shift text

fj8283888 · #1 04-Nov-2005, 03:46

Hi:

I have a value char a[12]="123456";

If I have to add "0" before 123456 to become "000000123456" what should I do?

Paramesh · #2 04-Nov-2005, 03:58

Hi fj8283888,

Here is a hint how to solve it:
1. Find the length of the character array first.
2. We should use two loops.
3. One to shift the characters to the right and the other one to check how many times we should shift the characters.
4. To shift the characters to the right, use like this:

CPP / C++ / C Code:

          a[ i ] = a[ i - 1 ];

i.e make the character equal to the previous character.
Note that this should be done within the array length limit. i value should be between 1 and 11 only in your case.
5. After shifting them to the right, make the first character to be '0'.
6. Suppose the array length is 12(since a[12]) and the actual size is 6, then you must shift 12-6 times. i.e 6 times.

Hope this helps you a bit.

Regards,
Paramesh.

davekw7x · #3 04-Nov-2005, 10:56

Quote:

Originally Posted by fj8283888

Hi:

I have a value char a[12]="123456";

If I have to add "0" before 123456 to become "000000123456" what should I do?

That would be illegal:

The array a has memory enough for twelve chars.

The string "000000123456" takes 13 chars (one for each of the ascii values of the digits and one for the terminating zero byte).

This is a really, realllllly important point that even some experienced programmers forget sometimes.

CPP / C++ / C Code:

#include <stdio.h>
#include <string>

int main()
{
  char a[12] = "123456";
  char b[]   = "000000123456";

  int sizea;
  int sizeb;
  int lengtha;
  int lengthb;

  sizea = sizeof(a);
  lengtha = strlen(a);
  sizeb = sizeof(b);
  lengthb = strlen(b);

  printf("The size of array a is %d\n", sizea);
  printf("The length of the \"string\" in a is %d\n\n", lengtha);
  printf("The size of array b is %d\n", sizeb);
  printf("The length of the \"string\" in b is %d\n", lengthb);

  return 0;
}

Code:

The size of array a is 12
The length of the "string" in a is 6

The size of array b is 13
The length of the "string" in b is 12

If you want to shift, Paramesh has given you a way, but it is up to you, the programmer, to make sure there is enough room in the array.

Regards,

Dave

fj8283888 · #4 04-Nov-2005, 18:18

CPP / C++ / C Code:

#include <iostream>
#include <cstring>
using namespace std;
void Convert(int, char*);   

void main ()
{
char test[14]="1222349.10";

td=strlen(test);
int l=0;
char string[5];

do
{

l++;

}while (test[l]!='.');
test[13]=test[l+2];
test[12]=test[l+1];
test[11]=test[l];
int tes=0;
tes=12-l;//empty space
int lp=10;
int diff=0;
l--;
do
{
test[lp]=test[l--];
diff=lp-l;
lp--;

}while (0<=l);
cout<<diff;
int a=0;
diff--;
do
{
test[a]='0';
a++;
}while (diff>a);
cout << test;




}

I guess it should be a better way to perform

frug · #5 04-Nov-2005, 23:57

Here are a couple of examples that might help you.

CPP / C++ / C Code:

	// char buffers
	char szA[13] = _T("123456");
	int nPos  = lstrlen(szA) - 1;
	int nInsertPos = sizeof(szA) - 2;
	memset(szA + nPos + 1, '0', nInsertPos - nPos);
	while(nPos >= 0)
	{
		szA[nInsertPos--] = szA[nPos];
		szA[nPos--] = '0';
	}
	TRACE1(_T("%s\n"), szA);

	// If you are initialy dealing with a number, then it is trivial
	char szB[13] = {};
	int n = 123456;
	wsprintf(szB, _T("%12.12d"), n);
	TRACE1(_T("%s\n"), szB);

Note that the memset is not important for the example string we are shifting, but if the length of string we are shifting is less then half, then the memset is critical. If not used the resulting string would replace the original string with '0's.

I hope the examples helped.

Felix

Paramesh · #6 05-Nov-2005, 00:07

Hi fj8283888 ,

There is no need to check for a '.' inside the string.
You can simply shift the numbers and add zero without checking for any other things.

Here is how we do it:
The array is like this:

CPP / C++ / C Code:

char test[14]="1222349.10";

1. Find the length of the character array first.

CPP / C++ / C Code:

    int length = strlen(test);

2. We should use two loops.
3. One to shift the characters to the right and the other one to check how many times we should shift the characters.
4. To shift the characters to the right, use like this:

CPP / C++ / C Code:

          a[ i ] = a[ i - 1 ];

i.e make the character equal to the previous character.
Note that this should be done within the array length limit. i value should be between 1 and 13 only in your case.
5. After shifting them to the right, make the first character to be '0'.

CPP / C++ / C Code:

    test[0] = '0';

6. Since the array length is 14(since a[12]) and the actual size is 10, then you must shift 13-10 times. i.e 3 times.
Dave explained us why it is 13 and not 14.

Here is the loop:

CPP / C++ / C Code:

    for(int j = 0; j < 13 - length; j++)
    {
        for(int i = 13; i > 0 ; i--)
        {
            test[ i ] = test[ i - 1 ];
        }
    test[0] = '0';
    }

We can also find the length of the array by using a loop like this instead of including the cstring header.

CPP / C++ / C Code:

    for( length = 0; test[ length ] != NULL ; length++ )
    ;

And also note that main must return an int rather than being void.

Summarising, here is the complete code:

CPP / C++ / C Code:

#include<iostream>
using namespace std;

int main()                                      //main MUST return an int.
{
    char test[ 14 ] = "1222349.10";             //a string test
    int length;                                 //length is used to hold the length of the string. 

    for( length = 0; test[ length ] != NULL ; length++ )
    ;                                           //calculate length using a loop.       
    
    
    for(int j = 0; j < 13 - length; j++)        //repeat the shifting 13 - length times.       
    {
         
        for(int i = 13; i > 0 ; i--)            //shift the characters.
        {
            test[ i ] = test[ i - 1 ];          //make previous character equal to the present character
        }
        
    test[0] = '0';                              //make first character to be 0
    }
    
    cout << test << endl;                       //print the string now and see the output
                
    return 0;                                   //return 0 indicates successful execution of the program.
}

You can also do this program using do while loops and while loops instead of for loops.

Regards,
Paramesh.

Dominic · #7 06-Nov-2005, 09:28

Seems a lot of user code, where simple formating by sprintf might work. Try this example :

CPP / C++ / C Code:

double Number = 123456.235;
	char Buff[50];
	sprintf(Buff,"%f",Number);
	printf("\nNumber = %s\n",Buff);

	sprintf(Buff,"%d",static_cast <int> (Number));
	printf("\nNumber = %s\n",Buff);


	sprintf(Buff,"% 020d",static_cast <int> (Number));
	printf("\nNumber = %s\n",Buff);


	sprintf(Buff,"% 020f",Number);
	printf("\nNumber = %s\n",Buff);


	sprintf(Buff,"% 020.3f",Number);
	printf("\nNumber = %s\n",Buff);

I used several examples, and the printf statement is just there if you want to see the result. The sprintf function alone will format the number into the character array. Note the size of Buff. Its always better to be bigger than too small. With modern computers, there is usaully plenty of memory. But over run one array bounds, just once and anything could happen, including a very long time to debug the problem.

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