Is there a better way to do the following?

purpleburple · 07-02-2002, 09:59 PM

Hi. New to C and loving it. I'm mostly learning from a few books and there's alot of cool websites on Beginning C. I just began learning about arrays and was wondering if the following was a good way to check for matches between two arrays? Or is there some easier or better way to do it?

/* compare users input to int array; print results to standard out */

for(i=0; i<=4; i++)
{
while(a[j] != b[i] )
{
j++;
}
printf("b[%d] matches a[%] they both contain %d\n",i, j, b[i] );

/* set variable j back to 0 */
j = 0;

}

i have an array (a[10] containing 10 ints from 1-10 )
then the user enters 5 ints into b[5] and the above searches thru the array and prints out the matches

thanks for your time

New to C and Linux (RedHat 7.2)

tundra · 07-03-2002, 01:29 AM

depends on what you mean by good

actually, as long as it gets the job done, don't worry about it too much. you'll find ways to improve whatever code you're doing later. especially if you have nothing else much to do

. bask in your new found achievement and when you've had enough, move on to other topics on C!

biosx · 07-03-2002, 07:25 AM

I would rather do it this way, it makes more sense if you think about it:

Code:

for(i = 0; i < 5; i++) 
    for(j=0; j < 5; j++) 
        if( a[i] == b[j] )
            printf("a[%d] and b[%d] both contain %d\n", i, j, a[i]);

purpleburple · 07-03-2002, 10:50 AM

thanks I didnt know you could put for loops inside each other like that. Cool!

tundra · 07-04-2002, 03:09 AM

you could include error checking...

kervin · 07-07-2002, 05:29 PM

I know you're just starting off, but get a book on data structures and algorithms, if you plan to eventually use this code in something serious.

A "good" solution to your search problem is going to depend on the type of data you are search for ( integers?, strings?), the size of the array, are you looking for space ( memory ) or time ( CPU ) optimizations.

One integer approach is to make the array index the number. This would work if you have a bounded range or do not mind reallocating member. Therefore if the user enters '5' you do,
arrayOne[5]=1;
etc., That way your search loop becomes
for(int i=0;i<userInputSize;i++)
{
currentNumber = arrayTwo[i];
if(arrayOne[currentNumber]==1)
/* We have a match. Do stuff here */
}

The above can have *big* advantages if you make arrayOne a bitfield ( ie. an array of 'bits' instead of integers which would be 32 bits instead of 1 ).

But more importantly the algorithm is of order N instead of N^2 as the one you posted was. With your algorithm, the time it takes to execute will grow expontentially. The code will be twice as slower for every additional element in the array.

The second algorithm should grow linear.

So check out algorithms and data structures. Interesting stuff