Chnage the position of element!

[engartte]

Hi all,
Let consider the following solution:

#include <stdio.h>

int main ()
{
int a[10], temp;
int i;
for(i=0; i<10; i++){
a[i] = i+1;
printf("%d ",a[i]);

}

printf("\nType a number:");
scanf("%d", &temp);
if(temp>10);
temp = a[0];
for (i=0;i<temp;i++)
a[i] = a[i+1];
for(i=0; i<10; i++)
printf("%d ",a[i]);
printf("\n");
return 0;
}

Of Course,at the follow-up solution I am looking to response
to the following question:
Declare an array a[10],then read the integers into the elements
of the array from keyboard.but,insert the first element of the array
into thwe position specified by a following integer and print the
contents as follows:
1 2 3 4 5 6 7 8 9 10
4
2 3 4 1 5 6 7 8 9 10
but,my solution is not satisfy the condition.If you know some other
on,please show me.
tanx in advnce
engartte

[Perceptron]

Following is the simple solution for desired output :

int main()
{
int a[10], i, n, temp;

/* Read all elements */
for( i = 0 ; i < 10 ; ++i )
scanf( "%d", &a[i] );

/* Read desired position */
scanf( "%d", &n );

if( n > 10 )
n = 10;
if( n < 1 )
n = 1;

/* Shift elements */
temp = a[0];
for( i = 0 ; i < n-1 ; ++i )
a[i] = a[i+1];

/* Store first element at desired position */
a[n-1] = temp;

/* Print array */
for( i = 0 ; i < 10 ; ++i )
printf( "%d ", a[i] );

return 1;
}

If you can provide n before array elements the solution can be optimized
to :

int main()
{
int a[10], i, n, temp;

/* Read desired position */
scanf( "%d", &n );

if( n > 10 )
n = 10;
if( n < 1 )
n = 1;

/* Read fisrt element */
scanf( "%d", &temp );

/* Read other elements in corresponding positions */
for( i = 1 ; i < 10 ; ++i )
if( i < n )
scanf( "%d", &a[i-1] );
else
scanf( "%d", &a[i] );

/* Store first element in desired position */
a[n-1] = temp;

/* Print array */
for( i = 0 ; i < 10 ; ++i )
printf( "%d ", a[i] );

return 1;
}

[Barry Schwarz]

On 4 Jan 2005 22:20:44 -0800, "engartte" <na*****@hotmail.com> wrote:

Hi all,
Let consider the following solution:

#include <stdio.h>

int main ()
{
int a[10], temp;
int i;
for(i=0; i<10; i++){
a[i] = i+1;
printf("%d ",a[i]);

}

printf("\nType a number:");
scanf("%d", &temp);
Assume the user types in 10.
if(temp>10);
This extra semicolon transforms the if statement to a no-op. It has
no effect (except to possibly consume time). I assume you meant the
next statement to be conditional on the if.
temp = a[0];
for (i=0;i<temp;i++)
a[i] = a[i+1];
temp is 10. When i is 9, this statement invokes undefined behavior
since a[i+1] does not exist.


for(i=0; i<10; i++)
printf("%d ",a[i]);
printf("\n");
return 0;
}

Of Course,at the follow-up solution I am looking to response
to the following question:
Declare an array a[10],then read the integers into the elements
You didn't read integers into the array. You assigned values you
decided to the elements. The user had no choice.
of the array from keyboard.but,insert the first element of the array
into thwe position specified by a following integer and print the
After initializing the array, you did slide a number of elements to
the left but you never filled in the "vacated" element.
contents as follows:
1 2 3 4 5 6 7 8 9 10
4
2 3 4 1 5 6 7 8 9 10
but,my solution is not satisfy the condition.If you know some other
on,please show me.
tanx in advnce

<<Remove the del for email>>

[Barry Schwarz]

On Wed, 05 Jan 2005 22:24:37 -0500, "Perceptron"
<<undisclosed>@<undisclosed>.<undisclosed>> wrote:

Following is the simple solution for desired output :

int main()
{ snip return 1;
Why return a non-portable value from main??
}

snip
<<Remove the del for email>>