Is cin.get() reading multiple digit characters at once? - c++

We were asked to write a simple C++ program by our teacher to add two numbers in the following format:
input: 12 14
output: m+n = 26
The program must also work for other inputs in the form:
input: Hello please add 12 and 14 !
output: m+n = 26
The solution that was given was:
#include <iostream>
using namespace std;
int main(){
int m,n;
char ch;
while(cin.get(ch)){
if(isdigit(ch))
{
cin.putback(ch);
cin>>m;
break;
}
}
//cin.putback() restores the last character
//read by cin.get() back to the input stream
while(cin.get(ch)){
if(isdigit(ch))
{
cin.putback(ch);
cin>>n;
break;
}
}
cin.ignore(80,'\n');
cout<<"m + n = "<<m+n<<endl;
return 0;}
But now I need to know why this program also works for numbers that are not single digits. Shouldn't cin.get(char) just read one digit and cin.putback() return the same? Please help me I am a beginner.

Yes, cin.get() will read only one character at a time.
The important part, where the number is actually read, is 4 lines below: cin>>m;. This will consume as many digits as possible and store the resulting integer in m.
Some more details:
// example with input 524 42
while(cin.get(ch)) // extract one character at a time
{ // ch='5', remaining input="24 42"
if(isdigit(ch)) // if it's a digit (0-9), then:
{
cin.putback(ch); // Put it back into the stream, "unread" it
// remaining input="524 42"
cin >> m; // extract an integer from stream
// m=524, remaining input="42"
break;
}
}
The reason for the loops seems to be to skip over any non-numeric input before a number appears. Note that there is a little bug here, since it will also ignore leading signs. (e.g. input -4 4 will output m + n = 8)

Related

how to ignore n integers from input

I am trying to read the last integer from an input such as-
100 121 13 ... 7 11 81
I'm only interested in the last integer and hence want to ignore all
previous integers.
I thought of using cin.ignore but that won't work here due to
unknown integers (100 is of 3 digits, while 13 is of 2 digits & so on)
I can input integer by integer using a loop and do nothing with them. Is there a better way?
It all depends on the use case that you have.
Reading a none specified number of integers from std::cin is not as easy at it may seem. Because, in contrast to reading from a file, you will not have an EOF condition. If you would read from a file stream, then it would be very simple.
int value{};
while (fileStream >> value)
;
If you are using std::cin you could try pressing CTRL-D or CTRL-Z or whatever works on your terminal to produce an EOF (End Of File) condition. But usually the approach is to use std::getline to read a complete line until the user presses enter, then put this line into a std::istringstream and extract from there.
Insofar, one answer given below is not that good.
So, next solution:
std::string line{};
std::getline(std::cin, line);
std::istringstream iss{line};
int value{};
while (iss >> value)
;
You were asking
Is there a better way?
That also depends a little. If you are just reading some integers, then please go with above approach. If you would have many many values, then you would maybe waste time by unnecessarily converting many substrings to integers and loose time.
Then, it would be better, to first read the complete string, then use rfind to find the last space in the string and use std::stoi to convert the last substring to an integer.
Caveat: In this case you must be sure (or check with more lines of code) that there are no white space at the end and the last substring is really a number. That is a lot of string/character fiddling, which can most probably avoided.
So, I would recommend the getline-stringstream approach.
You can try this simple solution for dynamically ignoring rest of the values except the last given in this problem as shown:
int count = 0;
int values, lastValue; // lastValue used for future use
std::cout << "Enter your input: ";
while (std::cin >> values) {
lastValue = values; // must be used, otherwise values = 0 when loop ends
count++;
}
std::cout << lastValue; // prints
Note: A character must be required to stop the while(), hence it's better put a . at last.
Output example
Enter your input: 3 2 4 5 6 7.
7
Try this:
for( int i=0; i<nums_to_ignore; i++) {
int ignored;
std::cin >> ignored;
}

standard input stream cin failing to prompt for input

1 #include<iostream>
2 using namespace std;
3
4 int main()
5 {
6 const double yen_to_euro=0.007215;
7 const double euro_to_dollar=1.12;
8 char currency;
9 double x;
10
11 while(currency!='q')
12 {
13 cout << "enter currency and unit(y , e, or d)";
14 cin >> x >>currency;
15
16 switch(currency){
17
18 case 'y':
19 cout <<"euro:"<< x*yen_to_euro<<" dollar:"<<x*yen_to_euro*euro_to_dollar<<'\n';
20 break;
21 case 'e':
22 cout <<"yen:"<< (x*(1.0/yen_to_euro))<<" dollar:"<<(x*euro_to_dollar)<<'\n';
23 break;
24 case 'd':
25 cout <<" yen:"<< x*(1.0/yen_to_euro)*(1.0/euro_to_dollar)<<" euro:"<<x*(1.0/euro_to_dollar)<<'\n';
26 break;
27 case 'q':
28 currency='q';
29 break;
30 default:
31 cout << "invalid";
32 break;
33
34 }
35
36 }
37
38
39 }
~
The intended function of the code above is to convert a selected currency( y for Japanese Yen, e for Euro , and d for US dollar) into the other currencies.
For instance, if I want to convert to 12 Japenese Yen, I would enter:
12y
which then the program would output
euro:0.08658 dollar:0.0969696
However, if I were to input 12e, I would receive an infinite loop. Double checking the code, there doesn't seem to be any logical errors.
Nevertheless, I sense the source of the trouble has to do with the cin at line 14 because if I take the amount x and the currency type separately like this:
cin>> x;
cin>> currency;
The code works fine, but I need to input the amount , followed by pressing enter, and then pressing the character that represents the currency-type. Is there a way to just enter all in one line, no spaces?
Furthermore, why is it behaving this way? This unusual behavior of going through infinite loop only happens when I enter e for Euro and q for quit.
12e is interpreted as the begin of a floating point number like 12e03. But the end is missing, so your stream is put on error (failbit). This state causes all subsequent input to fail until the failstate is cleared.
You could have an inner loop that checks such formatting errors:
while (...) {
...
while ( (cin >> x >> currency).fail() && !cin.eof()) { // loops as long as there's input and it's invalid
cout << "invalid format";
cin.clear();
}
if (cin.eof()) // if there's no longer input stop looping
break;
...
}
live demo
Note that you should initialize currency to something different than 'q' if you want to make sure that your loop is executed in every circumstance.
By the way, if you enter 12 e (with a space), 12 will be interpreted as the number and put to x and e to currency, as you expect.
As explained by other answers 12e refers to the scientific notation of floating point numbers and hence currency does not store 'e'.
To get the input in a single line without space, you should use std::getline and then parse the input string.
Since you know that the last character is always indicative of currency, the front characters can be converted to a number using std::stoi.
Replace cin >> x >>currency;
with
try{
std::string myLine;
std::getline (std::cin, myLine); // store line into myLine.
currency = myLine.back(); // get last character from line.
myLine.pop_back(); // remove last character from myLine.
x = std::stod(myLine); // try to conver the rest of the string to a double.
}
catch (...) {
//The user typed in something invalid.
currency= 'i';
}
Also, make sure to #include <string>
Note this solution assumes you are using C11.

End array input with a newline?

Not sure if the title is properly worded, but what I am trying to ask is how would you signify the end of input for an array using newline. Take the following code for example. Not matter how many numbers(more or less) you type during the input for score[6], it must take 6 before you can proceed. Is there a method to change it so that an array can store 6 or 100 variables, but you can decide how many variables actually contain values. The only way I can think of doing this is to somehow incorporate '\n', so that pressing enter once creates a newline and pressing enter again signifies that you don't want to set any more values. Or is something like this not possible?
#include <iostream>
using namespace std;
int main()
{
int i,score[6],max;
cout<<"Enter the scores:"<<endl;
cin>>score[0];
max = score[0];
for(i = 1;i<6;i++)
{
cin>>score[i];
if(score[i]>max)
max = score[i];
}
return 0;
}
To detect "no input was given", you will need to read the input as a input line (string), rather than using cin >> x; - no matter what the type is of x, cin >> x; will skip over "whitespace", such as newlines and spaces.
The trouble with reading the input as lines is that you then have to "parse" the input into numbers. You can use std::stringstream or similar to do this, but it's quite a bit of extra code compared to what you have now.
The typical way to solve this kind of problem, however, is to use a "sentry" value - for example, if your input is always going to be greater or equal to zero, you can use -1 as the sentry. So you enter
1 2 3 4 5 -1
This would reduce the amount of extra code is relatively small - just check if the input is -1, such as
while(cin >> score[i] && score[i] >= 0)
{
...
}
(This will also detect end-of-file, so you could end the input with CTRL-Z or CTRL-D as appropriate for your platform)

Can't read first string

Here is the code
#include<iostream>
#include<cstring>
#define limit 25
using namespace std;
int main()
{
int te; //Number of test cases
cin>>te;
while(te)
{
char m[limit];
char w[limit];
cin.getline(m,limit); // This line is not getting executed for some reason
cin.getline(w,limit);
cout<<"m "<<m<<" "<<endl<<"w "<<w<<endl;
te--;
}
}
For god knows what reason, the machine refuses to read m for the first test case. It reads and prints values for both m and w in the other cases, but for the first case, it refuses to read m.
Sample:
INPUT
1
hello
m is
w is hello
2
hello
m
w hello
stack
overflow
m stack
w overflow
cin>>te;
This will extract the 1 from the input stream and then stop at but not extract the \n. You'll need to ignore() that character, otherwise the next line extraction you do will just read an empty line.
cin.ignore();
Or to ignore all characters up to and including the next \n character (in case somebody inputs 1foo or something), you can do:
cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');

Getting input from external files?

I need to get very basic input from an external file in C++. I tried searching the internet a few times but nothing really applied to what I need. This would be a .txt file that the input it coming from, and it would be filled with lines like this:
131
241
371
481
I have code already to manually get this input, and it looks like this:
using namespace std;
//Gets the initial values from the user.
int control=0;
while (rowb!=0){
cout << "Row: ";
cin >> rowb;
cout << "Column: ";
cin >> columnb;
cout << "Number: ";
cin >> numb;
row[control]=rowb-1;
column[control]=columnb-1;
num[control]=numb;
control++;
}
This is part of a program that solves sudoko boards. The inputed numbers are the initial values that a sudoko board holds, and the user is inputing the row, column, and number that comes from a board.
What I need is to be able to create a .txt file with these numbers stored in rows so that I do not have to enter so many numbers. I have very little idea how to go about doing this. Mainly I'll only be using the txt file for testing my program as I move along with adding more code to it. It takes 150+ entered numbers within my program just to get a single board, and it takes a lot of time. Any accidentally wrong entered value is also a huge problem as I have to start again. So how would I get C++ to read a text file and use those numbers as input?
Aside from the other suggestions, you can simply redirect a file to standard input, like so (where $ is the command prompt):
$ myprogram < mytextfile.txt
That will run myprogram just as normal but take input from mytextfile.txt as if you had typed it in. No need to adjust your own program at all.
(This works on both Unix/Linux systems and on Windows.)
You can open a file for input with std::ifstream from the header <fstream>, then read from it as you would from std::cin.
int main()
{
std::ifstream input("somefile.txt");
int a;
input >> a; // reads a number from somefile.txt
}
Obviously, you can use >> in a loop to read multiple numbers.
Create an std::ifstream object, and read from it just like you would from std::cin. At least if I understand what you're trying to do, the 131 as the first input is really intended to be three separate numbers (1, 3, and 1). If so, it's probably easiest to change your input file a bit to put a space between each:
1 3 1
2 4 1
3 7 1
4 8 1
Personally, I would start with a different format of the file: enter a value for each cell. That is, each row in the input file would represent a row in the sudoko board. Empty fields would use a space character. The immediate advantage is that the input actually pretty much looks like the sudoko board. Also, you would enter at most 90 characters: 9 characters for the board and a newline for each line:
#include <iostream>
#include <fstream>
#include <algorithm>
#include <iterator>
int main(int ac, char* av[])
{
std::ifstream in(ac == 1? "sudoko.init": av[1]);
char board[9][9];
for (int i(0); i != 9; ++i)
{
in.read(board[i], 9).ignore();
}
if (!in)
{
std::cout << "failed to read the initial board\n";
}
else
{
typedef std::ostream_iterator<char> iterator;
std::fill_n(iterator(std::cout << "board:\n\n+", "+"), 9, '=');
for (int i(0); i != 9; ++i)
{
std::copy(board[i] + 0, board[i] + 9, iterator(std::cout << "\n|", "|"));
std::fill_n(iterator(std::cout << "\n+", "+"), 9, (i + 1) % 3? '-': '=');
}
std::cout << "\n";
}
}
This would take input like this:
4 5 3 8
71 3
16 7
6 4 7
6 8
1 9 5
6 42
5 94
4 7 9 3
Note that each of these lines uses 9 characters. You might want to use something more visible like ..