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');
Related
I was given a question where the input will be like:
10 8
4 9
6 12
5 4
3
1
Here I don't know the number of lines that contains 2 integers. Those sets of 2 integers will be taken into an array. But when the program encounters "3", it will start taking input in another array.
I have tried this with
while(cin>>a>>b){ //some porcess with a and b }
but it doesn't work because it recognizes 3 and 1 as another set of two integers. Please help me to solve this problem.
cin >> a >> b skips not only spaces, but any delimeter characters too ('\n', '\t', ' ').
Here you actually may want to read input line-by-line and then check if there are two integers or one. Consider use of std::getline for retrieving each line of text. Then you can use read string as std::istream (like in example in the link above) and read from it with counting, how many numbers you read totally.
So think about your problem. Essentially it is, read one line at a time, and if it contains two numbers do one thing, but if it contains one number do something else.
But the code you have written reads numbers not lines. That is where the problem is.
Instead write your code to read only line at a time, analyse that line to see if it contains one or two numbers (or something else) and then proceed from there.
What you need is the ability to read a line of text into a string, and then read from that string into your numbers. To do that you use an istringstream. Something like this
#include <iostream>
#include <sstream>
#include <string>
int a, b;
string s;
getline(cin, s); // read one line from standard input
istringstream line(s); // put that string to a stream we can read from
if (line >> a) // try and read the first number from the stream
{
// got the first number
if (line >> b) // try and read the second number from the stream
{
// got the second number
...
}
else
{
// only one number
...
}
}
else
{
// didn't get any numbers, some sort of error
...
}
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)
I have been doing exercises on some online judges, and I encounter this question with this default answer.
Question description:
Finally, Hansbug has finally reached the moment to do the last math problem, and there are a bunch of messy addition and subtraction equations in front of him. Obviously success is at hand. But his brain cells have been exhausted, so this important task is left to you.
Input format:
One line, containing a string of addition and subtraction polynomials (the range of each item is 0-32767).
Output format:
An integer, which is the result of the calculation (guarantee that the result of the calculation will not exceed the range of the long integer).
Input and output sample:
Enter #1:
1+2-3
Output #1:
0
And the default answer is :
#include<bits/stdc++.h>
using namespace std;
int ans;
int c;
int main() {
while (cin >> c)
ans += c;
cout << ans;
return 0;
}
How is this even possible!?
All right, let's try it, let's put there some expression with addition and subtraction of ints only, after that press Return, then ctrl+D (end of input):
$ ./a.out
111-222+1
-110$
The loop while (cin >> c) will parse integers including the sign one by one using iostream capabilities until an end of input (you also have to terminate the last number by pressing Return, effectively putting the newline there and triggering the last cin >> c).
Consider this small piece of code:
#include <iostream>
#include<stdio.h>
using namespace std;
int main() {
int a;
while(true){
cin>>a;
cout<<a;
}
return 0;
}
Input
1 2 3 5 7 23
Output
125723
How I thought it will run is:
First iteration
1. Reads the first input ie '1' and stops reading further, right after reading the whitespace.
2.Prints the value 1.
Second iteration
1. Again asks for new input
2. Print that in the second line
But that doesn't happen instead it reads the elements we gave after space
First iteration:
Peek at next character in the stream. It's a digit ('1'), so read it.
Peek at next character in the stream. It's not a digit (' '), so don't read it; store 1 in a and return from >>.
(Output 1.)
Second iteration:
Peek at next character in the stream. It's whitespace (' '), so read and ignore it.
Peek at next character in the stream. It's a digit ('2'), so read it.
Peek at next character in the stream. It's not a digit (' '), so don't read it; store 2 in a and return from >>.
(Output 2.)
And so on ...
The point is that >> does not care about lines. cin is one long input stream of characters (some of which may be '\n'). The only thing you can do is read more characters (and then maybe decide that you don't want to do anything with them).
cin is not necessarily connected to a keyboard. The program that started you gets to decide where cin reads from. It can be a file, a network socket, or interactive user input. In the latter case, reading from cin may block until the user types more input, but it will never cause input to just be dropped.
If you want a sane user interface, always read whole lines and process them afterwards:
std::string line;
while (std::getline(std::cin, line)) {
// do stuff with line
}
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)