topic Re: problem with codes size(MCF5213) in ColdFire/68K Microcontrollers and Processors

problem with codes size(MCF5213)

doom — Wed, 05 Nov 2008 16:00:57 GMT

I built a stationary project with CW6.4. It is said that it can download 128k codes for free.But I can only write about 3,000 lines in my c source.If I write more,it overflows.
Can anyone tell me how many lines can I write with in the 128k limitition or how can I reduce the size of my output file,because I need about 10,000 lines in my project . It rather bothers me...

Thanks.

Re: problem with codes size(MCF5213)

RichTestardi — Wed, 05 Nov 2008 23:24:07 GMT

Hi,

It really depends on your code, but for my project of 8000 lines of source (excluding comments and blanks), I compile to between 60-80 kbytes, depending on the compiler optimization level, using CW7.1 (CW7.0 was more bloated than CW6.4, but I believe CW7.1 is comparable).

The things you want to check for are:

1. compiler optimization levels, parameter passing mechanism, etc.

2. large constant data structures/strings

3. excessive library usage

4. is linker deadstripping disabled?

The best place to look for your code bloat is in your xMAP file -- you can see where every single byte of code comes from -- whether from libraries or from your routines. Then you can shrink things appropriately. The xMAP file should be in your "bin" directory.

I'd expect you could get 10,000 lines of reasonable source to fit into 128 kbytes.

-- Rich

Re: problem with codes size(MCF5213)

doom — Fri, 07 Nov 2008 11:10:54 GMT

HI Rich
     Thanks for your reply.
      Ur.. well.
      1.I did not disable the linker deadstripping
      2.I changed compiler optimization levels fromm level 1 to level 4 and choosed the small size code

But it seems not good .After that, the output file shrinked fome 30k to 28 k.Do you think it has something
to do with the license?Maybe I am using an improper license?
I will try again,and thanks all the same.

--doom

Re: problem with codes size(MCF5213)

RichTestardi — Fri, 07 Nov 2008 12:39:44 GMT

Hi,

I may have misunderstood... I thought you had a 3000 source line project that was compiling to over 128k bytes, and you were getting a license error as a result...

If your code is compiling larger than you expect, it might help if you can post the corresponding xMAP file -- then I can probably tell you exactly why... (You also might want to select "Always Keep Map" under ColdFire Linker options.)

Also, can you tell us what the exact error is you are receiving? It might help to post your Linker Command File (.lcf) as well, depending on when you are getting the error.

Thanks!

-- Rich

Re: problem with codes size(MCF5213)

doom — Sat, 08 Nov 2008 11:18:50 GMT

HI
    Blame it on me that I did not explain clearly.
    Well,
1.(optimization off ) three days ago ,in order to test the maximal lines in a project ,I copied the codes as follows up to 3000 lines and then it was compiling to over 128k bytes.

       if(ADCNT>=4)
               {

               ADresult0[4]=(ADresult0[0]+ADresult0[1]+ADresult0[2]+ADresult0[3])/4;
               FRE1=(unsigned int)(ADresult0[4]*100*238.7/167.4346);
               DISPBF[0]=((unsigned int)FRE1)/10000;
               DISPBF[1]=(((unsigned int)FRE1)%10000)/1000;
               DISPBF[2]=((unsigned int)FRE1%1000)/100;
               DISPBF[3]=((unsigned int)FRE1%100)/10;
               DISPBF[4]=((unsigned int)FRE1%10);

               ADresult1[4]=(ADresult1[0]+ADresult1[1]+ADresult1[2]+ADresult1[3])/4;
               FRE1=(unsigned int)(ADresult1[4]*100*238.7/167.4346);
              DISPBF[6]=((unsigned int)FRE1)/10000;
              DISPBF[7]=(((unsigned int)FRE1)%10000)/1000;
              DISPBF[8]=((unsigned int)FRE1%1000)/100;
              DISPBF[9]=((unsigned int)FRE1%100)/10;
              DISPBF[10]=((unsigned int)FRE1%10);

              ADresult2[4]=(ADresult2[0]+ADresult2[1]+ADresult2[2]+ADresult2[3])/4;
              FRE1=(unsigned int)(ADresult2[4]*100*238.7/167.4346);
              DISPBF[12]=((unsigned int)FRE1)/10000;
              DISPBF[13]=(((unsigned int)FRE1)%10000)/1000;
              DISPBF[14]=((unsigned int)FRE1%1000)/100;
              DISPBF[15]=((unsigned int)FRE1%100)/10;
              DISPBF[16]=((unsigned int)FRE1%10);

             ADCNT=0;
            }

Then I test with copying this code"i=2+3*5-4".like this:
          main()
        {
             double i;
             i=sqrt(9.0);
             i=i=2+3*5-4;
             i=i=2+3*5-4;
             i=i=2+3*5-4;
             .
             .
             .
        }
I can write about 6000 lines

2 .(optimization level 4)After take your advice.I choose the optimiziton level 4 and then I can repeat "i=2+3*5-4" 16,000 times in the main function.

But even when I choose optimization level 4 ,if I copy the codes ：                                                                                    ADresult0[4]=(ADresult0[0]+ADresult0[1]+ADresult0[2]+ADresult0[3])/4;
                  FRE1=(unsigned int)(ADresult0[4]*100*238.7/167.4346);
                 DISPBF[0]=((unsigned int)FRE1)/10000;
                 DISPBF[1]=(((unsigned int)FRE1)%10000)/1000;
                 DISPBF[2]=((unsigned int)FRE1%1000)/100;
                 DISPBF[3]=((unsigned int)FRE1%100)/10;
                 DISPBF[4]=((unsigned int)FRE1%10);
I can still write 3000 lines at most.

So I agree it really depends on the codes.

3.The attachments are my Installed Plugins file , lcf file and xMAPfile . I use none libraries except support_common.h
stdio.h
and sqrt in math.h
I will appreciate it,if you guide me to remove the unused libraries.

Re: problem with codes size(MCF5213)

RichTestardi — Sat, 08 Nov 2008 12:28:56 GMT

Ah, that makes more sense...

From your xMAP file, you can see main() is taking 100k bytes, and the rest looks to be floating point and printf support libraries.

00000C10 00018ECC .text main (main.c)

If you right-click on your source file and click "Disassemble", you can see how many bytes each line is compiling to (by searching for the source line in the disassembly listing).

Can you tell me how these variables (ADresult0, FRE1, and DISPBF) are declared -- are they floating point or integer or unsigned integer?

               ADresult0[4]=(ADresult0[0]+ADresult0[1]+ADresult0[2]+ADresult0[3])/4;
               FRE1=(unsigned int)(ADresult0[4]*100*238.7/167.4346);
               DISPBF[0]=((unsigned int)FRE1)/10000;
               DISPBF[1]=(((unsigned int)FRE1)%10000)/1000;
               DISPBF[2]=((unsigned int)FRE1%1000)/100;
               DISPBF[3]=((unsigned int)FRE1%100)/10;
               DISPBF[4]=((unsigned int)FRE1%10);

It looks like you're doing a lot of floating point ops to get integer results...

I'm *guessing* ADresult0 might be a result from the a/d converter, and you're trying to do some kind of a digital filter? If so, the a/d results are 15 bit values (actually, 13 bit, I think, but multiplied by 4), so you might be able to use a 32 bit integer and replace the big line with an approximation:

FRE1=(unsigned int)(ADresult0[4]*14256/10000*100);

But if you can tell me the types of the variables, we'll know for sure what kind of optimization might be possible, hopefully avoiding floating point. Also, is your compiler set to size integers as 32 bit values?

-- Rich

Re: problem with codes size(MCF5213)

doom — Mon, 10 Nov 2008 08:57:05 GMT

THANKS

I did use lots of floating point to get precise datas.
variables (ADresult0, FRE1, and DISPBF) are declared as follows:
double result=0,ADresult0[5],ADresult1[5]={0},ADresult2[5]={0},ADresult3[5]={0},ADresult4[5]={0},ADresult5[5]={0};
int AD0[128],AD1[128],AD2[128],AD3[128],AD4[128],AD5[128];

unsigned char j=0,ADCNT=0,ADFLG=0,_100MS=0,i=0,cnt1;

unsigned int FRE1=0;
unsigned char DISPBF[18]={0,1,2,3,4,0x27,5,6,7,8,9,0x27,0x0a,0x0b,0x0c,0x0d,0x0e,0x00};
char cnt=0;

ADx[128] are result from the a/d converter.
I use double sqrt();in math.h and "ADresult0[x]=sqrt(result/128);"and use "ADresult0[4]=(ADresult0[0]+ADresult0[1]+ADresult0[2]+ADresult0[3])/4;"to do a digital filter.

I do not care about RAM.What I care about most is Flash.By Disassembling I found that:

    DISPBF[4]=((unsigned int)FRE1%10);
; 529:
;
0x00002700 0x223900000000           move.l   _FRE1,d1
0x00002706 0x4C431000               remul.l d3,d0:d1
0x0000270A 0x13C000000000           move.b   d0,_DISPBF+4
0x00002710 0x41F900000000           lea      _ADresult1+8,a0
0x00002716 0x2010                   move.l   (a0),d0
0x00002718 0x2F6800040010           move.l   4(a0),16(a7)
0x0000271E 0x41F900000000           lea      _ADresult1,a0
0x00002724 0x2F40000C               move.l   d0,12(a7)
0x00002728 0x2010                   move.l   (a0),d0
0x0000272A 0x22280004               move.l   4(a0),d1
0x0000272E 0x41EEDF28               lea      -8408(a6),a0
0x00002732 0x2F410008               move.l   d1,8(a7)
0x00002736 0x2F400004               move.l   d0,4(a7)
0x0000273A 0x2E88                   move.l   a0,(a7)
0x0000273C 0x4EB900000000           jsr      __d_add
0x00002742 0x2010                   move.l   (a0),d0
0x00002744 0x2F6800040010           move.l   4(a0),16(a7)
0x0000274A 0x41F900000000           lea      _ADresult1+16,a0
0x00002750 0x2F40000C               move.l   d0,12(a7)
0x00002754 0x2010                   move.l   (a0),d0
0x00002756 0x22280004               move.l   4(a0),d1
0x0000275A 0x41EEDF20               lea      -8416(a6),a0
0x0000275E 0x2F410008               move.l   d1,8(a7)
0x00002762 0x2F400004               move.l   d0,4(a7)
0x00002766 0x2E88                   move.l   a0,(a7)
0x00002768 0x4EB900000000           jsr      __d_add
0x0000276E 0x2010                   move.l   (a0),d0
0x00002770 0x2F6800040010           move.l   4(a0),16(a7)
0x00002776 0x41F900000000           lea      _ADresult1+24,a0
0x0000277C 0x2F40000C               move.l   d0,12(a7)
0x00002780 0x2010                   move.l   (a0),d0
0x00002782 0x22280004               move.l   4(a0),d1
0x00002786 0x41EEDF10               lea      -8432(a6),a0
0x0000278A 0x2F410008               move.l   d1,8(a7)
0x0000278E 0x2F400004               move.l   d0,4(a7)
0x00002792 0x2E88                   move.l   a0,(a7)
0x00002794 0x4EB900000000           jsr      __d_add
0x0000279A 0x203C40100000           move.l   #1074790400,d0        ; '@...'

As you see, this code is so inefficient.Maybe I should do some changes besides optimizing variables.

--DOOM

Re: problem with codes size(MCF5213)

RichTestardi — Mon, 10 Nov 2008 11:24:03 GMT

Hi,

I'm pretty sure you can get the same resulting precision with integer operations, since the raw a/d data only has 13 bits of significance, and each order of filter adds only about a bit of accuracy.

You can make a trivial integer square root function using successive approximation -- even at its slowest, it'll probably be quite a bit faster than floating point... Something like (you can tune the details for your range of integers -- I assumed 0..32767):

Code:

#define SUCCESSIVE  10static intisqrt(int x){    int i;    int n;        if (! x) {        return 0;    }        n = 128;  // initial guess for sqrt(mean 0..32767)    for (i = 0; i < SUCCESSIVE; i++) {        if (n == x/n) {            break;        }        n = (n + x/n + 1)/2;    }    return n;}

Do you have the parameters for the digital filter (order, coefficients, and IIR/FIR)? I used a simple low-pass filter on the a/d converter below in the past (hooked to the accelerometer) for input values 0..32767 -- notice how I multiply by ~10000 to efectively keep 4 decimal places in an integer:

Code:

// our low pass filter coefficientsstatic shortcoef[] = {      67,     449,     874,    1261,    1531,    1628,    1531,    1261,     874,     449,      67,};// our low pass filter order#define ORDER  (sizeof(coef)/sizeof(coef[0]))intlowpass(int x){    int i;    static int s;    static int xs[ORDER];    // low pass filter    xs[s] = x;    x = 0;    for (i = 0; i < ORDER; i++) {        x += coef[i]*xs[(s+i)%ORDER];    }    x /= 9992;  // sum of coef's    s = (s+1)%ORDER;    return x;}

-- Rich

Re: problem with codes size(MCF5213)

Everton — Wed, 10 Mar 2010 19:47:56 GMT

Hi, Rich T

I want to know if this algorithm can be used to filter high-pass and band-pass? Replacing only the coefficient of the filter.

Thanks advance

Everton Danilo

Re: problem with codes size(MCF5213)

RichTestardi — Wed, 10 Mar 2010 23:47:33 GMT

There are lots of sites on the web for computing the coefficients and getting guidance in digital filter design...

You might try something like: http://www.dsptutor.freeuk.com/FIRFilterDesign/FIRFilterDesign.html

Or just search for "fir digital filter design" or "iir digital filter design".

Good luck.

-- Rich