Solved

loop merging

Posted on 2004-09-09
5
248 Views
Last Modified: 2010-04-15
hello everybody
     i have got a question for u , its very simple, can u suggest me different ways to merge the loop in the below programm. i want to elinimate the       z[320] or do indexing with three loops given



const double* C = Coefficients;  \*constant coeff*/
int i,j;
float   X[320];
double   Z[320];
double   realAccu;
for ( i=0; i<320; i++ )
{
    Z[i] = X[i] * C[2*i];
}
      
      /* create array Y */
      for ( i=0; i<64; i++ ) {
            realAccu = 0.0;
      }
            for ( j=0; j<5; j++ ) {
                  realAccu = realAccu + Z[i + j * 64];
            
            Y[i] = realAccu;
      }



0
Comment
Question by:san_ys
5 Comments
 
LVL 11

Expert Comment

by:avizit
ID: 12023938
for ( i=0; i<64; i++ ) {
          realAccu = 0.0;
}

the above is unnecessary you are just assigning 0.0 to realAccu 64 times
you can just replace the above for loop with

realAccu = 0.0
0
 
LVL 45

Expert Comment

by:sunnycoder
ID: 12024779
Hi san_ys,

Your requirements are not very clear. Arrays Z and X are uninitialized when you use them. As Abhijit pointed out, one of the loops is redundant.

>i want to elinimate the       z[320] or do indexing with three loops given
You mean you wish to eliminate array Z from calculations or the just the loop which initializes Z (the initialization might be incorrect since X is uninitialized). Also, I am not sure what you mean by "do indexing with three loops given"
0
 
LVL 12

Accepted Solution

by:
stefan73 earned 50 total points
ID: 12024833
Hi san_ys,
It looks like the loop scopes are not correct. You probably meant to have the last two loops nested, like:

     /* create array Y */
     for ( i=0; i<64; i++ ) {
          realAccu = 0.0;
          for ( j=0; j<5; j++ )
               realAccu = realAccu + Z[i + j * 64];
         
          Y[i] = realAccu;
     }

The inner loop should be eliminated (loop unrolling) by the compiler, but you can also do that manually:

     /* create array Y */
     for ( i=0; i<64; i++ ) {
          Y[i] = Z[i] + Z[64+i] + Z[128+i] + Z[192+i] + Z[256+i];
     }

...that's a matter of personal preference. If you feel that the algorithm is expressed more clearly with two nested loops, fine. A good compiler should create identical code in both cases.

You can also eliminate the creation of your Z array - its values are only used once. That's probably done best with a macro:

#define Z(i) (X[i] * C[2*(i)])

Then you can write

     /* create array Y */
     for ( i=0; i<64; i++ ) {
          Y[i] = Z(i) + Z(64+i) + Z(128+i) + Z(192+i) + Z(256+i);
     }
...ending up with a single loop.

Did you initialize the X array with some values?


Cheers!

Stefan
0

Featured Post

Simplifying Server Workload Migrations

This use case outlines the migration challenges that organizations face and how the Acronis AnyData Engine supports physical-to-physical (P2P), physical-to-virtual (P2V), virtual to physical (V2P), and cross-virtual (V2V) migration scenarios to address these challenges.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Have you thought about creating an iPhone application (app), but didn't even know where to get started? Here's how: ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ Important pre-programming comments: I’ve never tri…
Preface I don't like visual development tools that are supposed to write a program for me. Even if it is Xcode and I can use Interface Builder. Yes, it is a perfect tool and has helped me a lot, mainly, in the beginning, when my programs were small…
The goal of this video is to provide viewers with basic examples to understand opening and writing to files in the C programming language.
The goal of this video is to provide viewers with basic examples to understand how to create, access, and change arrays in the C programming language.

777 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question