array of series

I need to take as input positive integers 1-100. Each of those numbers will be the first element in series. The series will be formed by a specified formula, in case of pair integer- set 1 & case of uneven integer- set 2. Then of course print it.

I'm not quit sure about the syntax of this code & also compiler through me with many errors & warnings. Cannot seem to find why!

Please review it and have your notes.

Code:

:



#include<stdio.h>
#define N 5
#define M 5
   void seri(int A[][M],int *,int *);
  int main ( void )
{    
                    int A[N][M],*p,*n=0;
                    p=A;   
      printf ("\nenter numbers from 1 to 100\n");
            scanf ("%d", p);
         if(*p<1||*p>100)
          {
           puts(" Number is too big/ too small. Try again!");
          }
        else
          {  
                              while (*p!=-1)      
                                        {
                                        for (p=A;p-A<M;p++)
                                                  {
                                        puts ("next number:");
                scanf ("%d", p++);
                seri(A,n,p);
                }
}



          void seri(int A[N][M],int *n,int *p)
           {                                     
            int t;
                     for (p=A;p-A>=1;p++)
                   { 
                      if (*p%2==0)
                       *p= *p/2;
                     else  
                       {
                      *p=3*(*p)+1;
                        }
               n=p-A;       
                    }                                     

              printf("n= %d. seri is :", n);

                for (t=0;t<=n;t++)

                        printf("  %d\n", *p);

                               }            

            }

TIA


Ronen

Why is it that whenever anyone says they get compiler errors and warnings, they never think to post them?

Well for starters, use [code] [/code] tags around any posted code. Had you actually taken the time to read the Announcements, you'd have known this already.

Next off, you have an unmatched number of { } pairs. Fix your indenting, and you'd have seen that.

Quzah.

Geee!!! Do you know what a pointer is?!?! Or an array?
First error:

Code:

int A[N][M],*p,*n=0;
p=A;

The problem is you're assigning a double pointer (the two-dimension array A) to a single pointer (the var *p)
Do like this

Code:

int A[N][M],*p,*n=0;
int x=0, y=0;
p=&(A[x][y]);
//where x and y are the indexes of the number you want to edit

This goes on for the entire source.

Plus.. this is awfull: for (p=A;p-A<M;p++)
More, you don't use *n for nothing... But I'll keep it
In this assignment int *n=0, you saying that you have a pointer that point to memory adress 0, or in other word NULL. If you want to point something you must first allocate some memory to that pointer:

Code:

n=malloc(sizeof(int))

A detail: if you write int *n=0, your defining the variable value.
If you then write *n=0, after n being defined, then you're altering the pointed int.

Specify better want you want to do..

Where's another one....

Code:

#include<stdio.h>
#define N 5
#define M 5

void seri(int A[N][M],int x);

int main(){    
	int A[N][M],*p, n=0;
	int x = 0, y = 0;
	p=&(A[x][y]);
	
	printf ("\nenter numbers from 1 to 100\n");

	while(x<N){
		scanf("%d", &(A[x][0]));
		if(A[x][0]<1||A[x][0]>100)
			puts(" Number is too big/ too small. Try again!");
		else{
			seri(A,x);
			x++;
			if(x<N)	puts ("next number:");
		}
	}
}

void seri(int A[N][M],int x){                                     
	int i, n;
	
	for(i=0;i<M-1;i++){
		if (A[x][i]%2==0)
			A[x][i+1]= A[x][i]/2;
		else
			A[x][i+1]=3*A[x][i]+1;
		n=A[x][i+1]-A[x][0];
	}                                     

	printf("n = %d. seri is :\n", n);
	for(i=0;i<M;i++)
		printf("\t%d\n", A[x][i]);
}

Is this the seri function you wanted??

Appologize for the uncomplete info regarding the drill:
the series needs to begin in the entered integer at the beginnig & end with 1. Delimeter for the end of input is -1.
like:
input - 10
output
10 5 16 8 4 2 1.

all input no. has to be entered together & all output data has to come out at the end as a bulk.

like:

10 5 16 8 4 2 1
11 34 17 51 26 13 40 20 10 5 16 8 4 2 1
.
.
.
& finaly n reperents the index of elements in each line. In other words: how many numbers in line.

hope I have cleared everything now...

TIA.

what do you think about this one?

Code:

:

#include<stdio.h>
#define N 5
#define M 900

void seri(int A[N][M],int x,int y,int n);

int main()
{    
	int A[N][M],*p, n=0;
	int x = 0, y = 0;
	p=&(A[x][y]);
	
	printf ("\nenter numbers from 1 to 100\n");

	while(x<N)
	{
		scanf("%d", &(A[x][0]));
		if(A[x][0]<1||A[x][0]>100)
			puts(" Number is too big/ too small. Try again!");
		else
		{
			seri(A,x,y,n);
			x++;
			if(x<N)	puts ("next number:");
		}
	}
}

void seri(int A[N][M],int x,int y,int n)
{                                    
	
	for(A[x][0];A[x][y]>1;y++)
	{
		if (A[x][y]%2==0)
			A[x][y+1]= A[x][y]/2;
		else
			A[x][y+1]=3*A[x][y]+1;
		n=y;
		printf("\t%d\n", A[x][y]);
	}                                     

	printf("n = %d. seri is :\n", n);
	
}

Now I miss
1. the 1 at the end. If I write:

Code:

for(A[x][0];A[x][y]=>1;y++)

I get compiler error:
error C2059: syntax error : '>'
error C2143: syntax error : missing ';' before ')'
error C2181: illegal else without matching if
error C2143: syntax error : missing ')' before 'string'
error C2143: syntax error : missing '{' before 'string'
error C2059: syntax error : '<Unknown>'
error C2059: syntax error : ')'
error C2059: syntax error : '}'


2. the bulk printing at output mentioned above.

Switch them around. You don't put the equal sign first on the compound operators like that:

Code:

*=
/=
+=
-=
%=
|=
&=
~=
<<=
>>=
^=
!=
<=
>=

Quzah.
*Yeah, I know it's not there, but read the first sentence to figure out why I didn't list it.
**If you don't know what I'm talking about, ignore these.
***If you're really confused now close your browser.

tried this too before"

Code:

for(A[x][0];A[x][y]>=1;y++)

Then there is some kind of bug, or I dont know what...
I get very long list of:

enter numbers from 1 to 100
6


6
3
10
5
16
8
4
2
1
4
2
1
4
2
1
.
.

it is longer the array size, but with respect to it. For instance on a 2*3 array, I got ~44 lines. On larger arrays- more lines etc...

>for(A[x][0];A[x][y]>=1;y++)
This will loop until A[x][y] is 0. Your series isn't very good at getting that particular result. Just use > and print the last number outside of the loop:

Code:

void seri(int A[N][M],int x,int y,int n)
{
  for(;A[x][y]>1;y++)
  {
    if (A[x][y]%2==0)
      A[x][y+1]= A[x][y]/2;
    else
      A[x][y+1]=3*A[x][y]+1;
    n=y;
    printf("\t%d\n", A[x][y]);
  }
  printf("\t%d\n", A[x][y]);
  printf("n = %d. seri is :\n", n);
}

How about the "bulk" printing in output mentioned above?

>How about the "bulk" printing in output mentioned above?
Here's that feature along with a bit of cleaning up I added for fun:

Code:

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

#define NSERIES    5
#define SERIES_LEN 900

int  in_range(int value, int low, int high);
void seri(int series[SERIES_LEN]);

int main(void)
{    
  int series_list[NSERIES][SERIES_LEN];
  int x = 0;
  int y;

  while (x < NSERIES)
  {
    printf("Enter a number from 1 to 100: ");
    fflush(stdout);
    /* Always error check scanf */
    if (scanf("%d", &series_list[x][0]) != 1)
    {
      fprintf(stderr, "Invalid number\n");
      return EXIT_FAILURE;
    }
    if (!in_range(series_list[x][0], 1, 100))
      fprintf(stderr, "Number is too big/small. Try again!\n");
    else
      seri(series_list[x++]);
  }
  /* Bulk print */
  for (x = 0; x < NSERIES; x++)
  {
    for (y = 0; y < SERIES_LEN - 1 && series_list[x][y] > 1; y++)
      printf("%-3d ", series_list[x][y]);
    printf("%-3d ", series_list[x][y++]);
    printf("\nn = %d\n", series_list[x][y]);
  }

  return EXIT_SUCCESS;
}

int in_range(int value, int low, int high)
{
  return value >= low && value <= high;
}

void seri(int series[SERIES_LEN])
{
  int y;
  int n = 0;

  /* Place n in the last spot */
  for (y = 0; y < SERIES_LEN - 1 && series[y] > 1; y++)
  {
    if (series[y] % 2 == 0)
      series[y + 1] = series[y] / 2;
    else
      series[y + 1] = 3 * series[y] + 1;
    n = y;
  }
  series[++y] = n;
}

Wow, cool!