs) to be any size. Thats why i tried inputs[x]. I set x just because i wanted to get the gist of the program out the way before putting a section in calculate x.

The second point is a mistake on my behalf. The for loop should go to x and i should be computing to the power of 2. Messed up there.

I'm sorry, i don't know of another way to do the exponent calculation. how does 2**x work and i don't understand what you said about 1 << x. Is that a bitwise shift x bits to the left?

I thought C didn't deal with binary very well?

** is not an operator in C -- it's just another notation for exponent (from back before they invented the ^ key). (Yes, I learned Fortran once upon a time. Why do you ask?) And yes, << is bit shifting. Things are binary, so shift by 1 == multiply by 2, just like in decimal notation that we all use, shift decimal point by 1 == multiply by 10.

Anyway, if you really want the number of inputs to be arbitrary, it will take a slight bit of work, but if you're looking at an expression parser, then it is probably likely that you won't see more than 26 variables (unless you allow arbitrary subscripting like v1, ..., v87132498, etc.), and 26 bits is certainly smaller than an int these days. So you should be able to get away with a single int variable inputs.