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C problem, likely simple but I can't see it!

P: n/a
sam
hi, i am new to C/C++ and have just finished a program based on the UK
national lottery, where you enter 6 numbers, six are generated by the
computer and there are appropriate messages and a compare function to decide
your winnings. I am using Microsoft's C++ version 6 standard, the program
is as follows:

/* * * * * * * * * * * * * * * * * * * *
* *
* Lottery Simulation By Sam Halston *
* *
* * * * * * * * * * * * * * * * * * * * */


#include <stdio.h>
#include <stdlib.h>
#include <time.h>

#define NMAX 6
int a[NMAX];
int b[NMAX];
void getIntArray(int x[]);
void bubbleSort(int x[]);
void printArrays(int x[], int y[]);
void getrandomarray(int x[]);
void winnings();
int wins = 0;

int main(void) {
//welcome
printf("\t Welcome To The National Lottery!\n");
printf("\tPlease Enter Six Whole Numbers Between 1 And 49\n\n");

getIntArray(a);
getrandomarray(b);
bubbleSort(a);
bubbleSort(b);
printf("Your Numbers In Numerical Order Are: \n");
printArrays(a, b);
winnings();
return 0;
}


void getIntArray(int x[])

{
int n = 0;
int temp;

here://I know goto loops are not liked in C, but this is the best way to
achieve this feature in my opinion!
;

do {
printf("Choose A Number\n");
scanf("%d", &temp);
if (temp==0) break;
if (n==5)
{
printf("All Six Numbers Have Been Entered, Press Zero To Confirm\n");

}

if ((temp<01) || (temp>49))
{
printf("%s\n", "Sorry, That Is An Unacceptable Number. Please Re Enter");
goto here;
}

if (temp==a[0])
{
printf("%s\n", "Sorry, You Have Already Selected That Number. Please Make
Another Choice");
goto here;
}
if (temp==a[1])
{
printf("%s\n", "Sorry, You Have Already Selected That Number. Please Make
Another Choice");
goto here;
}
if (temp==a[2])
{
printf("%s\n", "Sorry, You Have Already Selected That Number. Please Make
Another Choice");
goto here;
}
if (temp==a[3])
{
printf("%s\n", "Sorry, You Have Already Selected That Number. Please Make
Another Choice");
goto here;
}
if (temp==a[4])
{
printf("%s\n", "Sorry, You Have Already Selected That Number. Please Make
Another Choice");
goto here;
}
if (temp==a[5])
{
printf("%s\n", "Sorry, You Have Already Selected That Number. Please Make
Another Choice");
goto here;
}

else
a[n++] = temp;
}while (1);
}
void bubbleSort(int x[])
/* It sorts in non-decreasing order the six positions of a. It uses
* a technique called the bubble sort method.
*/
{
int lcv;
int limit = NMAX-1;
int temp;
int lastChange;

while (limit) {
lastChange = 0;
for (lcv=0;lcv<limit;lcv++)

if (x[lcv]>x[lcv+1]) {
temp = x[lcv];
x[lcv] = x[lcv+1];
x[lcv+1] = temp;
lastChange = lcv;
}
limit = lastChange;
}
}
void printArrays(int x[], int y[])

/* These values are printed out, six per line. */
{
//print user array
for (int i=0; i<NMAX;)
{
printf("\t%d ", x[i++]);
if (i%6==0)
printf("\n");
}
printf("\n");

printf("The Drawn Numbers In Numerical Order Are: \n");
//print random array

for (i=0; i<NMAX;){
printf("\t%d ", y[i++]);
if (i%6==0)
printf("\n");
}
printf("\n");

}

void getrandomarray(int x[])
{

//make random numbers
srand((unsigned int)time((time_t *)NULL));

for (int i=0; i<NMAX; i++)
{
x[i] = (rand()%49)+1; //This safety feature makes the
generator regenerate should any
for(int j=0; j<i; j++) //of its numbers be the same as previous
numbers.
{
if(x[i] == x[j])
i--;
}
}

}
void bubblesort(int x[])
/* It sorts in non-decreasing order the first 6 positions of b. It uses
* the bubble sort method.
*/
{
int lcv;
int limit = NMAX-1;
int temp;
int lastChange;

while (limit)
{
lastChange = 0;
for (lcv=0;lcv<limit;lcv++)
{
/* Notice that the values in positions LIMIT+1 .. in
* their final position, i.e. they are sorted right */
if (a[lcv]>a[lcv+1])
{
temp = x[lcv];
x[lcv] = x[lcv+1];
x[lcv+1] = temp;
lastChange = lcv;
}
limit = lastChange;
}

}
}
void winnings()
{

for(int i=0; i<6; i++)
{
for(int j=0; j<6; j++)
{
if(a[i]==b[j])
{
wins++;
}
}

if(wins==0)
{
printf("You Haven't Matched Any Numbers, Sorry\n");
return;
}
if(wins==1)
{
printf("You Have Matched One Number, Sorry\n");
return;
}

}
if(wins==2)
{
printf("You Have Matched Two Numbers, Sorry\n");
return;
}
if(wins==3)
{
printf("You Have Matched 3 Numbers. Well Done, You Have Won 10\n");
return;
}
if(wins==4)
{
printf("You Have Matched 4 Numbers. Well Done, You Have Won 500\n");
return;
}
if(wins==5)
{
printf("You Have Matched 5 Numbers. Well Done, You Have Won 300,000!\n");
return;
}
if(wins==6)
{
printf("You Have Matched 3 Numbers. Congratulations, You Have Won
5,000,000!!\n");
return;
}

}
ok-so i guess its a mess to most of you, it compiles and runs no bother,
just doesnt give a correct readout at the end of how many numbers have
matched! iv looed at this for 4 days now, seems to me that only one or no
numbers are being matched almost as if its doing a true and false as to
whether any numbers match. trouble is, the lottery rules are 0, 1 or 2
numbers matched = 0, 3 =10, 4=500, 5=300,000, 6= 5,000,000. any quick
fixes or suggestions as to an alternative end will be greatly appreciated,
but if i could stick to this structure it would be good as i am trying hard
to understand even what i've written! did anyone find this or am i just
doomed to not be a programmer!?

thanks in advance

sam
Jul 22 '05 #1
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10 Replies


P: n/a

"sam" <sam@home> wrote in message
news:q4********************@brightview.com...
hi, i am new to C/C++ and have just finished a program based on the UK
national lottery, where you enter 6 numbers, six are generated by the
computer and there are appropriate messages and a compare function to decide your winnings. I am using Microsoft's C++ version 6 standard, the program
is as follows:

/* * * * * * * * * * * * * * * * * * * *
* *
* Lottery Simulation By Sam Halston *
* *
* * * * * * * * * * * * * * * * * * * * */


#include <stdio.h>
#include <stdlib.h>
#include <time.h>
in C++, these should be

#include <cstdio>
#include <cstdlib>
#include <ctime>

#define NMAX 6
int a[NMAX];
int b[NMAX];
void getIntArray(int x[]);
void bubbleSort(int x[]);
void printArrays(int x[], int y[]);
void getrandomarray(int x[]);
void winnings();
int wins = 0;

int main(void) {
//welcome
printf("\t Welcome To The National Lottery!\n");
printf("\tPlease Enter Six Whole Numbers Between 1 And 49\n\n");

getIntArray(a);
getrandomarray(b);
bubbleSort(a);
bubbleSort(b);
printf("Your Numbers In Numerical Order Are: \n");
printArrays(a, b);
winnings();
return 0;
}
good short program. Often newcomers make hidiously long programs, good one!


void getIntArray(int x[])

{
int n = 0;
int temp;

here://I know goto loops are not liked in C, but this is the best way to
achieve this feature in my opinion!
;

do {
printf("Choose A Number\n");
scanf("%d", &temp);
if (temp==0) break;
if (n==5)
{
printf("All Six Numbers Have Been Entered, Press Zero To Confirm\n");

if the last number is entered invalid, then you are asking them to press
zero, which is invalid, infinite loop!

}

if ((temp<01) || (temp>49))
{
printf("%s\n", "Sorry, That Is An Unacceptable Number. Please Re Enter");
goto here;
}

if (temp==a[0])
{
printf("%s\n", "Sorry, You Have Already Selected That Number. Please Make
Another Choice");
goto here;
}
if (temp==a[1])
{
printf("%s\n", "Sorry, You Have Already Selected That Number. Please Make
Another Choice");
goto here;
}
why didnt you do:
for (int loop=0; loop<NMAX; ++loop)
if (temp==a[loop])
// ...

For two reasons. One its easier to read as its far shorter. Two, its more
maintainable. What would you do if you wanted to do a lottery with 8
numbers? Or consider 1000, what would you do then? (i know its extreme, bit
its meant to make a point)

if (temp==a[2])
{
printf("%s\n", "Sorry, You Have Already Selected That Number. Please Make
Another Choice");
goto here;
}
if (temp==a[3])
{
printf("%s\n", "Sorry, You Have Already Selected That Number. Please Make
Another Choice");
goto here;
}
if (temp==a[4])
{
printf("%s\n", "Sorry, You Have Already Selected That Number. Please Make
Another Choice");
goto here;
}
if (temp==a[5])
{
printf("%s\n", "Sorry, You Have Already Selected That Number. Please Make
Another Choice");
goto here;
}

else
a[n++] = temp;
}while (1);
}
void bubbleSort(int x[])
/* It sorts in non-decreasing order the six positions of a. It uses
* a technique called the bubble sort method.
*/
{
int lcv;
int limit = NMAX-1;
int temp;
int lastChange;

while (limit) {
lastChange = 0;
for (lcv=0;lcv<limit;lcv++)

if (x[lcv]>x[lcv+1]) {
temp = x[lcv];
x[lcv] = x[lcv+1];
x[lcv+1] = temp;
lastChange = lcv;
}
limit = lastChange;
}
}
there are shorter ways of doing this, e.g.

for (int i=0; i<NMAX; ++i)
for (int j=i; j<NMAX; ++j)
{
// compare and swap items
}



void printArrays(int x[], int y[])

/* These values are printed out, six per line. */
{
//print user array
for (int i=0; i<NMAX;)
{
printf("\t%d ", x[i++]);
if (i%6==0)
this prints a newline after the first number only, why? Again, you should
use NMAX for maintainability
printf("\n");
}
printf("\n");

printf("The Drawn Numbers In Numerical Order Are: \n");
//print random array

for (i=0; i<NMAX;){
printf("\t%d ", y[i++]);
if (i%6==0)
printf("\n");
}
printf("\n");

}

void getrandomarray(int x[])
{

//make random numbers
srand((unsigned int)time((time_t *)NULL));

for (int i=0; i<NMAX; i++)
{
x[i] = (rand()%49)+1; //This safety feature makes the
generator regenerate should any
the rand() function requires you to seed the random number generator, do
something like
srand(time(NULL)); // seed rand gen based on time

also, the lower bits returned by rand() are not very random, consider using
something like:
rand() / (RAND_MAX / 49);

for(int j=0; j<i; j++) //of its numbers be the same as previous
numbers.
{
if(x[i] == x[j])
i--;
}
}

}
void bubblesort(int x[])
this has already been declared!
/* It sorts in non-decreasing order the first 6 positions of b. It uses
* the bubble sort method.
*/
{
int lcv;
int limit = NMAX-1;
int temp;
int lastChange;

while (limit)
{
lastChange = 0;
for (lcv=0;lcv<limit;lcv++)
{
/* Notice that the values in positions LIMIT+1 .. in
* their final position, i.e. they are sorted right */
if (a[lcv]>a[lcv+1])
{
temp = x[lcv];
x[lcv] = x[lcv+1];
x[lcv+1] = temp;
lastChange = lcv;
}
limit = lastChange;
}

}
}
void winnings()
{

for(int i=0; i<6; i++)
{
for(int j=0; j<6; j++)
{
if(a[i]==b[j])
{
wins++;
}
}
this will only work if the numbers are in the same index of the array, e.g.
if you have
1,2,3,4,5,6
and
2,3,4,5,6,7

then this will report no numbers the same! Even though 5 are the same

you need to use two loops to do this.

if(wins==0)
{
printf("You Haven't Matched Any Numbers, Sorry\n");
return;
}
if(wins==1)
{
printf("You Have Matched One Number, Sorry\n");
return;
}

}
if(wins==2)
{
printf("You Have Matched Two Numbers, Sorry\n");
return;
}
if(wins==3)
{
printf("You Have Matched 3 Numbers. Well Done, You Have Won 10\n");
return;
}
if(wins==4)
{
printf("You Have Matched 4 Numbers. Well Done, You Have Won 500\n");
return;
}
if(wins==5)
{
printf("You Have Matched 5 Numbers. Well Done, You Have Won 300,000!\n"); return;
}
if(wins==6)
{
printf("You Have Matched 3 Numbers. Congratulations, You Have Won
5,000,000!!\n");
return;
}
i would use something like:
static char[NMAX][]sWins = {"0", "0", "10", "300,000", "5,000,000"};
printf("You have won %s", sWins[wins]);

which I am sure you agree, is far neater

}
ok-so i guess its a mess to most of you, it compiles and runs no bother,
just doesnt give a correct readout at the end of how many numbers have
matched! iv looed at this for 4 days now, seems to me that only one or no
numbers are being matched almost as if its doing a true and false as to
whether any numbers match. trouble is, the lottery rules are 0, 1 or 2
numbers matched = 0, 3 =10, 4=500, 5=300,000, 6= 5,000,000. any quick fixes or suggestions as to an alternative end will be greatly appreciated,
but if i could stick to this structure it would be good as i am trying hard to understand even what i've written! did anyone find this or am i just
doomed to not be a programmer!?

thanks in advance

sam


you have made a good attempt. The program seems well structured apart from
your 'goto'. Instead of the goto have your structure do something like:

int ValidNums=0;
while(1)
{
// get a number
// check valid, if valid then ValidNums++
if (ValidNums == NMAX-1)
break;
}

but consider a quit option to allow the user to quit gracefully.
Also, scanf is not brilliant for getting numbers, try typing a letter and
find out! In C, I used to do scanf("%s"); and then use strtol() to
check/convert it to a number. However there are better C++ solutions that
somebody may point out.

Hope this helps
Allan
Jul 22 '05 #2

P: n/a
sam writes:
hi, i am new to C/C++ and have just finished a program based on the UK
national lottery, where you enter 6 numbers, six are generated by the
computer and there are appropriate messages and a compare function to decide your winnings. I am using Microsoft's C++ version 6 standard, the program
is as follows:

/* * * * * * * * * * * * * * * * * * * *
* *
* Lottery Simulation By Sam Halston *
* *
* * * * * * * * * * * * * * * * * * * * */


#include <stdio.h>
#include <stdlib.h>
#include <time.h>

#define NMAX 6
int a[NMAX];
int b[NMAX]; <snip>

Try to use more descriptive names than a and b. One is a client and the
other is the government, is it not? It would be easier to help if these had
some meaning.
void getIntArray(int x[]);
void bubbleSort(int x[]);
void printArrays(int x[], int y[]);
void getrandomarray(int x[]);
void winnings();
int wins = 0;

int main(void) {
// welcome
printf("\t Welcome To The National Lottery!\n");
printf("\tPlease Enter Six Whole Numbers Between 1 And 49\n\n");

getIntArray(a);
getrandomarray(b);
bubbleSort(a);
bubbleSort(b);
printf("Your Numbers In Numerical Order Are: \n");
printArrays(a, b);


Are you happy with the results of that print? Knowing the answer to that
divides the problem into two major "hunks".

I scanned around but didn't spot anything noteworthy. I suppose there may
well be an answer by the time I post this.

<snip>
Jul 22 '05 #3

P: n/a
I've added som comments to your code, hope that helps.

/Andreas
"sam" <sam@home> skrev i meddelandet
news:q4********************@brightview.com...
hi, i am new to C/C++ and have just finished a program based on the UK
national lottery, where you enter 6 numbers, six are generated by the
computer and there are appropriate messages and a compare function to decide your winnings. I am using Microsoft's C++ version 6 standard, the program
is as follows:

/* * * * * * * * * * * * * * * * * * * *
* *
* Lottery Simulation By Sam Halston *
* *
* * * * * * * * * * * * * * * * * * * * */


#include <stdio.h>
#include <stdlib.h>
#include <time.h>

#define NMAX 6
int a[NMAX];
int b[NMAX];
You definately should initiate these, like
init a[NMAX] = null;
or so.
void getIntArray(int x[]);
void bubbleSort(int x[]);
void printArrays(int x[], int y[]);
void getrandomarray(int x[]);
void winnings();
int wins = 0;

int main(void) {
//welcome
printf("\t Welcome To The National Lottery!\n");
printf("\tPlease Enter Six Whole Numbers Between 1 And 49\n\n");

getIntArray(a);
getrandomarray(b);
bubbleSort(a);
bubbleSort(b);
printf("Your Numbers In Numerical Order Are: \n");
printArrays(a, b);
winnings();
return 0;
}


void getIntArray(int x[])

{
int n = 0;
int temp;

here://I know goto loops are not liked in C, but this is the best way to
achieve this feature in my opinion!
;

do {
printf("Choose A Number\n");
scanf("%d", &temp);
if (temp==0) break;
if (n==5)
{
printf("All Six Numbers Have Been Entered, Press Zero To Confirm\n");
}

if ((temp<01) || (temp>49))
{
printf("%s\n", "Sorry, That Is An Unacceptable Number. Please Re Enter");
goto here;
}

if (temp==a[0])
{
printf("%s\n", "Sorry, You Have Already Selected That Number. Please Make
Another Choice");
goto here;
}
if (temp==a[1])
{
printf("%s\n", "Sorry, You Have Already Selected That Number. Please Make
Another Choice");
goto here;
}
if (temp==a[2])
{
printf("%s\n", "Sorry, You Have Already Selected That Number. Please Make
Another Choice");
goto here;
}
if (temp==a[3])
{
printf("%s\n", "Sorry, You Have Already Selected That Number. Please Make
Another Choice");
goto here;
}
if (temp==a[4])
{
printf("%s\n", "Sorry, You Have Already Selected That Number. Please Make
Another Choice");
goto here;
}
if (temp==a[5])
{
printf("%s\n", "Sorry, You Have Already Selected That Number. Please Make
Another Choice");
goto here;
}

else
a[n++] = temp;
}while (1);
}
void bubbleSort(int x[])
/* It sorts in non-decreasing order the six positions of a. It uses
* a technique called the bubble sort method.
*/
{
int lcv;
int limit = NMAX-1;
int temp;
int lastChange;

while (limit) {
lastChange = 0;
for (lcv=0;lcv<limit;lcv++)

if (x[lcv]>x[lcv+1]) {
temp = x[lcv];
x[lcv] = x[lcv+1];
x[lcv+1] = temp;
lastChange = lcv;
}
limit = lastChange;
}
}
void printArrays(int x[], int y[])

/* These values are printed out, six per line. */
{
//print user array
for (int i=0; i<NMAX;)
{
printf("\t%d ", x[i++]);
if (i%6==0)
printf("\n");
}
printf("\n");

printf("The Drawn Numbers In Numerical Order Are: \n");
//print random array

for (i=0; i<NMAX;){
printf("\t%d ", y[i++]);
if (i%6==0)
printf("\n");
}
printf("\n");

}

void getrandomarray(int x[])
{

//make random numbers
srand((unsigned int)time((time_t *)NULL));

for (int i=0; i<NMAX; i++)
{
x[i] = (rand()%49)+1; //This safety feature makes the
generator regenerate should any
for(int j=0; j<i; j++) //of its numbers be the same as previous
numbers.
{
if(x[i] == x[j])
i--;
}
}

}
void bubblesort(int x[])
/* It sorts in non-decreasing order the first 6 positions of b. It uses
* the bubble sort method.
*/
{
int lcv;
int limit = NMAX-1;
int temp;
int lastChange;

while (limit)
{
lastChange = 0;
for (lcv=0;lcv<limit;lcv++)
{
/* Notice that the values in positions LIMIT+1 .. in
* their final position, i.e. they are sorted right */
if (a[lcv]>a[lcv+1])
{
temp = x[lcv];
x[lcv] = x[lcv+1];
x[lcv+1] = temp;
lastChange = lcv;
}
limit = lastChange;
}

}
}
void winnings()
{

for(int i=0; i<6; i++)
{
for(int j=0; j<6; j++)
{
if(a[i]==b[j])
{
wins++;
}
}

You forgot to end the first for-loop! Add one '}'!
if(wins==0)
{
printf("You Haven't Matched Any Numbers, Sorry\n");
return;
}
if(wins==1)
{
printf("You Have Matched One Number, Sorry\n");
return;
}

}
The above '}' shouldn't be there!
if(wins==2)
{
printf("You Have Matched Two Numbers, Sorry\n");
return;
}
if(wins==3)
{
printf("You Have Matched 3 Numbers. Well Done, You Have Won 10\n");
return;
}
if(wins==4)
{
printf("You Have Matched 4 Numbers. Well Done, You Have Won 500\n");
return;
}
if(wins==5)
{
printf("You Have Matched 5 Numbers. Well Done, You Have Won 300,000!\n"); return;
}
if(wins==6)
{
printf("You Have Matched 3 Numbers. Congratulations, You Have Won
5,000,000!!\n");
return;
}

}
ok-so i guess its a mess to most of you, it compiles and runs no bother,
just doesnt give a correct readout at the end of how many numbers have
matched! iv looed at this for 4 days now, seems to me that only one or no
numbers are being matched almost as if its doing a true and false as to
whether any numbers match. trouble is, the lottery rules are 0, 1 or 2
numbers matched = 0, 3 =10, 4=500, 5=300,000, 6= 5,000,000. any quick fixes or suggestions as to an alternative end will be greatly appreciated,
but if i could stick to this structure it would be good as i am trying hard to understand even what i've written! did anyone find this or am i just
doomed to not be a programmer!?

thanks in advance

sam

Jul 22 '05 #4

P: n/a
You definately should initiate these, like
init a[NMAX] = null;
or so.

If you mean initialize, they are by default. What is

init a[NMAX] = null;

Suppose to mean? Do you mean:

int a[NMAX] = {0};

?


Jul 22 '05 #5

P: n/a

"Xenos" <do**********@spamhate.com> skrev i meddelandet
news:c7*********@cui1.lmms.lmco.com...
You definately should initiate these, like
init a[NMAX] = null;
or so.

If you mean initialize, they are by default. What is

init a[NMAX] = null;

Suppose to mean? Do you mean:

int a[NMAX] = {0};

?


Yes, i'm sorry. It was a long time ago since i coded C.

/Andreas
Jul 22 '05 #6

P: n/a
On Fri, 7 May 2004 18:55:04 +0100, "sam" <sam@home> wrote:
hi, i am new to C/C++ and have just finished a program based on the UK
national lottery, where you enter 6 numbers, six are generated by the
computer and there are appropriate messages and a compare function to decide
your winnings. I am using Microsoft's C++ version 6 standard, the program
is as follows:

"Before":
void winnings()
{

for(int i=0; i<6; i++)
{
for(int j=0; j<6; j++)
{
if(a[i]==b[j])
{
wins++;
}
}

if(wins==0)
{
printf("You Haven't Matched Any Numbers, Sorry\n");
return;
}
if(wins==1)
{
printf("You Have Matched One Number, Sorry\n");
return;
}

}
if(wins==2)
{
printf("You Have Matched Two Numbers, Sorry\n");
return;
}

...
"After":

void winnings()
{
printf("a: ");
for (int i = 0; i < 6; i++)
printf("%d ", a[i]);
printf("\nb: ");
for (int i = 0; i < 6; i++)
printf("%d ", b[i]);

for(int i=0; i<6; i++)
{
for(int j=0; j<6; j++)
{
if(a[i]==b[j])
{
wins++;
}
}

if(wins==0)
{
printf("You Haven't Matched Any Numbers, Sorry\n");
return;
}
if(wins==1)
{
printf("You Have Matched One Number, Sorry\n");
return;
}

}
if(wins==2)
{
printf("You Have Matched Two Numbers, Sorry\n");
return;
}
// ....

Notice anything different? ;-)
-leor
--
Leor Zolman --- BD Software --- www.bdsoft.com
On-Site Training in C/C++, Java, Perl and Unix
C++ users: download BD Software's free STL Error Message Decryptor at:
www.bdsoft.com/tools/stlfilt.html
Jul 22 '05 #7

P: n/a
"sam" <sam@home> wrote in message
news:q4********************@brightview.com...
hi, i am new to C/C++ and have just finished a program based on the UK
national lottery, where you enter 6 numbers, six are generated by the
computer and there are appropriate messages and a compare function to decide your winnings. I am using Microsoft's C++ version 6 standard, the program
is as follows:

<snip>

void winnings()
{

for(int i=0; i<6; i++)
{
for(int j=0; j<6; j++)
{
if(a[i]==b[j])
{
wins++;
A "break;" would work well here, as you are done processing once you find
the match, since they are sorted.

}
}

You need a } HERE, it was moved down below the following...
if(wins==1)
{
printf("You Have Matched One Number, Sorry\n");
return;
}

}
....to HERE. So fix that.

Also:
if(wins==6)
{
printf("You Have Matched 3 Numbers. Congratulations, You Have Won
5,000,000!!\n");
return;
}
wins == 6, but the message says 3.

any quick
fixes or suggestions as to an alternative end will be greatly appreciated,
but if i could stick to this structure it would be good as i am trying hard to understand even what i've written! did anyone find this or am i just
doomed to not be a programmer!?


Well.... the code, in general, sucks, but then we all start somewhere. ;-)

--
Mabden

Jul 22 '05 #8

P: n/a
sam
thanks for the help, it's looking a lot tidier now! i still cant get the
numbers to compare properly at the end though, only 0 and 1 numbers works,
as if im still doing a true or false comparison, why is it so hard to
compare numbers!? im trying to get at:

if the first number from the user inputted array equals or is equivalent to
any of the numbers from the random generator, add 1.

if the second.. etc

then we should have up to 5 (6) matches

i thought of covering all possible matches one at a time, ie if a[0] = b[0],
b[1] etc, but that would take ages and won't look so good!
Jul 22 '05 #9

P: n/a
sam wrote:

thanks for the help, it's looking a lot tidier now! i still cant get the
numbers to compare properly at the end though, only 0 and 1 numbers works,
as if im still doing a true or false comparison, why is it so hard to
compare numbers!?


It isn't.
Let me show you something. The beauty of consistent code indentation and how
it helps to identify some type of bugs.

Lets look at your winnings function(), but this time reformatted such
that the indentation is consistent, correct and stands out a little bit more
by using 2 spaces instead of just 1

void winnings()
{

for(int i=0; i<6; i++)
{
for(int j=0; j<6; j++)
{
if(a[i]==b[j])
{
wins++;
}
}

if(wins==0)
{
printf("You Haven't Matched Any Numbers, Sorry\n");
return;
}
if(wins==1)
{
printf("You Have Matched One Number, Sorry\n");
return;
}
}

if(wins==2)
{
printf("You Have Matched Two Numbers, Sorry\n");
return;
}
if(wins==3)
{
printf("You Have Matched 3 Numbers. Well Done, You Have Won 10\n");
return;
}
if(wins==4)
{
printf("You Have Matched 4 Numbers. Well Done, You Have Won 500\n");
return;
}
if(wins==5)
{
printf("You Have Matched 5 Numbers. Well Done, You Have Won 300,000!\n");
return;
}
if(wins==6)
{
printf("You Have Matched 3 Numbers. Congratulations, You Have Won 5,000,000!!\n");
return;
}
}

Do you notice something?
The if-s handling the case for 0 and 1 are not at the same indentation level
then the other cases! Also the if-s for cases 0 and 1 are at the complete
wrong indentation level. You need to compare *all* numbers, not just a[0] with
all b[j] to decide if there are 0 or 1 matches.

--
Karl Heinz Buchegger
kb******@gascad.at
Jul 22 '05 #10

P: n/a
"Karl Heinz Buchegger" <kb******@gascad.at> wrote in message
news:40***************@gascad.at...
sam wrote:

thanks for the help, it's looking a lot tidier now! i still cant get the
numbers to compare properly at the end though, only 0 and 1 numbers works, as if im still doing a true or false comparison, why is it so hard to
compare numbers!?
It isn't.
Let me show you something. The beauty of consistent code indentation and

how it helps to identify some type of bugs.

Lets look at your winnings function(), but this time reformatted such
that the indentation is consistent, correct and stands out a little bit more by using 2 spaces instead of just 1

void winnings()
{

for(int i=0; i<6; i++)
{
for(int j=0; j<6; j++)
{
if(a[i]==b[j])
{
wins++;
}
}

if(wins==0)
{
printf("You Haven't Matched Any Numbers, Sorry\n");
return;
}
if(wins==1)
{
printf("You Have Matched One Number, Sorry\n");
return;
}
}

if(wins==2)
{
printf("You Have Matched Two Numbers, Sorry\n");
return;
}
if(wins==3)
{
printf("You Have Matched 3 Numbers. Well Done, You Have Won 10\n");
return;
}
if(wins==4)
{
printf("You Have Matched 4 Numbers. Well Done, You Have Won 500\n");
return;
}
if(wins==5)
{
printf("You Have Matched 5 Numbers. Well Done, You Have Won 300,000!\n"); return;
}
if(wins==6)
{
printf("You Have Matched 3 Numbers. Congratulations, You Have Won 5,000,000!!\n"); return;
}
}

Do you notice something?
The if-s handling the case for 0 and 1 are not at the same indentation level then the other cases! Also the if-s for cases 0 and 1 are at the complete
wrong indentation level. You need to compare *all* numbers, not just a[0] with all b[j] to decide if there are 0 or 1 matches.


<Uncle voice>"And one more thing..."

Am I the only one who notices case 6: "You Have Matched 3 Numbers"?

--
Mabden
Jul 22 '05 #11

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