CODEWITHSUNDEEP
carrays
Shashwat Black
Shashwat Black

Reputation: 1002

What am I doing wrong (C arrays)?

I'm just a beginner at C.
I'm trying to make a simple program to arrange the user-entered digits in ascending order. I have figured out the solution but can't understand why my other code wouldn't work :(

-------------------------------------------------------------------------
working code:
-------------------------------------------------------------------------
#include <stdio.h>
int main()
{
    int i,j,num[10];
 printf("Enter 10 numbers\n");
 for (i=0;i<10;i++)
  {scanf("%d",&num[i]);}


    for (i=0;i<9;i++)
     {
         for (j=i+1;j<10;j++)
           {
               if (num[i]>num[j])
                 {
                    num[i]+=num[j];
                    num[j]=num[i]-num[j];
                    num[i]=num[i]-num[j];
                 }
           }
     }
    printf("The numbers in ascending order are:");
    for (i=0;i<10;i++)
      {
          printf(" %d",num[i]);
      }

    return 0;
}

-------------------------------------------------------------------------
code that won't work:
-------------------------------------------------------------------------
#include <stdio.h>
int main()
{
    int i,j,num[10];
 printf("Enter 10 numbers\n");
 for (i=1;i<=10;i++)
  {scanf("%d",&num[i]);}


    for (i=1;i<10;i++)
     {
         for (j=i+1;j<=10;j++)
           {
               if (num[i]>num[j])
                 {
                    num[i]+=num[j];
                    num[j]=num[i]-num[j];
                    num[i]=num[i]-num[j];
                 }
           }
     }
    printf("The numbers in ascending order are:");
    for (i=1;i<=10;i++)
      {
          printf(" %d",num[i]);
      }

    return 0;
}

In the latter program, numbers appear out of order, and there even are numbers that haven't been entered.

My question is, isn't it basically the same code? Just that in the latter program numbers would be stored from num[1] to num[10] instead of num[0] through num[9]?
Does it have something to do with array definitions?
It seems I have serious misconceptions, please help me out!

Upvotes: 1

Views: 196

Answers (6)

onurbaysan
onurbaysan

Reputation: 1258

for(i=0;i<9;i++)  //**i<9**    
   for (j=i+1 ...)

If i=8 then j=9 , everything is OK.

In second code snippet:

for(i=0;i<10;i++) //**i<10**
   for (j=i+1 ...)

If i=9 then j=10, so you try to access num[10] and it gives you error.

If you want to access num[10] then you must declare array int num[11] and then you can access num[10].

Array Basics

int num[10]

  • Capacity of array = 10
  • Every element of this array are integer.
  • First element index is 0. So If you want to access first element , num[0]
  • Last element index is 9. So If you want to access last element, index must be length of array - 1 , so num[9]
  • There are 10 elements in the array and they are :
    • num[0] , num[1] , num[2] , num[3] , num[4] , num[5] , num[6] , num[7] , num[8] and num[9]

You can learn further at http://www.cplusplus.com/doc/tutorial/arrays/

Upvotes: 0

PTBG
PTBG

Reputation: 595

Indexes in C go start from 0. so when you declare an array of size 10, and you try to get element at index 10, you're actually getting the 11th element. Since you haven't defined the 11th element, the array will most likely get some random numbers from memory, which is why you are noticing numbers you have note entered.

Since you are new to programming, I would suggest taking the time now to really learn about how C manages memory, and how different data structures access the memory. It might be a little boring now, but you'll save yourself some headaches in the future, and you will start to build good habits and good practices, which will lead to writing good, optimal code

Upvotes: 1

Jon Egeland
Jon Egeland

Reputation: 12613

In your other code example, you were doing this:

int num[10];

for(int i = 1; i <= 10; i++) {
    //...
}

That means you have an array with ten spaces [0-9], and for the last part of your for loop, you were trying to access num[10] which doesn't exist.

Your working example goes from 0-9 and never tries to read num[10], so it works.

Arrays in C, as in most languages, start with position 0 and count that as position one. So the last element of your array would be the size you entered when you declared the variable, minus one.

Upvotes: 0

Weaver
Weaver

Reputation: 619

Welcome to programming! I believe you are a bit confused about how arrays are indexed in C.

Arrays in most languages (including C) are known as zero-indexed arrays. This means the start of an array is always at position 0. So if you make int num[10], trying to access num[10] isn't actually a valid call at all, because the start is num[0]. Hence you can only access from num[0] to num[9].

It's an easy mistake to make, even though I've been programming for years sometimes when it's been a long night I'll still make silly array indexing issues.

Upvotes: 0

TJD
TJD

Reputation: 11896

In your non-working example, you have invalid memory accesses. For example, when i = 9 and j = 10, you access num[10], which is invalid memory.

Upvotes: 0

kfmfe04
kfmfe04

Reputation: 15327

In C, when you have int num[10];, your indexes need to go from 0 to 9, never to 10. So look over your code, if any i or j ends up with a value of 10 any time during the program run, that's bad news.

Upvotes: 8

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