Last teLoop issue using arrays

[DaemonCoder]

I am trying to write an array in c that stores the last ten command entered in a program. To do this i created an eleven element global array, as different functons need to act on it. i am tring to shif the elements in the array to the right by one posistion. what i end up with is the last two commands. the first one is in the right place but the next is repeated 9 times. here is the code.. can you help me figure out what im doing wrong..?

int CHOICE;
int LTENC[11];
 
void cArrayModify()
{
 
 
    for(int i=0;i<10;i++)
    {
        int tempValue=LTENC[i];
        int copyLocation = i + 1;
        LTENC[copyLocation]=tempValue;
        i++;
    }
 
 
    LTENC[0]=CHOICE;
}
 
void displayLastTenChoices()
{
    printf("\n \n Last ten choices: ");
    for(int i=0;i<10;i++)
    {
        printf(" %d", LTENC[i]);
    }    
    printf(" \n");
}

[weaknessforcats]

for(int i=0;i<10;i++)
{
int tempValue=LTENC[i];
int copyLocation = i + 1;
LTENC[copyLocation]=tempValue;
i++;
}

So, the loop starts with i=0. That means:
int tempValue=LTENC[0]; //original element 0
int copyLocation = 0 + 1; //target location
LTENC[1]=tempValue; //element 0 moves to element 1

OK but you didn't save element 1. It just got overwritten.

What you need is a circular array. That is, you need a counter that says where the array "starts". You add to the array by adding to the start-1 position.
The counter has to range from 0 to 10, and when 10 goes back to zero.

The last 10 commands are displayed by reading the array backwards from the start-1 location to the start location.

[Parul Bagadia]

M not so sure about it; but i think in the first for loop you have done right shifting; and incremented i by one; so i think you should check for the limit of i, what i mean is; you start from first element instead of 0th. So your condition in for loop should be i<11.....so that ull start with 1st element and end up with 10th......

[weaknessforcats]

True, but after 10, you have to wrap around to the start of the array. The 11th choice is in element 0. The 10th choice is in element 9.

If you don't write a cricular buffer, then you can't show the "last 10" choices.

[DaemonCoder]

So i should add code to store element 1 like so?
for(int i=0;i<10;i++)
{
int j = i+1;
int tempValue0=LTENC[i];
int tempValue1=LTENC[j];
int copyLocation = i + 1;
LTENC[copyLocation+1]=tempValue0;
LTENC[copyLocation]=tempValue1;
i++;
}

[DaemonCoder]

Im not sure what you mean by "circular buffer" all i want to do is have a place to store the choices. Each execution of the function will run ten times therefore adding one choice to the array and knocikng the " tenth" choice off.

[weaknessforcats]

"circular buffer"

A circular buffer is an endless buffer.

For example, you need to keep the last 3 commands. Let's call them A, B, C etc as they come in. Let's also assume you use a 3-element char array.

The start is element 0. You out the command A in element 0 and make the start element 1.
0 A
1 AB
2 ABC
Start is now 0 again.
0 DBC
1 DEC
2 DEF
Start is now 0 again
0 GEF
1 GHF
2 GHI

So, the number is the start of the buffer. To read the buffer, you begin at the start+1, read to end of the array by wrapping around to the beginning and reading to the start.

So GHF with a start at 1 reads FGH. DBC with a start at 0 reads BCD.


Once you figure this out, you put it in a function and it just works. You could even write a C++ circular buffer class based on a linked list.

[MACKTEK]

for(int i=0;i<10;i++)
{
int tempValue=LTENC[i];
int copyLocation = i + 1;
LTENC[copyLocation]=tempValue;
i++;
}

Logically I ask myself where is the first command? is it LTEN[0] or [9]:
lets say LTEN[0] = 100
now using your code and going thru the first iteration:
tempValue = 100
copyLocation = i+1 , which is 1
then you do LTENC[1]=tempValue;
therefore you assigned LTENC[1] the value that the old LTEN[0] had.

then you increment...
the problem is, that the next loop simply reassigns...

LTENC[1] = 100
now using your code and going thru the next iteration:
tempValue = 100
copyLocation = i+1 , which is 2
then you do LTENC[2]=tempValue;
therefore you assigned LTENC[2] the value that the old LTENC[1] had.

this problem will repeat itself.

What you would want to do is something like:

for(int i=0;i<9;i++) // NOTICE I lowered the max...
    {
        LTENC[i]=LTENC[i+1];
         i++;
    }
LTENC[9]= newval;
 

lets assume from 0 to 9 contains 10,11,12,13...etc.
iterating for the first few we get...
LTENC[0] = 11
LTENC[1] = 12
...
LTENC[9] = newval

the next time they enter a new value... each element in the array would move down 1.


I hope that helps, if I made a mistake please let me know.

[DaemonCoder]

Thank you MackTek.... You answered my question completely.. I was trying to do a right shift when all i need to do was change the loop and shift left. Thanks again for all who tried to help.

[DaemonCoder]

for(int i=0;i<9;i++)
    {
        LTENC[i]=LTENC[i+1];
 
    }
LTENC[9]=CHOICE;
 
}

I figured it out.... With the example above before the new val is assigned there is an extra increment resultion in the following ouput.

0,0,0,0,0,0,0,0,0,1

the first nine will remain zero.. with the increment removed we get the desired affect
1,2,3,4,5,6,7,8,9,0

Thanks for all your help guys. And thanks again MackTek for the information.. IO would have been chasing my tail for weeks trying to figure this one out.