理解/阐明C语言代码的逻辑

Question

Following is the code from www.tutorialspoint.com, where i am learning basics of algorithm & data structure. This tutorial code gives an ex. of insertion operation on array.

    #include <stdio.h>
    main() {
       int LA[] = {1,3,5,7,8};
       int item = 10, k = 3, n = 5;
       int i = 0, j = n;

       printf("The original array elements are :\n");

       for(i = 0; i<n; i++) {
          printf("LA[%d] = %d \n", i, LA[i]);
       }

       n = n + 1;

       while( j >= k){
          LA[j+1] = LA[j];
          j = j - 1;
       }

       LA[k] = item;

       printf("The array elements after insertion :\n");

       for(i = 0; i<n; i++) {
          printf("LA[%d] = %d \n", i, LA[i]);
       }
    }

I understand the code that it adds new item at give value of k as index of array. And it runs fine without any error when compiled & runs. What confuses me is logic of line LA[j+1] = LA[j] in while loop.

My understanding is that in c you have to declare the size of array. But in ex. code int LA[] = {1,3,5,7,8}; bracket [ ] is empty. So i am not sure if it's a fixed size or more elements can be added to array.

Now j is given the value of n which is 5(length of array). Array declaration has 5 element, and index of array is 0 to 4.

(On first iteration of while loop)So LA[j + 1] is LA[5 + 1] is LA[6] . Now array has only 5 elements indexed 0 to 4. So according to me even after assuming that more element can be added to array, how can it assign value of LA[j] which is 5 to LA[j + 1] . it's saying LA[6] = LA[5] but there is nothing at index 5 as last index is 4.

I have tried to search on google but I am not sure what to search except the code or part of code. And searching with code wasn't helpful.

Please help me understand. Thank You.

Answer 1

int LA[] = {1,3,5,7,8};

is equivalent to:

int LA[5] = {1,3,5,7,8};

The empty bracket just tells the compiler to automatically set the number of elements as the size of the array.

---

Yes, you are right, L[6] is out of bounds access, since L has size of 5. So, as you nicely said, the indices are in [0, 4], so L[5] is also out of bounds!

This code is wrong!

---

Read more on Undefined Behaviour : Undefined, unspecified and implementation-defined behavior

Answer 2

First of all it is a very bad code that is written by a very weak programmer and has undefined behaviour.

For example the array LA

int LA[] = {1,3,5,7,8};

has only 5 elements.

However in these loops

   while( j >= k){
      LA[j+1] = LA[j];
      ^^^^^^^
      j = j - 1;
   }

and

   for(i = 0; i<n; i++) {
              ^^^^ n is already set to 6
      printf("LA[%d] = %d \n", i, LA[i]);
   }

there are attempts to write to the memory beyond the array. Also there are some magic values as for example n = 5 . It would be much better to write at least

n = sizeof( LA ) / sizeof( *LA )

Take into account that function main without parameters in C should be declared like

int main( void )

The program can look for example the following way

#include <stdio.h>

int main( void ) 
{
    int a[] = { 1, 3, 5, 7, 9 };
    const size_t N = sizeof( a ) / sizeof( *a );

    while ( 1 )
    {
        printf( "The original array elements are:" );
        for ( size_t i = 0; i < N; i++ ) printf( " %d", a[i] );
        printf( "\n" );

        printf( "\nEnter a number to insert in the array (0 - exit): " );
        int value;

        if ( scanf( "%d", &value ) != 1 || value == 0 ) break;

        printf( "Enter a position in the array where to insert: " );
        size_t pos;

        if ( scanf( "%zu", &pos ) != 1 ) break;

        size_t j = N;

        if ( pos < N )
        {
            while ( --j != pos ) a[j] = a[j-1];
            a[j] = value;
        }           

        printf( "\nThe array elements after insertion:");

        for ( size_t i = 0; i < N; i++ ) printf( " %d", a[i] );
        printf( "\n\n" );
    }

    return 0;
}

Its output might look like

The original array elements are: 1 3 5 7 9

Enter a number to insert in the array (0 - exit): 2
Enter a position in the array where to insert: 1

The array elements after insertion: 1 2 3 5 7

The original array elements are: 1 2 3 5 7

Enter a number to insert in the array (0 - exit): 6
Enter a position in the array where to insert: 4

The array elements after insertion: 1 2 3 5 6

The original array elements are: 1 2 3 5 6

Enter a number to insert in the array (0 - exit): 0

The logic behind the algorithm is simple. If you have an array with N elements and want to insert a value in the position pos that less than N then you need to move all elements to the right starting from the position pos and write the new value in the element with index pos . The right most element of the original array will be lost because it will be overwritten by the preceding element. The program output shows this result.

As you correctly pointed out yourself the valid range of indices for the array defined as having 5 elements used in the program is [0, 4] .

When an array is declared without its dimension as in this example

int a[] = { 1, 3, 5, 7, 9 };

then the compiler determinates its dimension from the number of initializers. Thus the above declaration is equivalent to

int a[5] = { 1, 3, 5, 7, 9 };