I was looking at the latest parrot and saw in the examples, a mandelbrot generator, with the equivalent c source code. So I modified them slightly, and ran them thru 1000 generations. Parrot is fast, but it took 5 times the cpu, and roughly 3 times as long on my machine(Duron 800). It would be interesting to compare it to equivalent perl5 code, and if anyone can convert it, please post it. I'm almost there, but the C for statements are a little complicated for me to convert. :-)

Anyways, here is the code for you to try. I slightly modified the original code, to add timer variables.

#################################################
# mandel.pasm ...run with "parrot mandel.pasm"

        time N10
        set I0, 0

loop:
        set S1, " .:,;!/>)|&IH%*#"
        set I2, 30



YREDO:  #  C = y*0.1 - 1.5
        set N8, I2
        set N9, 0.1
        mul N7, N8, N9
        set N8, 1.5
        sub N7, N7, N8

        set I1, 0

XREDO:  # c = x*0.04 - 2
        set N8, I1
        set N9, 0.04
        mul N6, N8, N9
        set N8, 2.0
        sub N6, N6, N8
        set N3, 0
        set N4, 0

        set N1, N6
        set N2, N7

        set I3, 0

KREDO:  # t = z*z - Z*Z + r
        mul N8, N3, N3
        mul N9, N4, N4
        sub N5, N8, N9
        add N5, N5, N1

        # Z = 2*z*Z + i
        add N4, N4, N4
        mul N4, N4, N3
        add N4, N4, N2

        # z = t
        set N3, N5

        # if (z*z + Z*Z > 10) break;
        mul N8, N3, N3
        mul N9, N4, N4
        add N8, N8, N9
        gt N8, 10.0, PRINT

        inc I3
        lt I3, 112, KREDO

PRINT:  mod I4, I3, 16
        substr S2, S1, I4, 1
        print S2

        inc I1
        lt I1, 75, XREDO

        print "\n"
        dec I2
        gt I2, 0, YREDO


inc I0
lt I0, 1000, loop

print I0
print " count in "
time N11
sub N11,N10
print N11
print " seconds\n"

END:    end
############################################
[download]

And the c source

/* mandel.c  ... compile with gcc -o mandel mandel.c */
#include <stdio.h> 
#include <time.h>  

int main() {
  int x, y, k, counter;
  char *b = " .:,;!/>)|&IH%*#";
  float r, i, z, Z, t, c, C;
  time_t begin, end;
  counter = 0;

begin = time (NULL);

while(counter < 1000){
    for (y=30; puts(""), C = y*0.1 - 1.5, y--;) {
     for (x=0; c = x*0.04 - 2, z=0, Z=0, x++ < 75;) {
      for (r=c, i=C, k=0; t = z*z - Z*Z + r, Z = 2*z*Z + i, z=t, k<112
+; k++)
       if (z*z + Z*Z > 10) break;
        printf ("%c", b[k%16]);
      }
    }
counter++;
}

end = time (NULL);

printf ("%d count in %ld secs \n", counter, end - begin);

}
[download]

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