program not working properly. book example. program included.

[G G]

The program is an Example from Deitel and Deitel, chapter 6.
I typed it in carefully, I believe.

The program seems to freeze in the function median, where
median calls printArray(). printArray() prints only 80 elements
and then it seems to be in a loop of some type running
the cpu at 100%. well, freeze may not be the right terminology.
really it happens in the printArray(), because printArray()
does not finish printing.


output i get:

********
Mean
********
The mean is the average value of the data
items. The mean is equal to the of
all the data items divided by the number
of data items ( 99 ). The mean value for
this run is: 681 / 99 = 6.8788


*********
Median
*********
The unsorted array of responses is
6 7 8 9 8 7 8 9 8 9 7 8 9 5 9 8 7 8 7 8
6 7 8 9 3 9 8 7 8 7 7 8 9 8 9 8 9 7 8 9
6 7 8 7 8 7 9 8 9 2 7 8 9 8 9 8 9 7 5 3
5 6 7 2 5 3 9 4 6 4 7 8 9 6 8 7 8 9 7 8

seems to be in a loop here. it's still running.
after some time i get an exit code 9.

thanks everyone,

g.

/* Fig. 6.16: fig06_16.c
This program introduces the topic of survey data analysis.
It computes the mean, median and mode of the data */

#include <stdio.h>

#define SIZE 99

/* functtion prototypes */
void mean( const int answer[] );
void median( int answer[] );
void mode( int freq[], const int answer[] );
void bubbleSort( int a[] );
void printArray( const int a[] );

/* function main begins program execution */
int main( void )
{
int frequency[ 10 ] = { 0 }; /* initialize array frequency */

/* initialize array response */
int response[ SIZE ] =
{ 6, 7, 8, 9, 8, 7, 8, 9, 8, 9,
7, 8, 9, 5, 9, 8, 7, 8, 7, 8,
6, 7, 8, 9, 3, 9, 8, 7, 8, 7,
7, 8, 9, 8, 9, 8, 9, 7, 8, 9,
6, 7, 8, 7, 8, 7, 9, 8, 9, 2,
7, 8, 9, 8, 9, 8, 9, 7, 5, 3,
5, 6, 7, 2, 5, 3, 9, 4, 6, 4,
7, 8, 9, 6, 8, 7, 8, 9, 7, 8,
7, 4, 4, 2, 5, 3, 8, 7, 5, 6,
4, 5, 6, 1, 6, 5, 7, 8, 7 };

/* process reponses */
mean( response );
median( response );
mode( frequency, response );

return 0; /* indicates successful termination */

} /* end main */


/* calculate average of all response values */
void mean( const int answer[] )
{
int j; /* counter for totaling array elements */
int total = 0; /* variable to hold sum of array elements */

printf ( "%s\n%s\n%s\n", "********", " Mean", "********" );

/* total response values */
for ( j = 0; j < SIZE; j++ )
total += answer[ j ];

printf ( "The mean is the average value of the data\n"
"items. The mean is equal to the of\n"
"all the data items divided by the number\n"
"of data items ( %d ). The mean value for\n"
"this run is: %d / %d = %.4f\n\n",
SIZE, total, SIZE, ( double )total / SIZE );

} /* end function mean */


/*sort array and determine median element's value */
void median( int answer [] )
{
printf( "\n%s\n%s\n%s\n%s",
"*********", " Median", "*********",
"The unsorted array of responses is" );

printArray( answer ); /* output unsorted array */

bubbleSort( answer ); /* sort array */

printf( "\n\nThe sort array is" );
printArray( answer ); /* output sorted array */

/* display median element */
printf( "\n\nThe median is element %d of \n"
"the sorted %d element array.\n"
"For this run the median is %d\n\n",
SIZE / 2, SIZE, answer[ SIZE / 2 ] );

} /* end function median */


/* determine most frequent response */
void mode( int freq[], const int answer[] )
{
int rating; /* counter for accessing elements 1-9 of array freq */
int j; /* counter for summarizing elements 0-98 of array answer */
int h; /* counter for displaying histograms of elements in array freq */
int largest = 0; /* represents largest frequency */
int modeValue = 0; /* represents most frequenct response */

printf( "\n%s\n%s\n%s\n",
"*********", " Mode", "********" );

/* initalize frequencies to 0 */
for ( rating = 1; rating <= 9; rating++ )
freq[ rating ] = 0;

/* summarise frequencies */
for ( j = 0; j < SIZE; j++ )
++freq[ answer[ j ] ];

/* output headers for result columns */
printf( "%s%11s%19s\n\n%54s\n%54s\n\n",
"Response", "Frequency", "Histogram",
"1 2 2 2", "5 0 5 0 5");

/* output results */
for ( rating = 1; rating <= 9; rating++ )
{
printf( "%8d%11d ", rating, freq[ rating ] );

/* keep track of mode value and largest frequency value */
if ( freq[ rating ] > largest )
{
largest = freq[ rating ];
modeValue = rating;
} /* end if */

for ( h = 1; h <= freq[ rating ]; h++ )
printf( "*" );

printf( "\n" );
} /* end outer for */

/* display the mode value */
printf( "The mode is the most frequent value.\n"
"For this run the mode is %d which occured"
" %d times.\n", modeValue, largest );
} /* end function mode */

/* function that sorts an array with bubble sort algorithm */
void bubbleSort( int a[] )
{
int pass; /* pass counter */
int j; /* comparison */
int hold; /* temporary location used to swap elements */

/* loop to control number of passes */
for ( pass = 1; pass < SIZE; pass++ )
{
for ( j = 0; j < SIZE - 1; pass++ )
{
/* swap elements if out of order */
if ( a[ j ] > a[ j + 1 ] )
{
hold = a[ j ];
a[ j ] = a[ j + 1 ];
a[ j + 1 ] = hold;
} /* end if */

} /* end inner for */

} /* end outer for */

} /* end function bubbleSort */


/* output array contents (20 values per row) */
void printArray( const int a[] )
{
int j; /* counter */

/* output array contents */
for ( j = 0; j < SIZE; j++ )
{
if ( j % 20 == 0 )
printf( "\n" );

printf( "%2d", a[ j ] );

} /* end for */

} /* end function printArray */

 

[guinness.tony]

The program is an Example from Deitel and Deitel, chapter 6.
I typed it in carefully, I believe.

Possibly not carefully enough.

The program seems to freeze in the function median, where
median calls printArray(). printArray() prints only 80 elements
and then it seems to be in a loop of some type running
the cpu at 100%. well, freeze may not be the right terminology.
really it happens in the printArray(), because printArray()
does not finish printing.

printArray() outputs all 99 values, but does not issue a final
newline, so the last 19 values are still buffered in stdout,
waiting to be flushed out.

The problem lies in bubbleSort(). Have another look at the
loop variables: what they're initialised with, what their end
conditions are and, in particular, which variables are incremented
in which loop.

 

[Maxim Fomin]

The program is an Example from Deitel and Deitel, chapter 6.
I typed it in carefully, I believe.

The program seems to freeze in the function median, where
median calls printArray(). printArray() prints only 80 elements
and then it seems to be in a loop of some type running
the cpu at 100%. well, freeze may not be the right terminology.
really it happens in the printArray(), because printArray()
does not finish printing.

It hangs in bubbleSort() actually.

/* function that sorts an array with bubble sort algorithm */
void bubbleSort( int a[] )
{
int pass; /* pass counter */
int j; /* comparison */
int hold; /* temporary location used to swap elements */

/* loop to control number of passes */
for ( pass = 1; pass < SIZE; pass++ )
{
for ( j = 0; j < SIZE - 1; pass++ )
{
/* swap elements if out of order */
if ( a[ j ] > a[ j + 1 ] )
{
hold = a[ j ];
a[ j ] = a[ j + 1 ];
a[ j + 1 ] = hold;
} /* end if */

} /* end inner for */

} /* end outer for */

} /* end function bubbleSort */

In second for loop you should increment 'j', not 'pass'.

 

[BartC]

for ( pass = 1; pass < SIZE; pass++ )
{
for ( j = 0; j < SIZE - 1; pass++ )

Make that j++ instead of pass++, and try again.

(I've never liked C's for statement with having to write so much extra stuff
that you can get wrong. I now use a simple macro FOR(i,a,b).)

 

[Ben Bacarisse]

The program is an Example from Deitel and Deitel, chapter 6.
I typed it in carefully, I believe.

The program seems to freeze in the function median, where
median calls printArray(). printArray() prints only 80 elements
and then it seems to be in a loop of some type running
the cpu at 100%. well, freeze may not be the right terminology.
really it happens in the printArray(), because printArray()
does not finish printing.

No, it finishes. The problem is in the bubble sort function. On many
systems, output only appears when the line is complete (or the stream is
flushed) and that's what's confusing you. Print a final \n in
printArray and you'll see the last 19 elements.

The error in bubbleSort is easy to spot once you know to look there
(I'll leave you pleasure of finding it!).

<snip>

 

[James Kuyper]

The program is an Example from Deitel and Deitel, chapter 6.
I typed it in carefully, I believe.

The program seems to freeze in the function median, where
median calls printArray(). printArray() prints only 80 elements
and then it seems to be in a loop of some type running
the cpu at 100%. well, freeze may not be the right terminology.
really it happens in the printArray(), because printArray()
does not finish printing.

Actually, that's not the case. C streams can be either unbuffered, line
buffered, or fully buffered. Which of theses applies to a given stream
is up to the implementation, unless you override it by calling setbuf()
or setvbuf(). The standard prohibits stderr from being fully buffered,
and says that "When opened, a stream is fully buffered if and only if it
can be determined not to refer to an interactive device". It says the
same thing about stdin and stdout.

It would appear that on your system, as on mine, stdout is line
buffered. Since printArray() doesn't print a final newline, the last
line of output is incomplete, so the buffer never flushes, and that
incomplete line never appears on the screen. You should add printf("\n")
at the end of printArray().

Your program is actually stuck somewhere in bubbleSort(), which was my
first guess - that's the trickiest part of your program. I'll let you
have the fun of actually figuring out what the actual problem is.

 

[G G]

thanks everyone, i see the problem.
! how could i have missed that? !
yes, not careful enough.

i did not know about the buffering that takes place.
that's important to know. thanks.

you guys are great help.

g.