Thread: 1d array into 2d array

One last question and again thanks everyone for your help I'm just impatient and instead of waiting till Monday to hear back from my teacher I was hoping someone here might know. Is there a limitation of the number of integers C can display on one line? It seems that even though I want 4 rows it put a 2 more rows because it is not able to list 25 integers on one line. Or is this a limitation of visual studio? Thanks again. I put the full code in case anyone wants to see what it prints out.

Code:

void main()
{
 int change;
 int i = 0;
 int j = 0;
 int array [100] = {0};
  for(i = 0; i<100; i++)
  {
 
   array[i] = rand();
 
  } 
 
 
 change = changeA(array);
 
}
int changeA(int array[100])
{
 int i = 0;
 int j = 0;
 int array2[4][25] = {0};
  for(i = 0; i < 4; i++)
  {
  
   for(j = 0; j<24; j++)
    {
     array2[i][j] = array[i*25+j];
     printf("%2d %2d", i,j);
    
    }
  }
 printf("\n\n");
 
 
 system("pause");
}

Thanks Click_here changed that but I'm stuck lol. I'll wait to see when the Prof. gets back to me I'm just burnt out been messing with this since 3:00pm central time lol. Thanks again for everyone help def got me moving along I'll keep you guys updated

Code:

int changeA(int array[]);

int main()
{
 int change;
 int i = 0;
 int j = 0;
 int array [100] = {0};
  for(i = 0; i<100; i++)
  {
 
   array[i] = rand();
 
  } 
 
 
 change = changeA(array);
 system("pause");
 return 0;
}
int changeA(int array[100])
{
 int i = 0;
 int j = 0;
 int array2[4][25] = {{0}};
  for(i = 0; i < 4; i++)
  {
 
   for(j = 0; j<24; j++)
    {
     array2[i][j] = array[i*25+j];
     printf("%d %d", array2[i][j]);
    
    }
  }
 printf("\n\n");
 
 
 
}

Originally Posted by phillyflyers

Is there a limitation of the number of integers C can display on one line? It seems that even though I want 4 rows it put a 2 more rows because it is not able to list 25 integers on one line.

How many numbers there are on one line is dependent on the number of digits and the horizontal size of your terminal. It has nothing to do with C.

Code:

for(j = 0; j<24; j++)

The highest value of "j" will be 23 thus you aren't assigning a value to the 25th element (index 24) in each row.

Code:

printf("%d %d", array2[i][j]);

You should get a warning here (do you know how to turn on warnings for your compiler?). You have two format specifiers but only one value to print.

Bye, Andreas

I have visual studio 2012 and I am not get any warnings it just prints out nonsense. I tried doing what my Prof. replied with I'm not sure if the program is correct, because it still is printing on more than 4 rows. Is this just cause its the way I have it programmed wrong or just how visual studio 2012 shows it? Thanks.

Code:

#include <stdio.h>
#include <stdlib.h>
 
int changeA(int array[]);

int main()
{
 int change;
 int i = 0;
 int j = 0;
 int array [100] = {0};
  for(i = 0; i<100; i++)
  {
 
   array[i] = rand();
 
  } 
 
 
 change = changeA(array);
 system("pause");
 return 0;
}
int changeA(int array[100])
{
 int i = 0;
 int j = 0;
 int array2[4][25] = {{0}};
  for(i = 0; i < 4; i++)
  {
 
   for(j = 0; j<24; j++)
    {
     array2[i][j] = array[i*25+j];
     printf("%6d", array2[i][j]);
    
    }
  }
 printf("\n\n");
 return array2[i][j]

Hi Andi, thank you i was missing this one line it now prints it right. It is the terminal that makes it look like it isnt doing the numbers on 4 rows.

Code:

for(i = 0; i < 4; i++)
  {
   printf("\n");
   for(j = 0; j<24; j++)
    {
     array2[i][j] = array[i*25+j];
     printf("%6d", array2[i][j]);
     
    
    }