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

[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("\tPleas e 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<limi t;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<limi t;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

[Mabden]

"Karl Heinz Buchegger" <kb******@gasca d.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