Adding and deducting big numbers - c++

Can you please help me buy showing how to create substraction of two long numbers?
I found this code on the internet:
#include<iostream>
#include<string>
using namespace std;
int main(void)
{
// the two "numbers" to be added. Make them as long as you like.
string numStr1;
string numStr2;
cout << "Enter 1st number: "; cin >> numStr1;
cout << "Enter 2nd number: "; cin >> numStr2;
// keeping track of which string is longest using references
string& rLongStr = numStr1.length() > numStr2.length() ? numStr1 : numStr2;
string& rShortStr = numStr1.length() <= numStr2.length() ? numStr1 : numStr2;
// initialize the sum with the long string but with space for a final carry at the beginning
string numStrSum = '0' + rLongStr;// the '0' in case of a final carry
// must go through the strings backwards since the least
// significant digit is at the end, not the beginning.
string::reverse_iterator r_itShort, r_itSum;
r_itShort = rShortStr.rbegin();// point to last "digit" in the short string
r_itSum = numStrSum.rbegin();// same for sum string
// add the "digits" one by one from end to beginning of the short string
while( r_itShort != rShortStr.rend() )
{
*r_itSum += *r_itShort - '0';// "add" the digits
if( *r_itSum > '9' )// must carry a one to the next "digit"
{
*(r_itSum + 1) += 1;
*r_itSum -= 10;
}
++r_itShort;// move back 1 character
++r_itSum;// in each string
}
if( numStrSum.at(0) == '0' )// if 1st character is stiil '0'
numStrSum.erase(0,1);// erase it
// output result
cout << numStrSum;
cout << endl;
return 0;
}
So, I could not make substraction of this two numbers. I tried smt like that:
while( r_itShort != rShortStr.rend() )
{
*r_itSum -= *r_itShort + '0';
if( *r_itSum > '9' )
{
*(r_itSum + 1) -= 1;
*r_itSum += 10;
}
--r_itShort;
--r_itSum;
}
if( numStrSum.at(0) == '0' )
numStrSum.erase(0,1);
But it shows me the same. Can you please help me and tell what am I doing wrong?
Thank You!

You can both input and output float / double / integer types to the iostream with cin/cout so string parsing wouldn't be a problem. You can just get them as double type.
If you really want to get these numbers as strings, you could use atof to convert them to floats. Then simply calculate what you need and either use std::to_string or just print directly.

Related

What will I have to write in my program so that it would only accept commas in their proper location (i.e. cannot input 10,00.0 , 100,0.2334, etc)?

I am taking a summer course in computer science, and I get assigned two projects a week, so bear with me if I'm getting some terminology wrong.
This week, I was able to get the first one done, but not the second one. In the second assignment, we were required to rewrite the ReadDouble function so that it's more user-friendly; user-friendly as in allowing the user to input commas along with numbers. Additionally,
we were required to allow the first character to be a number, plus or minus sign, or a decimal point.
All other characters can be a number, comma, or decimal point (if there already wasn't one).
As mentioned, commas must be written in properly (we can't allow the user to input 10,00.244, 343,2.334, or some nonsensical number like that).
No commas should be allowed after the decimal point.
Only one decimal point should be allowed in the number.
So far, I was able to able to meet 1), 2), and 5). 3) and 4)? Not so much.
The underlying issue is that I don't know what classes, objects, and whatnot I should be using to make the program read the string input and determine if the commas were inserted properly. I have an idea that I would need to use something similar to the "input.length()" code, set it as a variable that can be compared in an if statement to ensure that the amount of digits until the next comma can be used is met.
I also tried writing a for loop that would check after the decimal place for commas or any other invalid character, but I didn't know what to write down as its initialization. How would I get the for loop to start looking from the decimal after it's aware that one decimal exists?
Another major issue I am encountering is that when I input something like 1.2, it is displayed as 12, meaning that "atof(convert.cstr())" has stripped the decimal from the return value. However, when I enter it as just .2, it comes out as 0.2.
I will provide the code of what I have written so far along with the code of what a friend has suggested to me.
My code:
#include <iostream>
#include <cstdlib>
#include <string>
#include <climits>
using namespace std;
// Main Menu Error Prototype
double ReadDouble(string prompt);
double ReadDouble(string prompt)
{
string input;
string convert;
bool isValid = true;
do {
// Reset the flag to valid input
isValid = true;
// Read input from user
cout << prompt;
cin >> input;
// Validate user input
// Check the first character is a number, + or -
int decimal = 0;
if (input[0] != '+' && input[0] != '-' && input[0] != '.' && isdigit(input[0]) == 0) {
cout << "Error! Input was not an integer.\n";
isValid = false;
}
else {
if (input[0] == '.') {
decimal++;
//cout << "." << endl;
}
convert = input.substr(0, 1);
}
// check that the remaining characters are numeric
long len = input.length();
for (long index = 1; index < len && isValid == true && decimal <= 1; index++) {
if (input[index] == ',') {
; // do nothing if character is a ','
}
else if (input[index] == '.') {
decimal++; // do nothing if character is a '.'
if (decimal > 1) {
cout << "Error! You can have only one decimal point.\n";
isValid = false;
}
}
else if (isdigit(input[index]) == 0) {
cout << "Error! Input was not an integer.\n";
isValid = false;
}
else {
convert += input.substr(index, 1);
}
}
// Start looking where the decimal starts
/*
long decimal=input.find('.');
for (decimal; decimal < len && isValid==true; decimal++) {
if (input[decimal] =='.') {
; // do nothing if character is a '.'
}
}
*/
//cout << "\nDecimal value is " << decimal << endl; -- Test Code
} while (isValid == false);
double returnvalue = atof(convert.c_str());
return returnvalue;
}
int main()
{
double x = ReadDouble("Enter a value: ");
cout << "Value entered was " << x << endl;
return 0;
}
My friend's incomplete code:
ReadDouble(){
isValid = true
do{
get user input and set it to a variable called input
set output variable to a variable called output
bool hasDecimal = false;
int digitsUntilNextComma = 3
for(i = 0; i < input length; i++){
if(input[i] == ','){
if((i < 3 && i > 0) || digitsUntilNextComma == 0){
digitsUntilNextComma = 3;
output += input[i];
}else{ //if it gets to here the comma was in a bad place like ,123 or 12,12,123
isValid = false;
i = input length //breaks out of for loop
}
} else if(input[i] == '.'){
if(i < 3 || digitsUntilNextComma == 0){
if(hasDecimal){ //if this is true then the input had 2 decimals
isValid = false;
i = input length //breaks out of for loop
}else{
hasDecimal = true;
digitsUntilNextComma = 3;
output += input[i];
}
}else{ //if it gets to here, a previous comma was in a bad place like 12,34.56
isValid = false;
i = input length //breaks out of for loop
}
}else{
output += input[i];
digitsUntilNextComma--;
}
}
}while(isValid == false)
}
I hope what I provided wasn't too vague or messy. Again, I had little exposure to programming in the past, so forgive me if I mess some terminology up.
This is a job for a FSM.
Start on + to IN0
Start on - to IN0
Start on Digit to IN1
Start on Decimal to D_prime
IN0 on Decimal to D_prime
IN0 on Digit to IN1
IN1 on Digit to IN2
IN1 on Decimal to D
IN1 on Comma to N0
IN2 on Digit to N3
IN2 on Comma to N0
IN2 on Decimal to D
IN# is "initial numbers, seen #".
D_prime on Digit to D
D on Digit to D
N0 on Digit to N1
N1 on Digit to N2
N2 on Digit to N3
N3 on Comma to N0
N3 on Decimal to D
Valid End states are N3, IN1, IN2, IN3 and D.
Any unexpected character is an input error.
For State, keep an accumulator, a sign, and a decimal count with initial state 0, positive, 0.
On Start(-) set sign to negative.
Whenever you see a digit, multiply accumulator by 10 and add it.
In D or D_prime state, also increase decimal count by 1. D_prime exists to eliminate . as a valid number.
Result is (sign is negative?-1:1)*accumulator/10 to the power of decimal count.
To test this, write a program that generates valid numbers, another that generates gibberish, and another that randomly adds, subtracts, shuffles and splices characters from valid number(s).

Checking if items from a particular txt file agree to constraints in c++ - Name That Number USACO

I have got some doubts while solving - Name That Number.
It goes like this -
Among the large Wisconsin cattle ranchers, it is customary to brand cows with serial numbers to please the Accounting Department. The cowhands don't appreciate the advantage of this filing system, though, and wish to call the members of their herd by a pleasing name rather than saying, "C'mon, #4734, get along."
Help the poor cowhands out by writing a program that will translate the brand serial number of a cow into possible names uniquely associated with that serial number. Since the cowhands all have cellular saddle phones these days, use the standard Touch-Tone(R) telephone keypad mapping to get from numbers to letters (except for "Q" and "Z"):
2: A,B,C 5: J,K,L 8: T,U,V
3: D,E,F 6: M,N,O 9: W,X,Y
4: G,H,I 7: P,R,S
Acceptable names for cattle are provided to you in a file named "dict.txt", which contains a list of fewer than 5,000 acceptable cattle names (all letters capitalized). Take a cow's brand number and report which of all the possible words to which that number maps are in the given dictionary which is supplied as dict.txt in the grading environment (and is sorted into ascending order).
For instance, brand number 4734 produces all the following names:
GPDG GPDH GPDI GPEG GPEH GPEI GPFG GPFH GPFI GRDG GRDH GRDI
GREG GREH GREI GRFG GRFH GRFI GSDG GSDH GSDI GSEG GSEH GSEI
GSFG GSFH GSFI HPDG HPDH HPDI HPEG HPEH HPEI HPFG HPFH HPFI
HRDG HRDH HRDI HREG HREH HREI HRFG HRFH HRFI HSDG HSDH HSDI
HSEG HSEH HSEI HSFG HSFH HSFI IPDG IPDH IPDI IPEG IPEH IPEI
IPFG IPFH IPFI IRDG IRDH IRDI IREG IREH IREI IRFG IRFH IRFI
ISDG ISDH ISDI ISEG ISEH ISEI ISFG ISFH ISFI
As it happens, the only one of these 81 names that is in the list of valid names is "GREG".
Write a program that is given the brand number of a cow and prints all the valid names that can be generated from that brand number or ``NONE'' if there are no valid names. Serial numbers can be as many as a dozen digits long.
Here is what I tried to solve this problem. Just go through all the names in the list and check which is satisfying the constraints given.
int numForChar(char c){
if (c=='A'||c=='B'||c=='C') return 2;
else if(c=='D'||c=='E'||c=='F') return 3;
else if(c=='G'||c=='H'||c=='I') return 4;
else if(c=='J'||c=='K'||c=='L') return 5;
else if(c=='M'||c=='N'||c=='O') return 6;
else if(c=='P'||c=='R'||c=='S') return 7;
else if(c=='T'||c=='U'||c=='V') return 8;
else if(c=='W'||c=='X'||c=='Y') return 9;
else return 0;
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
freopen("namenum.in","r",stdin);
freopen("namenum.out","w",stdout);
string S; cin >> S;
int len = S.length();
freopen("dict.txt","r",stdin);
string x;
while(cin >> x){
string currName = x;
if(currName.length() != S.length()) continue;
string newString = x;
for(int i=0;i<len;i++){
//now encode the name as a number according to the rules
int num = numForChar(currName[i]);
currName[i] = (char)num;
}
if(currName == S){
cout << newString << "\n";
}
}
return 0;
}
Unfortunately, when I submit it to the judge, for some reason, it says no output produced that is my program created an empty output file. What's possibly going wrong?
Any help would be much appreciated. Thank You.
UPDATE: I tried what Some Programmer Dude suggested by adding a statement else return 0; at the end of the numOfChar function in case of a different alphabet. Unfortunately, it didn't work.
So after looking further at the question and exploring the information for Name That Number. I realized that it is not a current contest, and just a practice challenge. Thus, I updated my answer and also giving you my version of a successful submission. Nonetheless, that is a spoiler and will be posted after why your code was not working.
First, you forgot a } after the declaration of your number function. Secondary, you did not implement anything to check whether if the input fail to yield a valid name. Third, when you use numForChar() on the character of currName, the function yielded an integer value. That is not a problem, the problem is that it is not the ASCII code but is a raw number. You then compare that against a character of the input string. Of which, is an ASCII's value of a digit. Thus, your code can't never find a match. To fix that you can just add 48 to the return value of the numForChar() function or xor the numForChar() return's value to 48.
You are on the right track with your method. But there is a few hints. If you are bored you can always skip to the spoiler. You don't need to use the numForChar() function to actually get a digit value from a character. You can just use a constant array. A constant array is faster than that many if loop.
For example, you know that A, B, C will yield two and A's ASCII code is 65, B's is 66, and C's equal to 67. For that 3, you can have an array of 3 indexes, 0, 1, 2 and all of them stores a 2. Thus, if you get B, you subtract B's ASCII code 65 will yield 1. That that is the index to get the value from.
For getting a number to a character you can have a matrix array of char instead. Skip the first 2 index, 0 and 1. Each first level index, contain 3 arrays of 3 characters that are appropriate to their position.
For dictionary comparing, it is right that we don't need to actually look at the word if the length are unequal. However, besides that, since their dictionary words are sorted, if the word's first letter is lower than the range of the input first letter, we can skip that. On the other hand, if words' first letter are now higher than the highest of the input first letter, there isn't a point in continue searching. Take note that my English for code commenting are almost always bad unless I extensively document it.
Your Code(fixed):
#include <iostream>
#include <fstream>
#include <string>
using namespace std;
int numForChar(char c){
if (c=='A'||c=='B'||c=='C') return 2;
else if(c=='D'||c=='E'||c=='F') return 3;
else if(c=='G'||c=='H'||c=='I') return 4;
else if(c=='J'||c=='K'||c=='L') return 5;
else if(c=='M'||c=='N'||c=='O') return 6;
else if(c=='P'||c=='R'||c=='S') return 7;
else if(c=='T'||c=='U'||c=='V') return 8;
else if(c=='W'||c=='X'||c=='Y') return 9;
else return 0;
}
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
ifstream fin("namenum.in");
ifstream dict("dict.txt");
ofstream fout("namenum.out");
string S;
fin >> S;
int len = S.length();
bool match = false;
string x;
while(dict >> x){
string currName = x;
if(currName.length() != S.length()) continue;
string newString = x;
for(int i=0;i<len;i++){
//now encode the name as a number according to the rules
int num = numForChar(currName[i]) ^ 48;
currName[i] = (char)num;
}
if(currName == S){
fout << newString << "\n";
match = true;
}
}
if ( match == false ){
fout << "NONE" << endl;
}
return 0;
}
Spoiler Code(Improved):
#include <fstream>
#include <string>
using namespace std;
// A = 65
// 65 - 0 = 65
const char wToN[] = {
// A ,B ,C ,D ,E ,F ,G ,H ,I ,
'2','2','2','3','3','3','4','4','4',
// J ,K ,L ,M ,N ,O ,P ,Q ,R ,S
'5','5','5','6','6','6','7','7','7','7',
// T ,U ,V ,W ,X ,Y ,Z
'8','8','8','9','9','9','9'
};
// 2 = {A, B, C} = 2[0] = A, 2[1] = B, 2[2] C
const char nToW[10][3] = {
{}, // 0 skip
{}, // 1
{'A','B','C'},
{'D','E','F'},
{'G','H','I'},
{'J','K','L'},
{'M','N','O'},
{'P','R','S'},
{'T','U','V'},
{'W','X','Y'}
};
int main(){
ifstream fin("namenum.in");
ifstream dict("dict.txt");
ofstream fout("namenum.out");
string S;
fin >> S;
// Since this will not change
// make this a const to make it
// run faster.
const int len = S.length();
// lastlen is last Index of length
// We calculate this value here,
// So we do not have to calculate
// it for every loop.
const int lastLen = len - 1;
int i = 0;
unsigned char digits[len];
unsigned char firstLetter[3];
// If not match print None
bool match = false;
for ( ; i < len; i++ ){
// No need to check upper bound
// constrain did not call for check.
if ( S[i] < '2' ) {
fout << "NONE" << endl;
return 0;
}
}
const char digit1 = S[0] ^ 48;
// There are 3 set of first letter.
// We get them by converting digits[0]'s
// value using the nToW array.
firstLetter[0] = nToW[digit1][0];
firstLetter[1] = nToW[digit1][1];
firstLetter[2] = nToW[digit1][2];
string dictStr;
while(dict >> dictStr){
// For some reason, when keeping the i = 0 here
// it seem to work faster. That could be because of compiler xor.
i = 0;
// If it is higher than our range
// then there is no point contineuing.
if ( dictStr[0] > firstLetter[2] ) break;
// Skip if first character is lower
// than our range. or If they are not equal in length
if ( dictStr[0] < firstLetter[0] || dictStr.length() != len ) continue;
// If we are in the letter range
// we always check the second letter
// not the first, since we skip the first
i = 1;
for ( int j = 1; j < len; j++ ){
// We convert each letter in the word
// to the corresponding int value
// by subtracting the word ASCII value
// to 65 and use it again our wToN array.
// if it does not match the digits at
// this current position we end the loop.
if ( wToN[dictStr[i] - 65] != S[j] ) break;
// if we get here and there isn't an unmatch then it is a match.
if ( j == lastLen ) {
match = true;
fout << dictStr << endl;
break;
}
i++;
}
}
// No match print none.
if ( match == false ){
fout << "NONE" << endl;
}
return 0;
}
I suggest you use c++ file handling. Overwriting stdin and stdout doesn't seem appropriate.
Add these,
std::ifstream dict ("dict.txt");
std::ofstream fout ("namenum.out");
std::ifstream fin ("namenum.in");
Accordingly change,
cin >> S --to--> fin >> S;
cin >> x --to--> dict >> x
cout << newString --to--> fout << newString

How to store in a string and convert to character array?

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;
}
}
}
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 (e.g. 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();
}
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;
}
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;
}

Integer to String: Without using stoi library

I found this question in the book where it asks us to transfer an Int to a string. Without using stoi library
so for example if x = 10, s="10"
the code should handle negative numbers.
I found this solution in the book. I typed it in my compiler however it is giving the string of the first number only
so if x= 45, it is giving "4"
I do not understand this line s = '0' + x%10; to be able to fix the code.
why is he adding '0' to the string. What is the best solution.
here is the code: I added comments to the parts I understand
#include<iostream>
#include<string>
using namespace std;
void IntToString(int x);
int main()
{
int num;
cout << "Please enter a number" << endl;
cin >> num;
IntToString(num);
}
void IntToString(int x)
{
bool isNegative = false;
if(x < 0) //if its negative make boolean true
{
x = -x;
isNegative = true;
}
string s;
do
{
s = '0' + x%10; //modulus for getting the last number
x = x/10; //shorten the number
}while(x);
reverse(s.begin(), s.end()); //reverse the string since it starts from end
if(isNegative)
s = '-' + s;
cout << s << endl;
}
s = '0' + x%10;
will grabs the last digit from x%10 and add ASCII of 0 i.e. 48, giving the ASCII of desired last digit, which get copy assigned to string s using its assignment operator.
And btw, you need :
s += '0' + x%10;
~~ // += operator
The problem with the do ... while loop is that you just extract the last digit of the changed x only to replace it with the second last digit and so on, until you get the first digit of x stored in s.
There is no effect of reverse(s.begin(), s.end()) since s practically holds only one character.
Also, we add '0' to s because s initially stores the integer value of the number, and adding '0' converts it into its ASCII form.
Example:
void IntToString(int x)
{
bool isNegative = false;
if(x < 0)
{
x = -x;
isNegative = true;
}
string s;
do
{
//The below code appends the new number at the beginning of s.
s = ('0' + x%10) + s; //modulus for getting the last number
x = x/10; //shorten the number
}while(x);
if(isNegative)
main_string = '-' + main_string;
cout << main_string << endl;
}

A cleaner way to convert a string to int after checking for hex prefix?

This little exercise is meant to get a string from the user that could be decimal, hexadecimal, or octal. 1st I need to identify which kind of number the string is. 2nd I need to convert that number to int and display the number in its proper format, eg:
cout <<(dec,hex,oct, etc)<< number;
Here's what I came up with. I'd like a simpler, cleaner way to write this.
string number = "";
cin >> number;
string prefix = "dec";
char zero = '0';
char hex_prefix = 'x';
string temp = "";
int value = 0;
for(int i =0; i<number.size();++i)
{
if(number[0] == zero)//must be octal or hex
{
if (number[0] == zero && number[1] == hex_prefix ) //is hex
{
prefix = "hex";
for(int i = 0; i < (number.size() - 2); ++i)
{
temp[i] = number[i+2];
}
value = atoi(temp.c_str());
}
//... code continues to deal with octal and decimal
You are checking number[0] twice, that's the first most obvious problem.
The inner if already checks both number[0] and number[1], I don't see the point of the outer one.
The outermost loop is also hard to understand, do you expect non-hex data before the number, or what? Your question could be clearer on how the expected input string looks.
I think the cleanest would be to ignore this, and push it into existing (library) code that can parse integers in any base. In C I would recommend strtoul(), you can of course use that in C++ too.
You have two inner loop with same value integer this could be a conflict problem in your code. I suggest you look at the isdigit and islower methods in the c++ library and take advantage of those methods to accomplish your task. isdigit & islower
Good Luck
This is prints the number after deleting the hex prefix, otherwise return 0:
#include<iostream>
#include<cmath>
#include<stdlib.h>
using namespace std;
int main(){
string number = "";
cin >> number;
string prefix = "dec";
char zero = '0';
char hex_prefix = 'x';
string temp = "";
int value = 0;
if (number.size()>=2 && number[0] == zero && number[1] == hex_prefix ) //is hex
{
prefix = "hex";
for(int i = 0; i < (number.size() - 2); ++i)
{
temp[i] = number[i+2];
}
value = atoi(temp.c_str());
}
cout<<value;
return 0;
}
This partial solution that I found is as clean as possible, but it doesn't report the format of the integer:
int string_to_int(std::string str)
{
std::istringstream stream;
stream.unsetf(std::ios_base::dec);
int result;
if (stream >> result)
return result;
else
throw std::runtime_error("blah");
}
...
cout << string_to_int("55") << '\n'; // prints 55
cout << string_to_int("0x37") << '\n'; // prints 55
The point here is stream.unsetf(std::ios_base::dec) - it unsets the "decimal" flag that is set by default. This format flag tells iostreams to expect a decimal integer. If it is not set, iostreams expect the integer in any base.