Solved

# array input loop

Posted on 2003-12-05
302 Views
enter 3 digits into each of the array value's and keep entering 3 digits until we enter a digit of 600 which then stops us entering to the array and will then print the array on screen, it looks like it could be a really simple solution but i cant see it.

here is what we have so far:

void main()
{

const int max = 100;

int a[ max ];

int instruction;

cin >> instruction;

a[ max ] = instruction;

while ( instruction!=600 )
{

for ( int i =0; i < max; i++ )

a[ max ] = instruction;

i++;

cin >> instruction;

}

for ( int j = 0; j < max; j++ )

cout << setw( 7 ) << j << setw( 13 ) << a[ j ] << endl;

}
0
Question by:a_migdal

LVL 11

Accepted Solution

ID: 9883652
(1) Look at the dimensionality of your loops:
while ...
for ...

for ...

The first pair of loops are nested and process a two dimensional structure. a is a one dimensional structure. There is a problem.

So, which loop should go?

Since you read until you see 600, you need a while loop.

You also need a counter to count the number of instructions that have been stored in the array.

You also need to store the 600 in the array so you really need a do ... while () loop. Something like:

do
// increment number of instructions counter
// store the instruction at a[number of instructions counter]
while (instruction != 600);

(Since this is homework I won't write the code for you).

Make sure you declare your counter before the do while.

(2) This does have one potential flaw in that if the number of instructions exceeds max you will have a buffer over run problem. I will ignore that for the moment unless you feel it is important.

-bcl
0

LVL 2

Assisted Solution

kmalhotra earned 20 total points
ID: 9884502
void main()
{

const int max = 100;
int a[ max ];
int instruction, count;
count = 0;

do {
cin>>instruction;
a[count] = instruction;
count++;
}while (instruction != 600);

cout<<"-----------------------"<<endl;
for (int ctr = 0; ctr < count; ctr++) {
cout << setw( 7 ) << ctr << setw( 13 ) << a[ ctr ] << endl;
}
}
0

LVL 7

Expert Comment

ID: 9884959
don't use an array for this. You will be running into memory problems like bcladd said.
I am going to edit kmalhotra's submission so  you won't have any issues with the memory.

void main()
{

string a = "";
int instruction, count;
count = 0;

do {
cin>>instruction;
a += instruction;
count++;
}while (instruction != 600);

cout<<"-----------------------"<<endl;
for (int ctr = 0; ctr < count; ctr++) {
cout << setw( 7 ) << ctr << setw( 13 ) << a[3*ctr] + a[3*ctr +1] + a[3*ctr + 2] << endl;
}
}

All this is is putting it into a string. It will automatically deal with any memory allocation needs because it is really a char* and it using pointers. Much simpler for your project then actually writing a linked list or something like that.
0

LVL 7

Expert Comment

ID: 9884967
Oh, kmalhotra
I hope you don't mind my working with the code you already posted.
0

LVL 11

Expert Comment

ID: 9885209
jj819430-

I think the original poster wants to be able to extract the instructions as units. Your suggestion to use an STL container is a good one: I think a vector of int would be more appropriate in this case (look at poster's other open questions for some background).

-bcl
0

LVL 3

Expert Comment

ID: 9894436
Just change the part of your code where you index.

You put max to index with. An array's first index is 0

a[ max ] = instruction;

Should be something like

a[current_index] = instruction;

Then you did it again.

while ( instruction!=600 )
{

for ( int i =0; i < max; i++ )

////////////////max....
a[ max ] = instruction;

Also, you should not copy other persons code here as you should know the code before you advance. Teachers will know you have no clue and give you the F.

Another thing you want to prompt inside your loop to get the values

while ( instruction!=600 && instruction < max )
{

clrscr();//conio.h
cout << "Enter value ";
cin   >> instruction;

a[ i ] = instruction;

the var i now has how many instructions got loaded. Hmmm, you could use that value in another loop to print them.
while(x < i)
{
print...
}

RJ
0

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