Thread: Counting number of TRUE or FALSE within an IF statement

Hi, new to these forums, intermediate self taught programmer, I have a question that could be considered an advanced use of the IF statement

Lets take the following example

if (a || b || c || d || e == 10)
then do something

In the above statement if any of 'a' 'b' 'c' 'd' 'e' are equal to 10, then the statement is TRUE. OK thats easy and straight forward.

However, what if more than one of 'a' 'b' 'c' 'd' 'e' are true, lets say a == 10 and c == 10, this statement again would still return TRUE, OK thats still easy and straight forward.

Now heres the trick i want to learn, I want to return TRUE only if a certain number of the inner variables meet the test, ie if two of the variables satisfy the test condition return TRUE otherwise return FALSE

Maybe something like this????

if ((a || b || c || d || e == 10) == 2)

ie if two of the variables == 10 then return TRUE otherwise return FALSE if 1 3 4 or all 5 are == 10,

I could certainly use individual IF statements for each parameter and count incrememmt each time one of them returns TRUE, but in the application i am developing this could mean quite extensive additional code and checking.

Is there a way of saying if 2 of the following 5 parameters meet the condition then return TRUE or FALSE. It could just as easily be if 3 or if 1 of the test variables, but the aim is to be able to stipulate for how many of the supplied variables is true or false

I hope this is clear(er than mud), many thanks for any help in advance

Hi thanks for the really quick reply, Yea i understand what you are saying, and yes, whilst the example solution code will compile as you say it will indeed not work. It was more just for illustration purpose.

What approach could you suggest to determine if X number of parameters in a multiparameter IF scenario are TRUE (or FALSE). To add context i have pasted a code snippet below.

This is part of an image processing line walking task, i wish to check if a focus pixel is ON (ie == 255), and if so, how many neighbour pixels are on. If more than one neighbour pixel is ON, then i am in the middle of a line. I only want to perform execution if i am at the end of a line (ie only one neighbour pixel is ON)

Code:

if (thispixel[u-(t->_imageWidth-1)] 	|| 
    thispixel[u+1]			                || 
    thispixel[u-1]  		        	|| 
    thispixel[u+(t->_imageWidth+1)] 	|| 
    thispixel[u+t->_imageWidth] 		|| 
    thispixel[u+(t->_imageWidth-1)]  == 255)
{
//find which neighbour is on
if (thispixel[u-(t->_imageWidth-1)] 	== 255 ) //NE
	{
		// compute trajectory
		pixelcount++; 	//increment pixel length counter
		u = u-(t->_imageWidth-1);
		v=v-1;		//adjust position of focus pixel pointer			
		goto label;		//continue walking using new focus pixel
	}
	else if (thispixel[u+1]		== 255 ) //E
	{
		// compute trajectory
		pixelcount++; 	//increment pixel length counter
		u = u+1;		//adjust position of focus pixel pointer			
		goto label;		//continue walking using new focus pixel
	}
	else if (thispixel[u-1]		== 255 ) //W
	{
		// compute trajectory
		pixelcount++; 	//increment pixel length counter
		u = u-1;		//adjust position of focus pixel pointer			
		goto label;		//continue walking using new focus pixel
	}
	else if (thispixel[u+(t->_imageWidth+1)]== 255 ) //SE 
	{
		pixelcount++;
		u = u+(t->_imageWidth+1);
		v=v+1;
		goto label;	
	}
	else if (thispixel[u+t->_imageWidth]	== 255 ) //S
	{
		pixelcount++;
		u = u+t->_imageWidth;
		v=v+1;
		goto label;	
	}
	else if (thispixel[u+(t->_imageWidth-1)]	== 255 ) //SW
	{
		pixelcount++;
		u = u+(t->_imageWidth-1);
		v=v+1;
		goto label;	
	}
}

Quzah, i was replying to CommonTater whilst you posted your answer. To CommonTater, perhaps i did not best describe my query. Quzah == abs(Legend) Thanks for that, you have put me right on the right track i thinks as me shivers me timbers.