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convert char*(number) to required unsigned char*

Sun Jul 06, 2008 1:47 pm

Given For e.g,
char input[] = "10011210582553796";

now Hexadecimal of 10011210582553796 is 2391249A8B74C4.

So output unsigned char* should have value,
unsigned char output[8] = {0x00,0x23,0x91,0x24,0x9A,0x8B,0x74,0xC4};

Any solution to convert given input char* to output char* with above mentioned requirement?

code should run on platform for which __int64/signed long long is not supported.

Thanx in advance...



Re: convert char*(number) to required unsigned char*

Mon Jul 07, 2008 5:09 pm

To convert from the decimal number system into hex number system .A way to do that , first convert the number you binary and from binary take every 4 digits and make 1 digit hex ,

example : 12 )decimal to 01010)binary to C ) hex

Re: convert char*(number) to required unsigned char*

Tue Jul 08, 2008 10:09 am

The number i have given is out of range of int datatype, so can not use printf or atoi operation which are supported for int datatype. Actually number can be stored in int64 type but system does not support int64 datatype. So i want to convert is to unsigned char output[8] = {0,35,145,36,154,139,116,196} = {0,0x23,0x91,0x24,0x9A,0x8B,0x74,0xC4}
Even i cannot use left shift or right shift (multiplication/devision) when number reaches above integer capacity.

Re: convert char*(number) to required unsigned char*

Tue Jul 15, 2008 5:14 pm

found solution
Code:
     int number = 0;
     char numberchars[] = "10011210582553796";
     int i = 0;
     int ans = 0;
     int carry = 0;

     char answerArray[100] = {0};

     char remainderArray[100] = {0};
     int remindex = 0;
     int ansindex = 0;
     int remainder = 0;

     while( numberchars[i] != '\0' )
     {
          while( numberchars[i] != '\0' )
          {
               char currentchar[2] = {0};
               currentchar[0]=numberchars[i];

               int num = atoi(currentchar);
               num += remainder;
               remainder = 0;

               if ( num < 2  )
               {               
                    remainder = num;
                    if(i>0)
                         answerArray[ansindex++] = '0';
               }
               else
               {
                    remainder = num % 2;
                    int answer = num / 2;

                    char a[2] = {0};
                    itoa(answer,a,10);

                    answerArray[ansindex++] = a[0];
               }

               i++;
               remainder *= 10;
          }

          char a[2] = {0};
          int rval = remainder / 10;
          itoa(remainder / 10,a,10);
         
          remainderArray[remindex++] = a[0];
          int size = sizeof(answerArray);
          memcpy(numberchars,answerArray,sizeof(answerArray));
           size = sizeof(answerArray);
          memset(answerArray,0,sizeof(answerArray));
          ansindex = 0;
          remainder = 0;
          i=0;
     }

     char int64[8] = {0};

     for(int k=0;remainderArray[k]!= '\0';k++)
     {
          int64[7-(k/8)] |= ((remainderArray[k]-'0') << (k%8));
     }


Re: convert char*(number) to required unsigned char*

Tue Jul 15, 2008 10:08 pm

thank you so much for posting the solution :gOOd: :gOOd:

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