如何存储在字符串中并转换为字符数组？

Question

Write a C++ program to perform addition of two hexadecimal numerals which are less than 100 digits long. Use arrays to store hexadecimal numerals as arrays of characters.the solution is to add the corresponding digits in the format of hexadecimal directly. From right to left, add one to the digit on the left if the sum of the current digits exceed 16. You should be able to handle the case when two numbers have different digits.

The correct way to get the input is to store as character array. You can either first store in a string and convert to character array, or you can use methods such as cin.getline(), getc() , cin.get() to read in the characters.

I don't know what is wrong with my program and it I don't know how to use the function getline() and eof()

 char a[number1],b[number1],c[number2],h;
int m,n,p(0),q(0),k,d[number1],z[number1],s[number2],L,M;
cout<<"Input two hexadecimal numerals(both of them within 100 digits):\n";
cin.getline(a,100);
cin.getline(b,100);
int x=strlen(a) ;
int y=strlen(b);
for(int i=0;i<(x/2);i++)
    {
      m=x-1-i;
      h=a[i];
      a[i]=a[m];
      a[m]=h;
    }
for(int j=0;j<(y/2);j++)
    {
      n=y-1-j;
      h=b[j];
      b[j]=b[n];
      b[n]=h;
    }
if(x>y)
{
    for(int o=0;o<x;o++)//calculate a add b
    {
        if(o>=(y-1))
            z[o]=0;//let array b(with no character)=0
        if(a[o]=='A')
        d[o]=10;
        else if(a[o]=='B')
        d[o]=11;
        else if(a[o]=='C')
        d[o]=12;
        else if(a[o]=='D')
        d[o]=13;
        else if(a[o]=='E')
        d[o]=14;
        else if(a[o]=='F')
        d[o]=15;
        else if(a[o]=='0')
        d[o]=0;
        else if(a[o]=='1')
        d[o]=1;
        else if(a[o]=='2')
        d[o]=2;
        else if(a[o]=='3')
        d[o]=3;
        else if(a[o]=='4')
        d[o]=4;
        else if(a[o]=='5')
        d[o]=5;
        else if(a[o]=='6')
        d[o]=6;
        else if(a[o]=='7')
        d[o]=7;
        else if(a[o]=='8')
        d[o]=8;
        else if(a[o]=='9')
        d[o]=9;
        if(b[o]=='A')
        z[o]=10;
        else if(b[o]=='B')
        z[o]=11;
        else if(b[o]=='C')
        z[o]=12;
        else if(b[o]=='D')
        z[o]=13;
        else if(b[o]=='E')
        z[o]=14;
        else if(b[o]=='F')
        z[o]=15;
        else if(b[o]=='0')
        z[o]=0;
        else if(b[o]=='1')
        z[o]=1;
        else if(b[o]=='2')
        z[o]=2;
        else if(b[o]=='3')
        z[o]=3;
        else if(b[o]=='4')
        z[o]=4;
        else if(b[o]=='5')
        z[o]=5;
        else if(b[o]=='6')
        z[o]=6;
        else if(b[o]=='7')
        z[o]=7;
        else if(b[o]=='8')
        z[o]=8;
        else if(b[o]=='9')
        z[o]=9;
        p=d[o]+z[o]+q;
        if(p>=16)//p is the remained number
        {
           q=1;
           p=p%16;
        }
        else
            q=0;

        if(p==0)
        c[o]='0';
        else if(p==1)
        c[o]='1';
        else if(p==2)
        c[o]='2';
        else if(p==3)
        c[o]='3';
        else if(p==4)
        c[o]='4';
        else if(p==5)
        c[o]='5';
        else if(p==6)
        c[o]='6';
        else if(p==7)
        c[o]='7';
        else if(p==8)
        c[o]='8';
        else if(p==9)
        c[o]='9';
        else if(p==10)
        c[o]='A';
        else if(p==11)
        c[o]='B';
        else if(p==12)
        c[o]='C';
        else if(p==13)
        c[o]='D';
        else if(p==14)
        c[o]='E';
        else if(p==15)
        c[o]='F';

    }
k=x+1;
if(q==1)//calculate c[k]
   {
    c[k]='1';
    for(int f=0;f<=(k/2);f++)
    {
      m=k-f;
      h=c[f];
      c[f]=c[m];
      c[m]=h;
    }
   }
else
   {
      for(int e=0;e<=(x/2);e++)
    {
      m=x-e;
      h=c[e];
      c[e]=c[m];
      c[m]=h;
    }
   }
}
if(x=y)
{
    for(int o=0;o<x;o++)//calculate a add b
    {
        if(a[o]=='A')
        d[o]=10;
        else if(a[o]=='B')
        d[o]=11;
        else if(a[o]=='C')
        d[o]=12;
        else if(a[o]=='D')
        d[o]=13;
        else if(a[o]=='E')
        d[o]=14;
        else if(a[o]=='F')
        d[o]=15;
        else if(a[o]=='0')
        d[o]=0;
        else if(a[o]=='1')
        d[o]=1;
        else if(a[o]=='2')
        d[o]=2;
        else if(a[o]=='3')
        d[o]=3;
        else if(a[o]=='4')
        d[o]=4;
        else if(a[o]=='5')
        d[o]=5;
        else if(a[o]=='6')
        d[o]=6;
        else if(a[o]=='7')
        d[o]=7;
        else if(a[o]=='8')
        d[o]=8;
        else if(a[o]=='9')
        d[o]=9;
        if(b[o]=='A')
        z[o]=10;
        else if(b[o]=='B')
        z[o]=11;
        else if(b[o]=='C')
        z[o]=12;
        else if(b[o]=='D')
        z[o]=13;
        else if(b[o]=='E')
        z[o]=14;
        else if(b[o]=='F')
        z[o]=15;
        else if(b[o]=='0')
        z[o]=0;
        else if(b[o]=='1')
        z[o]=1;
        else if(b[o]=='2')
        z[o]=2;
        else if(b[o]=='3')
        z[o]=3;
        else if(b[o]=='4')
        z[o]=4;
        else if(b[o]=='5')
        z[o]=5;
        else if(b[o]=='6')
        z[o]=6;
        else if(b[o]=='7')
        z[o]=7;
        else if(b[o]=='8')
        z[o]=8;
        else if(b[o]=='9')
        z[o]=9;
        p=d[o]+z[o]+q;
        M=p;
        if(p>=16)
        {
           q=1;
           p=p%16;
        }
        else
            q=0;
        s[o]=p;
        if(p==0)
        c[o]='0';
        else if(p==1)
        c[o]='1';
        else if(p==2)
        c[o]='2';
        else if(p==3)
        c[o]='3';
        else if(p==4)
        c[o]='4';
        else if(p==5)
        c[o]='5';
        else if(p==6)
        c[o]='6';
        else if(p==7)
        c[o]='7';
        else if(p==8)
        c[o]='8';
        else if(p==9)
        c[o]='9';
        else if(p==10)
        c[o]='A';
        else if(p==11)
        c[o]='B';
        else if(p==12)
        c[o]='C';
        else if(p==13)
        c[o]='D';
        else if(p==14)
        c[o]='E';
        else if(p==15)
        c[o]='F';

    }
k=x+1;
if(q==1)
   {
    c[k]='1';
    for(int f=0;f<=(k/2);f++)
    {
      m=k-f;
      h=c[f];
      c[f]=c[m];
      c[m]=h;
    }
   }
else
   {
      for(int e=0;e<=(x/2);e++)
    {
      m=x-e;
      h=c[e];
      c[e]=c[m];
      c[m]=h;
    }
   }
}

Answer 1

Lets look at what cin.getline does:

Extracts characters from stream until end of line. After constructing and checking the sentry object, extracts characters from *this and stores them in successive locations of the array whose first element is pointed to by s, until any of the following occurs (tested in the order shown):

• end of file condition occurs in the input sequence (in which case setstate(eofbit) is executed)
• the next available character c is the delimiter, as determined by Traits::eq(c, delim). The delimiter is extracted (unlike basic_istream::get()) and counted towards gcount(), but is not stored.
• count-1 characters have been extracted (in which case setstate(failbit) is executed).

If the function extracts no characters (eg if count < 1), setstate(failbit) is executed. In any case, if count>0, it then stores a null character CharT() into the next successive location of the array and updates gcount().

The result of that is in all cases, s now points to a null terminated string, of at most count-1 characters.

In your usage, you have up to 99 digits, and can use strlen to count exactly how many. eof is not a character, nor it is a member function of char .

You then reverse in place the inputs, and go about your overly repetitious conversions.

However, it's much simpler to use functions, both those you write yourself and those provided by the standard.

// translate from '0' - '9', 'A' - 'F', 'a' - 'f' to 0 - 15
static std::map<char, int> hexToDec { { '0', 0 }, { '1', 1 }, ... { 'f', 15 }, { 'F', 15 } };
// translate from 0 - 15 to '0' - '9', 'A' - 'F'
static std::map<int, char> decToHex { { 0, '0' }, { 1, '1' }, ... { 15, 'F' } };

std::pair<char, bool> hex_add(char left, char right, bool carry)
{
    // translate each hex "digit" and add them
    int sum = hexToDec[left] + hexToDec[right];
    // we have a carry from the previous sum
    if (carry) { ++sum; }
    // translate back to hex, and check if carry
    return std::make_pair(decToHex[sum % 16], sum >= 16);
}

int main()
{
    std::cout << "Input two hexadecimal numerals(both of them within 100 digits):\n";
    // read two strings
    std::string first, second;
    std::cin >> first >> second;

    // reserve enough for final carry
    std::string reverse_result(std::max(first.size(), second.size()) + 1, '\0'); 

    // traverse the strings in reverse
    std::string::const_reverse_iterator fit = first.rbegin();
    std::string::const_reverse_iterator sit = second.rbegin();
    std::string::iterator rit = reverse_result.begin();
    bool carry = false;

    // while there are letters in both inputs, add (with carry) from both
    for (; (fit != first.rend()) && (sit != second.rend()); ++fit, ++sit, ++rit)
    {
        std::tie(*rit, carry) = hex_add(*fit, *sit, carry);
    }

    // now add the remaining digits of first (will do nothing if second is longer)
    for (; (fit != first.rend()); ++fit)
    {
        // we need to account for a carry in the last place 
        // potentially all the way up if we are adding e.g. "FFFF" to "1"
        std::tie(*rit, carry) = hex_add(*fit, *rit++, carry);
    }
    // or add the remaining digits of second
    for (; (sit != second.rend()); ++sit)
    {
        // we need to account for a carry in the last place 
        // potentially all the way up if we are adding e.g. "FFFF" to "1"
        std::tie(*rit, carry) = hex_add(*sit, *rit++, carry);
    }

    // result has been assembled in reverse, so output it reversed
    std::cout << reverse_result.reverse();
}

Answer 2

This is a long answer. Because you have much bug in your code. Your using of getline is ok. But your are calling a eof() like e.eof() which is wrong. If you have looked at your compilation error, you would see that it was complaining about calling eof() on the variable e because it is of non-class type. Simple meaning it is not an object of some class. You cannot put the dot operator . on primitive types like that. I think what you are wanting to do, is to terminate the loop when you have reached the end of line. So that index1 and index2 can get the length of the string input. If I were you, I would just use C++ builtin strlen() function for that. And in the first place, you should use C++ class string to handle strings. Also strings have a null - terminating character '\\0' at the end of them. If you don't know about it, I suggest you take some time to read about strings.

Secondly, you have many bugs and errors in your code. The way you are reversing your string is not correct. Ask yourself, what are the contents of the arrays a and b at position which have higher index than the length of the string? You should use reverse() for reversing strings and arrays.

You have errors on adding loop also. Note, you are changing the arrays value when they are A , B , C , D , and so on for hexadecimal values with the corresponding decimal values 10,11,12,13 and so on. But you should change the values for the character '0' - '9' also. Because when the array holds '0' it is not integer 0. But is is ASCII '0' which has integer value of 48. And the character '1' has integer value of 49 and so on. You want to replace this values with corresponding integer values also. When you are also storing the result values in c , you are only handling only those values which are above 9 and replacing them with corresponding characters. You should also replace the integers 0 - 9 with there corresponding ASCII characters. Also don't forget to put a null terminating character at the end of the result.

Also, when p is getting larger than 15, you are only changing your carry, but you should also change p accordingly.

I believe you can reverse the result array c in a much more elegant way. By only reversing when the calculation has been performed totally. You can simple call reverse() for that.

I believe you can think hard a little bit more, and write the code in the right way. I have a few suggestions for you, don't use variable names like a,b,c,o. Try to name variables with what are they really doing. Also, you can improve your algorithm and shorten your code and headache with one simple change in the algorithm. First find the length of a and then find the length of b . If there lengths are unequal, find out which has lesser length. Then add 0s in front of it to make both lengths equal. Now, you can simply start from the back, and perform the addition. Also, you should use builtin methods like reverse() , swap() and also string class to make your life easier ;)

#include <iostream>
#include <algorithm>
#include <string>
using namespace std;



int main(){
    string firstVal,secondVal;

    cout<<"Input two hexadecimal numerals(both of them within 100 digits):\n";

    cin >> firstVal >> secondVal;

    //Adjust the length.

    if(firstVal.size() < secondVal.size()){
        //Find out the number of leading zeroes needed
        int leading_zeroes = secondVal.size() - firstVal.size();

        for(int i = 0; i < leading_zeroes; i++){
            firstVal = '0' + firstVal;
        }
    }
    else if(firstVal.size() > secondVal.size()){
        int leading_zeroes = firstVal.size() - secondVal.size();

        for(int i = 0; i < leading_zeroes; i++){
            secondVal = '0' + secondVal;
        }
    }

    // Now, perform addition.
    string result;

    int digit_a,digit_b,carry=0;

    for(int i = firstVal.size()-1; i >= 0; i--){

        if(firstVal[i] >= '0' && firstVal[i] <= '9')  digit_a = firstVal[i] - '0';
        else                                          digit_a = firstVal[i] - 'A' + 10;

        if(secondVal[i] >= '0' && secondVal[i] <= '9')  digit_b = secondVal[i] - '0';
        else                                            digit_b = secondVal[i] - 'A' + 10;


        int sum = digit_a + digit_b + carry;
        if(sum > 15){
            carry = 1;
            sum = sum % 16;

        }
        else{
            carry = 0;
        }

        // Convert sum to char.
        char char_sum;
        if(sum >= 0 && sum <= 9)    char_sum = sum + '0';
        else                        char_sum = sum - 10 + 'A';

        //Append to result.
        result = result + char_sum;

    }
    if(carry > 0)   result = result + (char)(carry + '0');

    //Result is in reverse order.
    reverse(result.begin(),result.end());

    cout << result << endl;

}

Answer 3

Regarding the text of your problem: “add one to the digit on the left if the sum of the current digits exceed 16” is wrong; it should be 15, not 16.

Regarding your code : I did not have the patience to read all your code, however:

• I have noticed one long if/else . Use a switch (but you do not need one).
• To find out if a character is a hex digit use isxdigit ( #include <cctype> ).
• The user might input uppercase and lowercase characters: convert them to the same case using toupper / tolower .
• To convert a hex digit to an integer:
  • if the digit is between '0' and '9' simply subtract '0'. This works because the codes for '0', '1'… are 0x30, 0x31... (google ASCII codes).
  • if the digit is between 'A' and 'F', subtract 'A' and add 10.

Solving the problem :

• “less than 100 digits long” This is a clear indication regarding how your data must be stored: a simple 100 long array, no std::string , no std::vector :

   #define MAX_DIGITS 100 typedef int long_hex_t[MAX_DIGITS]; 

  In other words your numbers are 100 digits wide, at most.

  Decide how you store the number: least significant digit first or last? I would chose to store the least significant first. 123 is stored as {3,2,1,0,…0}

• Use functions to simplify your code. You will need three functions: read , print and add :

   int main() { long_hex_t a; read( a ); long_hex_t b; read( b ); long_hex_t c; add( c, a, b ); print( c ); return 0; } 

• The easiest function to write is add followed by print and read .

• For read use get and putback to analyze the input stream: get extracts the next character from stream and putback is inserting it back in stream (if we do not know how to handle it).

Here it is a full solution ( try it ):

#include <iostream>
#include <cctype>

#define MAX_DIGITS 100

typedef int long_hex_t[MAX_DIGITS];

void add( long_hex_t c, long_hex_t a, long_hex_t b )
{
  int carry = 0;

  for ( int i = 0; i < MAX_DIGITS; ++i )
  {
    int t = a[i] + b[i] + carry;
    c[i] = t % 16;
    carry = t / 16;
  }
}

void print( long_hex_t h )
{
  //
  int i;

  // skip leading zeros
  for ( i = MAX_DIGITS - 1; i >= 0 && h[i] == 0; --i )
    ;

  // all zero
  if ( i < 0 )
  {
    std::cout << '0';
    return;
  }

  // print remaining digits
  for ( i; i >= 0; --i )
    std::cout << char( h[i] < 10 ? h[i] + '0' : h[i] - 10 + 'A' );
}

void read( long_hex_t h )
{
  // skip ws
  std::ws( std::cin );

  // skip zeros
  {
    char c;
    while ( std::cin.get( c ) && c == '0' )
      ;
    std::cin.putback( c );
  }

  //
  int count;
  {
    int i;
    for ( i = 0; i < MAX_DIGITS; ++i )
    {
      char c;
      if ( !std::cin.get( c ) )
        break;
      if ( !std::isxdigit( c ) )
      {
        std::cin.putback( c );
        break;
      }
      c = std::toupper( c );
      h[i] = c <= '9'
        ? ( c - '0' )
        : ( c - 'A' + 10 );
    }
    count = i;
  }

  // reverse
  for ( int i = 0, ri = count - 1; i < count / 2; ++i, --ri )
  {
    int t = h[i];
    h[i] = h[ri];
    h[ri] = t;
  }

  // fill the rest with zero
  for ( int i = count; i < MAX_DIGITS; ++i )
    h[i] = 0;
}

int main()
{
  long_hex_t a;
  read( a );

  long_hex_t b;
  read( b );

  long_hex_t c;
  add( c, a, b );

  print( c );

  return 0;
}