i have the next code which asks the user for a really long number like 100000000 and then it prints how many times a given digit appears on that number, the code works fine and does everything correctly, but the professor told me that i dont have to use strings or chars, but when the code asks the user for a number it necessarily needs a string and i don´t know how to modify it, i used the gmp library
#include <iostream>
#include <stdio.h>
#include <gmp.h>
#define MAX 40
using namespace std;
void searchDigit(FILE *fd);
int NewNumber();
int main()
{
FILE *fd;
int otherNumber;
string text;
mpz_t num;
do
{
if((fd = fopen("File.txt","w+"))!= NULL)
{
mpz_init(num);
cout << "Give me the number: " << endl;
cin >> text;
mpz_set_str(num,text.c_str(),10);
mpz_out_str(fd,10,num);
fclose(fd);
searchDigit(fd);
otherNumber = NewNumber();
}
else
cout << "Fail!!" << endl;
}while(otherNumber);
return 0;
}
void searchDigit(FILE *fd)
{
int car,continue = 1,r;
char answer,digit;
if((fd = fopen("File.txt","r"))!= NULL)
{
do
{
r = 0;
fseek(fd,0,SEEK_SET);
cout << "What digit do you want to search? " << endl;
cin >> digit;
while((car = fgetc(fd))!= EOF)
{
if(car == digit)
r++;
}
cout << "The digit x=" <<digit<< " appears " << r << " times" << endl;
cout << "Do you want to search any other digit? " << endl;
cin >> answer;
if(answer != 'S')
continue = 0;
}while(continue);
}
else
cout << "Fail!!" << endl;
}
int NewNumber()
{
char answer;
cout << "DO you wish to work with a new number? " << endl;
cin >> answer;
if(answer == 'S' || answer == 's')
return 1;
else
return 0;
}
Thanks in advance
Depends on how big your input might actually be... but for retrieving digits you could do something like:
#include <iostream>
using namespace std;
typedef unsigned long long UINT64;
int main() {
UINT64 i;
std::cin >> i;
while (i >= 1) {
int digit = i % 10;
std::cout << digit << " ";
i /= 10;
}
}
input: 18446744073709551614
outputs: 4 1 6 1 5 5 9 0 7 3 7 0 4 4 7 6 4 4 8 1
Related
I wanted to use only 1 and 0 for the binary. But instead the answer keep giving me the 2nd option with whatever number I typed. I had tried where did I programmed wrongly but unfortunately I still can't find it. So I hoped that I could get some help here.
#include<iostream>
#include<cmath>
using namespace std;
int DualzahlZuDezimal(long long n)
{
int dez = 0;
int i = 0, rem;
while (n != 0)
{
rem = n % 10;
n /= 10;
dez += rem * pow(2, i);
++i;
}
return dez;
}
string a;
int main()
{
long long n;
int dez;
cout << "Test Ein- und Ausgabe : \n";
cout << "----------------------- \n";
cout << "Eingabe einer Dualzahl : ";
cin >> n;
if ((n == '1') && (n == '0'))
{
cout << "Dual : " << n << endl;
cout << "Dezimal : " << DualzahlZuDezimal(n) << endl;
cout << "cin ok ? : ja-ok" << endl;
return 0;
}
else
{
cout << "Dual : 0" << endl;
cout << "Dezimal : 0" << endl;
cout << "cin ok ? : nein-nicht ok" << endl;
return 0;
}
}
If I understand this right, you want the user to enter a binary number, like 10001101001, and you will show the decimal equivalent (1129 in this case).
There are 2 general ways to do that yourself:
You can read the value as a number, as you do, and then apply your conversion
process, except that you check that rem is either 0 (in which case you do
nothing), or 1 (in which case you add the power of 2). If it's another value,
you report the error, and return 0.
You can read the value as a std::string instead. Then you can use
std::find_first_not_of()
to check for contents other than 0 or 1:
if (n.find_first_not_of("01") != string::npos) { /* complain */ }
but then you need to do the conversion based on characters.
But the best approach is not to reinvent the wheel and instead let the standard library handle it for you via stol():
#include <cstddef>
#include <iostream>
#include <string>
using namespace std;
int
main()
{
string text;
cout << "Enter a binary number: " << flush;
cin >> text;
size_t endpos = 0;
long decimal_number = stol(text, &endpos, 2); // base 2 == binary
if (endpos != text.size()) {
cerr << "'" << text << "' is not a valid binary number!" << endl;
return 1;
}
else {
cerr << "binary number: " << text << endl;
cerr << "decimal number: " << decimal_number << endl;
return 0;
}
}
Keep in mind that input from the console is text. If you need to check that the text matches a particular format (in this case, consists entirely of 1's and 0's), the simplest approach is to look at that text:
std::string input;
std::cin >> input;
bool input_is_valid = true;
for (int i = 0; input_is_valid && i < input.length(); ++i) {
if (input[i] != '0' && input[i] != '1')
input_is_valid = false;
}
then, if the input is valid, convert the text to a numeric value:
long long n = std::stoll(input);
I am coding a program that converts a binary number into decimal number by doubling (link to wikihow article).
If the user input is something other than 1 or 0, then its not a binary number, under that circumstance I want the loop to "break" and say something like:
"Oops! Binary numbers have only 1 or 0".
If not "then" the loop should continue.
That is I want to code something like
for(int digits = 0; digits != digitsINbinNum; ++digits){
if(a condition that checks if user input is anything else than 1 or 0){
coût << ""Oops! Binary numbers have only 1 or 0" << endl;
break;
}else{
cin >> binArray[digits];/*<-----------Here's the part where I am trying to do that*/
}
}
Refer to the code given below for more info:
#include <iostream>
#include <iterator>
using namespace std;
int main(){
int digitsINbinNum;
cout << "If you don't mind. Please enter the number of digits in your binary number: ";
cin >> digitsINbinNum;
int binArray[digitsINbinNum];
cout << "Enter the binary number: ";
for(int digits = 0; digits != digitsINbinNum; ++digits){
cin >> binArray[digits];/*<-----------Here's the part where I am trying to do that*/
}
/*using the doubling method as found in wikihow.com*/
int total = 0;
for(int posiOFdigit = 0; posiOFdigit != sizeof(binNum[noOFdigits]); posiOFdigit++){
total = total * 2 + binNum[posiOFdigit];
}
/*Printing the number*/
cout << "Decimal form of ";
for(int n = 0; n != noOFdigits; n++){
cout << binNum[n];
}
cout << " is " << total;
return 0;
}
The logic for converting a binary number into decimal number by the doubling method can be referred from the given link in the question.
Modifying the given code to keep it as close as possible to the question's reference code.
Note: As ISO C++ forbids variable length array, I am changing
int binArray[digits] to
int *binArray = (int *)malloc(sizeof(int) * digitsINbinNum);.
This modification makes it an integer pointer and it gets the memory of required size allocated at runtime.
#include <iostream>
using namespace std;
int main(){
int digitsINbinNum,
/* variable to keep decimal conversion of given binary */
decimal_val = 0;
bool is_binary = true;
cout << "If you don't mind. Please enter the number of digits in your binary number: ";
cin >> digitsINbinNum;
/*
ISO C++ forbids variable length array,
making it int pointer and allocating dynamic memory
*/
int *binArray = (int *)malloc(sizeof(int) * digitsINbinNum);
if (binArray == NULL)
{
cout << "Memory allocation failure" << endl;
exit -1;
}
cout << "Enter the binary number: ";
for(int digits = 0; digits != digitsINbinNum; ++digits){
cin >> binArray[digits];
/*<-----------Here's the part where I am trying to do that*/
/* doubling method logic for conversion of given binary to decimal */
if ((binArray[digits] == 0) ||
(binArray[digits] == 1))
{
decimal_val = (decimal_val * 2) + binArray[digits];
}
else /* not a binary number */
{
is_binary = false;
cout << "Oops! Binary numbers have only 1 or 0" << endl;
break;
}
}
/* if conversion is successful: print result */
if (is_binary)
{
cout << "Decimal Value for given binary is: " << decimal_val << endl;
}
if (binArray)
{
free(binArray);
}
return 0;
}
You don't need an array for this. Here is a simple solution:
#include <iostream>
int main(){
int digitsINbinNum;
std::cout << "If you don't mind. Please enter the number of digits in your binary number: ";
std::cin >> digitsINbinNum;
std::cout << "Enter the binary number: ";
int ret = 0;
for(int digits = 0; digits != digitsINbinNum; ++digits) {
int bin;
std::cin >> bin;
if (bin == 1 || bin == 0) {
ret = 2 * ret + bin;
} else {
std::cout << "Oops! Binary numbers have only 1 or 0" << std::endl;
return -1;
}
}
std::cout << ret << std::endl;
return 0;
}
I would like to read numbers into a static array of fixed size 10, but the user can break the loop by entering character E.
Here's my code:
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
int myArray[10];
int count = 0;
cout << "Enter upto 10 integers. Enter E to end" << endl;
for (int i = 0; i < 10; i++)
{
cout << "Enter num " << i + 1 << ":";
cin >> myArray[i];
if (myArray[i] != 'E')
{
cout << myArray[i] << endl;
count++;
}
else
{
break;
}
}
cout << count << endl;
system("PAUSE");
return 0;
}
However, I get the following results while entering E:
Enter upto 10 integers. Enter E to end
Enter num 1:5
5
Enter num 2:45
45
Enter num 3:25
25
Enter num 4:2
2
Enter num 5:E
-858993460
Enter num 6:-858993460
Enter num 7:-858993460
Enter num 8:-858993460
Enter num 9:-858993460
Enter num 10:-858993460
10
Press any key to continue . . .
How can I fix this code in the simplest way?
cin fails for parsing character 'E' to int. The solution would be to read string from user check if it is not "E" (it is a string not a single char so you need to use double quotes) and then try to convert string to int. However, this conversion can throw exception (see below).
Easiest solution:
#include <iostream>
#include <cmath>
#include <string> //for std::stoi function
using namespace std;
int main()
{
int myArray[10];
int count = 0;
cout << "Enter upto 10 integers. Enter E to end" << endl;
for (int i = 0; i < 10; i++)
{
cout << "Enter num " << i + 1 << ":";
std::string input;
cin >> input;
if (input != "E")
{
try
{
// convert string to int this can throw see link below
myArray[i] = std::stoi(input);
}
catch (const std::exception& e)
{
std::cout << "This is not int" << std::endl;
}
cout << myArray[i] << endl;
count++;
}
else
{
break;
}
}
cout << count << endl;
system("PAUSE");
return 0;
}
See documentation for std::stoi. It can throw exception so your program will end suddenly (by termination) that is why there is try and catch blocks around it. You will need to handle the case when user puts some garbage values in your string.
Just use:
char myArray[10];
because at the time of taking input console when get character then try to convert char to int which is not possible and store default value in std::cin i.e. 'E' to 0 (default value of int).
Use below code:
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
char myArray[10];
int count = 0;
cout << "Enter upto 10 integers. Enter E to end" << endl;
for (int i = 0; i < 10; i++)
{
cout << "Enter num " << i + 1 << ":";
cin >> myArray[i];
if (myArray[i] == 'E')
{
break;
}
else
{
cout << myArray[i] << endl;
count++;
}
}
exitloop:
cout << count << endl;
system("PAUSE");
return 0;
}
Output:
Enter upto 10 integers. Enter E to end
Enter num 1:1
1
Enter num 2:E
1
sh: 1: PAUSE: not found
If you debug this, you will find all your myArray[i] are -858993460 (=0x CCCC CCCC), which is a value for the uninitialized variables in the stack.
When you put a E to an int variable myArray[i]. std::cin will set the state flag badbit to 1.
Then when you run cin >> myArray[i], it will skip it. In other words, do nothing.
Finally, you will get the result as above.
The problem is that attempting to read E as an int fails, and puts the stream in an error state where it stops reading (which you don't notice because it just doesn't do anything after that) and leaves your array elements uninitialized.
The simplest possible way is to break on any failure to read an integer:
for(int i = 0; i < 10; i++)
{
cout << "Enter num " << i + 1 << ":";
if (cin >> myArray[i])
{
cout << myArray[i] << endl;
count++;
}
else
{
break;
}
}
If you want to check for E specifically, you need to read a string first, and then convert that to an int if it's not E.
As a bonus, you need to handle everything that's neither int nor E, which complicates the code a bit.
Something like this:
int count = 0;
string input;
while (cin >> input && count < 10)
{
if (input == "E")
{
break;
}
istringstream is(input);
if (is >> myArray[count])
{
cout << myArray[count] << endl;
count++;
}
else
{
cout << "Please input an integer, or E to exit." << endl;
}
}
I cannot figure out why my getchar() function is not working the way I want it to work. I am getting 10 not 2. Please take a look.
Main():
#include <cstdlib>
#include <iostream>
#include <fstream>
using namespace std;
int main() {
int var, newvar;
cout << "enter a number:" << endl;
cin >> var;
newvar = getchar();
cout << newvar;
return 0;
}
Here is my output:
enter a number:
220
10
Ultimately though I need to be able to distinguish between a '+' '-' or letter or number.
This is maybe not the cleanest way to do it but you can get every char one by one :
#include <iostream>
using namespace std;
int main()
{
int var;
cout << "enter a number:" << endl;
cin >> var;
std::string str = to_string(var);
for(int i=0; i < str.length();++i)
cout << str.c_str()[i] << endl;
return 0;
}
If you enter for example: "250e5" it will get only 250 and skip the last 5.
Edit:
This is just a simple parser and does not do any logic.
If you want to make a calculator I would recommend you to look at what Stroustrup did in his book the c++ programming language.
int main()
{
string str;
cout << "enter a number:" << endl;
cin >> str;
for(int i=0; i < str.length();++i) {
char c = str.c_str()[i];
if(c >= '0' && c <= '9') {
int number = c - '0';
cout << number << endl;
}
else if(c == '+') {
// do what you want with +
cout << "got a +" << endl;
} else if(c == '-')
{
// do what you want with -
cout << "got a -" << endl;
}
}
return 0;
}
I'm needing help in adding commas to the number the user enters, some guidance or help would be appreciated. So far I have it where i store the first three digits and the last six digits and then simply format it.
#include<iostream>
using namespace std;
int main ( int argc, char * argv[] )
{
unsigned long long userInput;
int fthreeDigit;
cout << "Enter a long long number: " << endl;
cin >> userInput;
fthreeDigit = ( userInput / 1000 );
userInput %= 1000;
cout << "Your Number: " << fthreeDigit << "," << userInput << endl;
system("pause");
return 0;
}
Is this what you need? The locale will do this for you correctly.
#include <iostream>
using namespace std;
int main ( int argc, char * argv[] )
{
unsigned long long userInput;
int fthreeDigit;
cout << "Enter a long long number: " << endl;
cin >> userInput;
std::cout.imbue(std::locale(""));
std::cout << userInput << std::endl;
return 0;
}
EDIT:
I have two solutions. first without playing with numbers (recommended) and second (division).
first solution is:
#include <cstdlib>
#include <iostream>
#include <locale>
#include <string>
using namespace std;
struct my_facet : public std::numpunct<char>{
explicit my_facet(size_t refs = 0) : std::numpunct<char>(refs) {}
virtual char do_thousands_sep() const { return ','; }
virtual std::string do_grouping() const { return "\003"; }
};
/*
*
*/
int main(int argc, char** argv) {
cout<<"before. number 5000000: "<<5000000<<endl;
std::locale global;
std::locale withgroupings(global, new my_facet);
std::locale was = std::cout.imbue(withgroupings);
cout<<"after. number 5000000: "<<5000000<<endl;
std::cout.imbue(was);
cout<<"and again as before. number 5000000: "<<5000000<<endl;
return 0;
}
before. number 5000000: 5000000
after. number 5000000: 5,000,000
and again as before. number 5000000: 5000000
RUN SUCCESSFUL (total time: 54ms)
and second (not recommended) is :
double f = 23.43;
std::string f_str = std::to_string(f);
or this
int a = 1;
stringstream ss;
ss << a;
string str = ss.str();
Then you can use string::substr() string::find() string::find_first_of() and similar methods to modify and format your string.
a similar topic
If you really want (have to) divide: (I think my version is cleaner & more efficient than the others)
unsigned long long userInput;
std::stringstream ss,s0;
std::string nr;
std::cout << "Enter a long long number: " << std::endl;
std::cin >> userInput;
int input=userInput;
int digits;
while(input>999){
input=input/1000;
digits=userInput-input*1000;
int mdigits=digits;
while(mdigits<100){s0<<"0";mdigits*=10;}
std::string s=ss.str();
ss.str("");
ss<<","<<s0.str()<<digits<<s;
userInput=input;
s0.str("");
}
std::string sf=ss.str();
ss.str("");
ss<<input<<sf;
std::cout << "Your Number: " << userInput << ";" << digits <<";"<<ss.str()<<std::endl;
Enter a long long number: 12345678 Your Number: 12;345;12,345,678
Here is the brute force but may be easiest to understand way to get every thousand digits with the help of a vector.
#include<iostream>
#include <algorithm>
#include <vector>
using namespace std;
int main ( int argc, char * argv[] )
{
long long userInput;
int fthreeDigit;
cout << "Enter a long long number: " << endl;
cin >> userInput;
vector <int> res; //use vector to store every 3 digits
while (userInput !=0)
{
fthreeDigit = userInput %1000;
res.push_back(fthreeDigit);
userInput = userInput / 1000 ;
}
std::reverse(res.begin(), res.end());
for (size_t i = 0; i < res.size()-1; ++i)
{
if (res[i] ==0)
{
cout << "000"<<",";
}
else
{
cout << res[i] << ",";
}
}
if (res[res.size()-1] == 0)
{
cout << "000";
}
else{
cout << res[res.size()-1];
}
cout <<endl;
cin.get();
return 0;
}
I tested this code with the following case:
Input: 123456 Output: 123,456
Input: 12 Output: 12
Input: 12345 Output: 12,345
Input: 1234567 Output: 1,234,567
Input: 123456789 Output: 123,456,789
Input: 12345678 Output: 12,345,678
I guess this is what you want according to your response to comments.
You could do this:
#include <iostream>
#include <string>
using namespace std;
string commify(unsigned long long n)
{
string s;
int cnt = 0;
do
{
s.insert(0, 1, char('0' + n % 10));
n /= 10;
if (++cnt == 3 && n)
{
s.insert(0, 1, ',');
cnt = 0;
}
} while (n);
return s;
}
int main()
{
cout << commify(0) << endl;
cout << commify(1) << endl;
cout << commify(999) << endl;
cout << commify(1000) << endl;
cout << commify(1000000) << endl;
cout << commify(1234567890ULL) << endl;
return 0;
}
Output (ideone):
0
1
999
1,000
1,000,000
1,234,567,890
// Accepts a long number, returns a comma formatted string
CString num_with_commas(long lnumber)
{
CString num;
num.Format(%d",lnumber);
if(num.GetLength() > 3) {num.Insert(num.GetLength()-3, ',');}
if(num.GetLength() > 7) { num.Insert(num.GetLength()-7, ','); }
if (num.GetLength() > 12) { num.Insert(num.GetLength()-12, ','); }
return(num);
}