How can I have an output of
Sample Input No.1:
9
Sample Output No.1:
1.2.3.4.5.6.7.8.9
If you input numbers less than or equal to 9, the output should be (1.2.3.4.5.6.7.8.9)
And if you input numbers greater than 9, for example:
Sample Input No.2:
20
Sample Output No.2:
01.02.03.04.05.06.07.08.09.10
11.12.13.14.15.16.17.18.19.20
My code below is for Sample Input & Output No.2. I tried adding another for loop for SAMPLE NO.1 but it still reads Sample No.2 code. What should I do?
#include <iostream>
#include <iomanip>
using namespace std;
int main ()
{
int a, num;
cin >> num;
if (num > 100 || num <= 1){
cout << "OUT OF RANGE";
}
else {
for (int a = 1; a < num; a++){
cout << setfill('0') << setw(2) << a << ".";
}
cout << num;
}
}
kind of new to programming, don't know much🥲
As a possible solution, you could read the input as a string, then convert it to an integer.
Use the string length as the field width for the setw manipulator.
This should be able to handle values of (theoretically) arbitrary length.
Related
I want to ask 6 digit pins from user, I tried this code, it does work. The problem is that it cannot read pin starting from 0. The objective of the code is to read 6 digit pin no matter what the start is, as long as it is an integer it will read the correct output.
using namespace std;
int main(){
int pin[0];
cin>>pin[0];
if (pin[0] >= 100000 && pin[0] <= 999999) {
cout << pin[0];
}
else {
cout << "Invalid input!";
}
}```
On the one hand you want to read an integer (you are comparing it to 100000 and 999999) on the other hand you want to read individual digits into an array. It cannot be both. And you cannot have an array of size 0.
Just stay with the single integer. If you want to access individual digits you can convert it to a string and access characters (you already made sure that it has 6 digits):
#include <iostream>
#include <string>
int main(){
int pin;
std::cin>>pin;
if (pin >= 100000 && pin <= 999999) {
std::cout << pin << "\n";
std::string pin_string = std::to_string(pin);
for (size_t i=0; i<6; ++i){
std::cout << pin_string[i] << "\n";
}
}
else {
std::cout << "Invalid input!";
}
}
Heres one way to do it.
Read CIN as std::string
Regex recv string for digit only
Confirm recv string len is == 6
Do w/e you want with it from there
Another way could be getChar()
I have used a loop for that:
int number1;
int sum=0;
for(int i =1; i<6; i++){
cout<<"Enter number:\n";
cin>>number1;
sum+=number1;
}
cout<<sum;
cout<<"Total Sum is = "<<sum<<"\n";
return 0;
}
My question is how can I print first statement like this ...
"Enter first number"
Enter Second number" and so on
Whenever you are reading numbers (or any value for that matter), you must check the stream-state (see: std::basic_istream State Functions). You have four stream states you must test following every input:
.bad() or .eof(). If badbit is set an unrecoverable error occurred, and if eofbit is set, there is nothing more to read (you can combine both into a single test that exits if either are set)
.fail() is set when a read error occurs, such as the user entering "FIVE" instead of 5 where integer input is expected. You handle failbit being set by calling .clear() to clear failbit and then call ignore() to empty the characters causing the failure before your next read attempt, and finally
.good() - valid input was received from the user, you can proceed to the next input.
By validating your input here, you can Require the user provide 5 valid integer values for you to sum. Do not use a for loop, instead use a while (or do .. while();) and only increment your counter when good input is received.
Putting that altogether, you can do:
#include <iostream>
#include <limits>
int main (void) {
int number = 0,
sum = 0;
const char *label[] = { "first", "second", "third", "fourth", "fifth" };
while (number < 5) /* loop continually until 5 int entered */
{
int tmp; /* temporary int to fill with user-input */
std::cout << "\nenter " << label[number] << " number: ";
if (! (std::cin >> tmp) ) { /* check stream state */
/* if eof() or bad() exit */
if (std::cin.eof() || std::cin.bad()) {
std::cerr << " (user canceled or unreconverable error)\n";
return 1;
}
else if (std::cin.fail()) { /* if failbit */
std::cerr << " error: invalid input.\n";
std::cin.clear(); /* clear failbit */
/* extract any characters that remain unread */
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
}
}
else { /* on succesful read of int, add to sum, increment number */
sum += tmp;
number++;
}
}
std::cout << "\nsum: " << sum << '\n';
}
Now your code will gracefully handle an invalid input without exiting just because a stray character was entered.
Example Use/Output
When you write an input routine, go try and break it. Enter invalid data and make sure you handle all error cases correctly. If something doesn't work right, go fix it. Repeat until you input routine can handle all corner-cases as well as the cat stepping on the keyboard:
$ ./bin/sumintlabel
enter first number: 3
enter second number: four five six seven!!
error: invalid input.
enter second number: 4
enter third number: 5
enter fourth number: 6
enter fifth number: 7
sum: 25
Form good habits now regarding handling input, it will pay dividends for the rest of your programming career. Let me know if you have questions.
if you need to print the words "first"... untill "fifth" then I'd do it like this:
#include <iostream>
#include <iomanip>
using namespace std;
int main() {
int number1;
int sum=0;
string positions[5] = {"first", "second", "third", "fourth", "fifth"};
for(int i = 0; i<5; i++){
cout<<"Enter the " << positions[i] << " number:" << endl;
cin>>number1;
sum+=number1;
}
cout<<"Total Sum is = "<<sum<<"\n";
return 0;
}
I used an array of strings to display the words and changed the for loop to start at 0 so that we can go through the array positions and add the 5 numbers as well. If you just want to use: 1st, 2nd, 3rd... then you could change the for loop to what it was and do:
cout<<"Enter the " << i << "st" << " number:" << endl;
But for this you would have to use the if statement to print the right endings("st", "rd", "nd"). I think it would take longer for it to run but its miliseconds so we wouldn't even notice hahaha.
Hope it helped :)
You can use switch():
#include <iostream>
using std::cin;
using std::cout;
using std::string;
int main() {
int number1;
int sum = 0;
for(int i = 1; i < 6; i++) {
string num;
switch(i) {
case 1:
num = "first";
break;
case 2:
num = "second";
break;
//and 3 4 5 like this
}
cout << "Enter " << num << " number:\n";
cin >> number1;
sum += number1;
}
cout << "Total Sum is = " << sum << "\n";
return 0;
}
or you can use struct or containers like vector (in fact you have to use containers if you want to get a huge number of data.)
I wrote code to convert decimal fraction number to its binary equivalent. It compiles fine but when executed hangs. The code here prints only first four digits of the binary conversion and if the number if with more than 4 digits, it shows '...' after it. On execution it hangs. Help!
#include <iostream>
using namespace std;
int main()
{
int i, x[10];
float num;
cout << "**PROGRAM TO CONVERT DECIMAL FRACTION INTO ITS EQUIVALENT BINARY**\n";
cout << "Enter a fraction in between 0 to 1 for conversion: ";
cin >> num;
if (num>=0 && num<=1)
{
i=1;
while (num!=1.000)
{
num*=2;
x[i]=num;
num=num-x[i];
i++;
}
if (i>4)
{
cout << "The binary equivalent is 0.";
for (i=1;i<=4;i++)
{
cout << x[i];
}
cout << "...";
}
else
{
cout << "The binary equivalent is 0.";
for (i=1;i<=4;i++)
{
cout << x[i];
}
}
}
else
{
cout << "The number entered is out of range.";
}
return 0;
}
The first obstacle is the infinite while loop:
Assuming input num=0.5
after first iteration, i=1, x[0]=1, num=0.0
after second iteration, i=2, x[1]=0, num=0.0
Continue forever, i=..., x[i-]1=0, num=0.0
With nothing to break the loop.
while (num!=1.000)
{
num*=2;
x[i]=num;
num=num-x[i];
i++;
}
To fix, consider few changes. There might be other issues.
put a limit on the while loop (i<10 should be a good condition, as this is the size of the x array), or i=4, as this is the maximum output.
The break condition for the while loop should probably be 'num != 0', or even better (num > 1e-7, or other small value).
float has 23 bit in mantissa, maybe it is because you are assign x[i] with i greater than 9.
try this:
//stop when you get four bits
while (i< 5)
Original code has several issues:
1 For input num=.5 and similar (really for all values) cycle never ends (dash-o suggested fix ideas)
2 array x[10] is overflowed with undefined behavior (Edney)
3 nitpicking: 1 is not a “fraction” and better check for a range 0 <= num < 1 instead of 0 <= num <= 1(see also OP printing code; 1 could be added); we could use x[4] with 0<=i <=3
4 string could also be used (PaulMcKenzie). Really “>>” uses string processing for parsing and calculating binary equivalent from which by multiplying by 2 (left shit) and truncation fractional part the code calculates target bits. Both approaches give correct identical results; implementing by string we need to add internal to operator “>>” implementation code to parsing valid formats for floats (decimals) such as 3.14e-1, .2718, 1e-1, etc.
This fix follows OP:
#include <iostream>
using namespace std;
int main()
{
int i, x[5];
float num;
cout << "**PROGRAM TO CONVERT DECIMAL FRACTION INTO ITS EQUIVALENT BINARY**\n";
cout << "Enter a fraction in between 0 to 1 for conversion: ";
cin >> num;
if (num>=0 && num<1)
{
i=1;
while (i<=4)
{
num*=2;
x[i]=num;
num=num-x[i];
i++;
}
cout << "The binary equivalent is 0.";
for (i=1;i<=4;i++)
{
cout << x[i];
}
if (num>0)
cout << "...";
}
else
{
cout << "The number entered is out of range.";
}
return 0;
}
This code is without cycles (they are in code implementing “>>”, bitset):
#include <iostream>
#include <bitset>
using namespace std;
int main () {
const int digits = 4;
int fraction;
float num;
cout << "**PROGRAM TO CONVERT DECIMAL FRACTION INTO ITS EQUIVALENT BINARY**\n";
cout << "Enter a fraction in between 0 to 1 for conversion: ";
cin >> num;
if (num >= 0 && num < 1) {
fraction = num = num * pow (2, digits);
cout << "The binary equivalent is 0.";
cout << bitset<digits> (fraction);
if (num - fraction > 0) cout << "...";
}
else cout << "The number entered is out of range.";
}
First off, I'll mark this as a homework problem I've been stuck on for a week as I can't seem to figure out what I'm doing wrong and I'm hoping the wonderful people at SO can come to my rescue yet again (I've searched SO and other C++ sites for the past week but the solutions offered didn't correct the issue - however it is possible I may have been setting the loop incorrectly.
The assignment: given a text file numbers.txt (which contains 9,999 numbers ranging from 1 to 10,000 randomly sorted with one number missing from the consecutive list) the assignment is to use a void function in order to determine what the missing integer is.
What I've tried: My last attempt of this contains the following code:
#include <iostream>
#include <iomanip>
#include <string>
#include <fstream>
using namespace std;
void find_number();
int main()
{
...
find_number();
}
void find_number();
{
int sum = 0;
int sum1 = 0;
int num;
for (int i = 1; i <= 10000; i++)
sum += i;
cout << "The sum of all the numbers between 1 and 10,000 is: " << sum << endl;
ifstream numbers;
numbers.open("numbers.txt");
if (!numbers.good()) {
return;
cout << "Error! Unable to open file!";
}
if (numbers) {
numbers >> num;
sum1 += num;
}
numbers.close();
cout << "The sum of all the numbers contained in the text file \"numbers.txt\" is: " << sum1 << endl;
cout << "By subtracting the sum of the text file from the sum of 1 to 10,000 the consecutive number missing from the text file is: " << sum - sum1 << endl;
}
What am I doing wrong? Thank you for any assistance.
There are at least two mistakes:
The return statement is executed before the diagnostic output
if (!numbers.good()) {
return;
cout << "Error! Unable to open file!";
}
The following lines will execute once instead of reading the whole file:
if (numbers) {
numbers >> num;
sum1 += num;
}
You can improve your code with the following suggestions:
Extract a number and check the stream status at the same time:
while(numbers >> num) sum1 += num;
You don't need to close the file stream it will do it automatically in its destructor.
You can open the file at the file stream initialization time:
ifstream numbers("numbers.txt");
Hint: you are not reading the whole file
First post! This is my second semester with "Advanced C & C++" so any help is GREATLY appreciated. I've already scoured as much of stackoverflow and a few other resources to try and help me understand what I'm doing (or not doing) with this slew of logically inept code.
The goal of this program is to recognize whether or not a 'number' given by the user is a palindrome. Sounds simple enough right?! Ugh...well this is what I have been stuck on:
#include <iostream>
using std::cout;
using std::cin;
#include <string>
using std::string;
#include <cstdlib>
int main()
{
//variable declarations
string buffer;
//user input
cout << "Enter a number to see if it is a palindrome[Q to quit]: ";
getline(cin, buffer);
//looooop
while(buffer != "Q" && buffer !="q")
{
int userNum, length, sum = 0, temp;
userNum = atoi(buffer.c_str());
for(temp = userNum; userNum !=0; userNum=userNum/10)
{
length = userNum % 10;
sum = sum*10+length;
}
if(temp==sum)
cout << temp << " is a palindrome!!\n\n";
else
cout << buffer << " is NOT a palindrome!\n\n";
cout << "Enter a number to see if it is a palindrome[Q to quit]: ";
getline(cin, buffer);
}
}
The problem arises when input of "010", or "400" is given. "400" is essentially "00400" in this case and both should be seen as a palindrome.
A better approach would be to get trailing zeros for the given number as below:
int noOfTrailingZeros = str.length;
while(str[--noOfTrailingZeros]=='0');
noOfTrailingZeros = str.length - noOfTrailingZeros;
Or the integer way as:
int noOfTrailingZeros = str.length;
while(num%10==0)
{
noOfTrailingZeros++;
num/=10;
}
Now, check for the input string whether it has the same number of zeros befire the number or not as:
int counterZeros = 0;
while(str[counterZeros++]=='0');
check these 2 numbers and if trailing zeros are more than the zeros at beginning, add that many at the beginning and pass that string to palindrome function.
First of all, to recognize a palindrome, you don't have to do atoi. Just pass from the start to the middle checking if
buffer[i] == buffer[length - i]
Second, use the atoi to make sure it is a number and you're done.
Other way is to compare the string with itself reversed:
string input;
cout << "Please enter a string: ";
cin >> input;
if (input == string(input.rbegin(), input.rend())) {
cout << input << " is a palindrome";
}