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{$/\ast$\it{}$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$\mbox{}\\
$\ast$ FILE: dotprod\_serial.c\mbox{}\\
$\ast$ DESCRIPTION:\mbox{}\\
$\ast$   This is a simple serial program which computes the dot product of two \mbox{}\\
$\ast$   vectors.  The threaded version can is dotprod\_mutex.c.\mbox{}\\
$\ast$\mbox{}\\
$\ast$ SOURCE: Vijay Sonnad, IBM\mbox{}\\
$\ast$ LAST REVISED: 9/20/98 Blaise Barney\mbox{}\\
$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast$$\ast/$}\mbox{}\\
\mbox{}\\
{\tt \#include} $<${\tt{}stdio.h}$>$\mbox{}\\
{\tt \#include} $<${\tt{}malloc.h}$>$\mbox{}\\
\mbox{}\\
{$/\ast$\it{}   \mbox{}\\
The following structure contains the necessary information  \mbox{}\\
to allow the function "dotprod" to access its input data and \mbox{}\\
place its output so that it can be accessed later. \mbox{}\\
$\ast/$}\mbox{}\\
\mbox{}\\
{\bf typedef} {\bf struct} \mbox{}\\
\hspace*{1\indentation}\{\mbox{}\\
\hspace*{3\indentation}{\bf double}      $\ast$a;\mbox{}\\
\hspace*{3\indentation}{\bf double}      $\ast$b;\mbox{}\\
\hspace*{3\indentation}{\bf double}     sum; \mbox{}\\
\hspace*{3\indentation}{\bf int}    veclen; \mbox{}\\
\hspace*{1\indentation}\} DOTDATA;\mbox{}\\
\mbox{}\\
{\tt \#define} VECLEN 100\mbox{}\\
\hspace*{3\indentation}DOTDATA dotstr; \mbox{}\\
\mbox{}\\
{$/\ast$\it{}\mbox{}\\
We will use a function (dotprod) to perform the scalar product. \mbox{}\\
All input to this routine is obtained through a structure of \mbox{}\\
type DOTDATA and all output from this function is written into\mbox{}\\
this same structure.  While this is unnecessarily restrictive \mbox{}\\
for a sequential program, it will turn out to be useful when\mbox{}\\
we modify the program to compute in parallel.\mbox{}\\
$\ast/$}\mbox{}\\
\mbox{}\\
{\bf void} $\ast$dotprod({\bf void})\mbox{}\\
\{\mbox{}\\
\mbox{}\\
{$/\ast$\it{} Define and use local variables for convenience $\ast/$}\mbox{}\\
\mbox{}\\
\hspace*{3\indentation}{\bf int} start, end, i; \mbox{}\\
\hspace*{3\indentation}{\bf double} mysum, $\ast$x, $\ast$y;\mbox{}\\
\mbox{}\\
\hspace*{3\indentation}start=0;\mbox{}\\
\hspace*{3\indentation}end = dotstr.veclen;\mbox{}\\
\hspace*{3\indentation}x = dotstr.a;\mbox{}\\
\hspace*{3\indentation}y = dotstr.b;\mbox{}\\
\mbox{}\\
{$/\ast$\it{}\mbox{}\\
Perform the dot product and assign result\mbox{}\\
to the appropriate variable in the structure. \mbox{}\\
$\ast/$}\mbox{}\\
\mbox{}\\
\hspace*{3\indentation}mysum = 0;\mbox{}\\
\hspace*{3\indentation}{\bf for} (i=start; i$<$end ; i++) \mbox{}\\
\hspace*{4\indentation}\{\mbox{}\\
\hspace*{6\indentation}mysum += (x[i] $\ast$ y[i]);\mbox{}\\
\hspace*{4\indentation}\}\mbox{}\\
\hspace*{3\indentation}dotstr.sum = mysum;\mbox{}\\
\mbox{}\\
\hspace*{3\indentation}{\bf return} ;\mbox{}\\
\}\mbox{}\\
\mbox{}\\
{$/\ast$\it{}\mbox{}\\
The main program initializes data and calls the dotprd() function.\mbox{}\\
Finally, it prints the result.\mbox{}\\
$\ast/$}\mbox{}\\
\mbox{}\\
{\bf int} main ({\bf int} argc, {\bf char} $\ast$argv[])\mbox{}\\
\{\mbox{}\\
\hspace*{3\indentation}{\bf int} i,len;\mbox{}\\
\hspace*{3\indentation}{\bf double} $\ast$a, $\ast$b;\mbox{}\\
\hspace*{3\indentation}\mbox{}\\
{$/\ast$\it{} Assign storage and initialize values $\ast/$}\mbox{}\\
\hspace*{3\indentation}\mbox{}\\
\hspace*{3\indentation}len = VECLEN;\mbox{}\\
\hspace*{3\indentation}a = ({\bf double}$\ast$) malloc (len$\ast${\bf sizeof}({\bf double}));\mbox{}\\
\hspace*{3\indentation}b = ({\bf double}$\ast$) malloc (len$\ast${\bf sizeof}({\bf double}));\mbox{}\\
\hspace*{2\indentation}\mbox{}\\
\hspace*{3\indentation}{\bf for} (i=0; i$<$len; i++)\mbox{}\\
\hspace*{4\indentation}\{\mbox{}\\
\hspace*{5\indentation}a[i]=1;\mbox{}\\
\hspace*{5\indentation}b[i]=a[i];\mbox{}\\
\hspace*{4\indentation}\}\mbox{}\\
\mbox{}\\
\hspace*{3\indentation}dotstr.veclen = len; \mbox{}\\
\hspace*{3\indentation}dotstr.a = a; \mbox{}\\
\hspace*{3\indentation}dotstr.b = b; \mbox{}\\
\hspace*{3\indentation}dotstr.sum=0;\mbox{}\\
\mbox{}\\
{$/\ast$\it{} Perform the  dotproduct $\ast/$}\mbox{}\\
\mbox{}\\
\hspace*{3\indentation}dotprod ();\mbox{}\\
\mbox{}\\
{$/\ast$\it{} Print result and release storage $\ast/$} \mbox{}\\
\mbox{}\\
\hspace*{3\indentation}printf ({\tt"Sum $=$  \%f $\backslash$n"}, dotstr.sum);\mbox{}\\
\hspace*{3\indentation}free (a);\mbox{}\\
\hspace*{3\indentation}free (b);\mbox{}\\
\}\mbox{}\\
\mbox{}\\
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