I want to make a program that input multiple integers in a single line without knowing the number of that integer itself. So I think I'll use sstream, but get stucked because this is my first time using sstream.
string zzz;
int i=0,current;
int main()
{
stringstream ss;
getline(cin,zzz);
while(stringstream(zzz)>> i)
{
cout << i<< endl;
}
}
example if I'm inputing 123 456 it will result in infinite loop of printing 123. How can that be happen?
The expression stringstream(zzz) >> i is evaluated in each iteration. That means each time the loop condition is checked, you are creating a new stream with the same content as before.
If you move the stringstream before the loop, it should work as expected:
getline(cin,zzz);
stringstream ss(zzz);
while(ss>> i)
{
cout << i<< endl;
}
While #nosid's answer does work, the use of std::getline() in this context would be ill-advised as this task can more simply be done with the direct use of the extractor:
while (std::cin >> i)
{
std::cout << i << std::endl;
}
Related
I am trying to write a code in C++ reading a text file contains a series of numerics. For example, I have this .txt file which contains the following series of numbers mixed with a character:
1 2 3 a 5
I am trying to make the code capable of recognizing numerics and characters, such as the 4th entry above (which is a character), and then report error.
What I am doing is like
double value;
while(in) {
in >> value;
if(!isdigit(value)) {
cout << "Has non-numeric entry!" << endl;
break;
}
else
// some codes for storing the entry
}
However, the isdigit function doesn't work for text file. It seems when I am doing in >> value, the code will implicitly type-cast a into double.
Can anyone give me some suggestion?
Thanks a lot!
Your while loop doesn't do what you think it does.
It only iterates one statement:
in >> value;
The rest of the statements are actually outside the loop.
Using curly braces for the while body is always recommended
I created a small mini script where I would be reading in a file through a standard fstream library object as I was a little unsure on what your "in" represented.
Essentially, try to read in every element as a character and check the digit function. If you're reading in elements that are not of just length 1, a few modifications would have to be made. Let me know if that's the case and I'll try to help!
int main() {
std::fstream fin("detect_char.txt");
char x;
while (fin >> x) {
if (!isdigit(x)) {
std::cout << "found non-int value = " << x << '\n';
}
}
std::cout << '\n';
return 0;
}
Try reading the tokens into string and explicitly parsing it
ifstream infile("data.txt");
string token;
while (infile >> token) {
try {
double num = stod(token);
cout << num << endl;
}
catch (invalid_argument e) {
cerr << "Has non-numeric entry!" << endl;
}
}
Since it looks like the Asker's end goal is to have a double value for their own nefarious purposes and not simply detect the presence of garbage among the numbers, what the heck. Let's read a double.
double value;
while (in) // loop until failed even after the error handling case
{
if (in >> value) // read a double.
{
std::cout << value; // printing for now. Store as you see fit
}
else // failed to read a double
{
in.clear(); // clear error
std::string junk;
in >> junk; // easiest way I know of to read up to any whitepsace.
// It's kinda gross if the discard is long and the string resizes
}
}
Caveat:
What this can't handle is stuff like 3.14A. This will be read as 3.14 and stop, returning the 3.14 and leave the A for the next read where it will fail to parse and then be consumed and discarded by in >> junk; Catching that efficiently is a bit trickier and covered by William Lee's answer. If the exception handling of stod is deemed to expensive, use strtod and test that the end parameter reached the end of the string and no range errors were generated. See the example in the linked strtod documentation
I am taking a course on C++ and I have been asked to compute the area of a circle by using getline. I was advised to refrain from using cin unless it is really necessary.
Below is my code..
#include <iostream>
#include <string>
int main()
{
std::string question ("Enter the radius: ");
std::string s_radius;
std::cout << question;
getline(std::cin, s_radius);
const double PI = 3.14159;
double n_radius = std::stod(s_radius);
double area = PI * n_radius * n_radius;
std::cout << "The area of the circle = " << area << std::endl;
return 0;
}
Is it really necessary to go through the process of accepting a string as input and convert it to a numeral to perform the calculation?
Is it really necessary to go through the process of accepting a string as input and convert it to a numeral to perform the calculation?
It isn't strictly necessary but it does enforce good habits. Let's assume you are going to ask for 2 numbers from the users and use something like
std::cout << "Enter first number: ";
std::cin >> first_number;
std::cout << "Enter second number: ";
std::cin >> second_number;
For the first number the user enters 123bob. operator>> is going to start reading it in and once it hits b it is going to stop and store 123 into first_number. Now the stream has bob still in it. When it asks for the second number since bob is still in the stream it will fail and second_number will get set to 0. Then the program will continue on and you will get garbage output because you accepted garbage input.
Now, if you read in as a std:string and then convert to what you want it makes it easier to catch these types of errors. getline(std::cin, some_string); would get all of 123bob out of the input buffer so you don't have to worry about having to clean it up. Then using stoi (or any of the stox functions) it will read the valid value of of it. The functions also have a second parameter which is a pointer to a size_t and if you pass it one it will store the position of the first unconverted character in the string and you can use that to tell if the entire input string is valid or not. So, if you had
std::string input;
std::cout << "Enter some number: ";
std::getline(std::cin, input);
std::size_t pos;
double number = std::stod(input, &pos);
if (pos == input.size())
std::cout << "valid input\n";
else
std::cout << "invalid input\n";
Then 1.23bob would cause invalid input to print where 1.23 cause valid input to print. You could even use this in a loop to make sure you get only valid input like
double number;
do
{
std::string input;
std::cout << "Enter some number: ";
std::getline(std::cin, input);
std::size_t pos;
number = std::stod(input, &pos);
if (pos != input.size())
std::cout << "invalid input\n";
else
break;
} while(true)
TL;DR: If you rely on the user to only input valid input eventually you are going to get burned. Reading in as a string and converting offers a consistent way to ensure that you are only getting good input from your user.
I was advised to refrain from using cin unless it is really necessary.
First of all, there is a misunderstanding, because you are still using std::cin. I will interpret your question as using std::cin::operator >> vs getline(std::cin,...). Next I would rather suggest you the opposite: Use operator>> directly if you can and fall back to getline if you have to.
A very contrived example:
Imagine you have integers stored in a file in this (unnecessarily complex) format,
3123212
^-------- number of digits of the first number
^^^----- first number
^---- number of digits of the second number
^^-- second number
You cannot use a bare std::cin >> to read those numbers. Instead you need to read the whole line and then parse the numbers. However, even for this I would highly recommend to provide an overload for operator>>, something along the line of
struct NumbersReadFromStrangeFormat {
std::vector<int> numbers;
};
std::isstream& operator>>(std::istream& in,NumbersReadFromStrangeFormat& num) {
// use getline and parse the numbers from the string
return in;
}
Such that you can again write:
NumbersReadFromStrangeFormat numbers;
std::cin >> numbers;
Conclusion:
Overloading the operator>> is the idomatic way to read something from an input stream. "Using std::cin" or "using getline" are not really alternatives. "Refraining from using std::cin" must be a misunderstanding. If there is no overload of operator>> that already does what you need, you can write your own and nicely hide any getline or other parsing details.
Why does trying to input two strings using cin.get() fails? I can successfully read the first string but the input fails for second string and subsequent operations.. See the code:
#include <iostream>
#include <stdlib.h>
int main(){
long int n,k;
char a[11],b[11];
cin.get(a,11);
n = atoi(a);
cin.get(b,11);
cout<<b;
k = atoi(b);
cout << "\ncin.rdstate(): " << cin.rdstate()
<< "\n cin.eof(): " << cin.eof()
<< "\n cin.fail(): " << cin.fail()
<< "\n cin.bad(): " << cin.bad()
<< "\n cin.good(): " << cin.good() << endl << endl;
}
I am trying to input two strings and store them into long int variables as shown in program, but the cin.get(b,11) fails and stack overflow occurs for
k= atoi(b) .Also, you may observe nothing is output for cout<<b .. And, at last cin.fail() is set to 1 , which means I am doing some kind of logical error.. Please help me in rectifying this!
Please suggest some method which is fast and meant for c++ only ..
(If you feel this question is too bad please mention in comments before down voting this, I am already struggling at 21 rep!)
\n will remain in the buffer after the first cin. You can solve this problem by adding an empty cin.get()
cin.get(a,11);
n = atoi(a);
cin.get();
cin.get(b,11);
cout<<b;
k = atoi(b);
cin.get() Does not extract the delimiter from the input (documentation).
If you are C++ with streams it makes sense to use the built in functionality. In particular, C++ offers formatted I/O. To read two numbers you should use:
long int a, b;
cin >> a;
cin >> b;
This will read two numbers from the standard input.
If speed is a concern, try to turn off C I/O synchronisation: std::ios::sync_with_stdio(false); There is an interesting benchmark here that shows that if you turn of synchronisation with C I/O, streams are actually pretty fast.
I am fairly new to programming and have to create a program which reads the prompt: "I have 8 dollars to spend." It then needs to print out with each word on a separate line, and then if any of the strings is numeric, it needs to be divided by 2. Therefore it should end up printing out as:
I
have
4
dollars
to
spend.
I have managed to do everything, except finding the numeric value and dividing it by 2. So far I have this:
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
int main()
{
string prompt;
string word;
cout << "Prompt: ";
getline(cin, prompt);
stringstream ss;
ss.str(prompt);
while (ss >> word)
{
cout << word << endl;
}
return 0;
}
After looking through various other posts, I cannot manage to get this to work. I'm assuming its an if/else statement within the while loop along the lines of, if numeric, set int num to num / 2 then cout << num << endl;, else cout << word << endl;, but I can't figure it out.
Thanks in advance.
You can use the stringstream class, which handles conversions between strings and other data types, to attempt to convert a given string to a number. If the attempt is successful, you know
The stringstream object allows you to treat a string as though it is a stream similar to cin or cout.
Incorporate this into your while loop, like so:
while (ss >> word)
{
int value = 0;
stringstream convert(word); //create a _stringstream_ from a string
//if *word* (and therefore *convert*) contains a numeric value,
//it can be read into an _int_
if(convert >> value) { //this will be false if the data in *convert* is not numeric
cout << value / 2 << endl;
}
else
cout << word << endl;
}
The strtol (C++11 version that works on std::string directly: std::stol) function is really good for testing whether a string holds a number, and if so, what the numeric value is.
Or you could continue using iostreams like you have been... try extracting a number (int or double variable), and if that fails, clear the error bit and read a string.
I dont have 50 rep so I cant comment, thats why I'm writing it as answer.
I think you can check it character by character, using Ascii value of each char, & if there are ascii values representing numbers between two spaces(two \n in this case as you've already seperated each word), then you have to divide the number by 2.
----------------- EDIT -----------------------
Based on juanchopanza's comment : I edit the title
Based on jrok's comment : I'm using ofstream to write, and ifstream to read.
I'm writing 2 programs, first program do the following tasks :
Has a vector of integers
convert it into array of string
write it in a file
The code of the first program :
vector<int> v = {10, 200, 3000, 40000};
int i;
stringstream sw;
string stringword;
cout << "Original vector = ";
for (i=0;i<v.size();i++)
{
cout << v.at(i) << " " ;
}
cout << endl;
for (i=0;i<v.size();i++)
{
sw << v[i];
}
stringword = sw.str();
cout << "Vector in array of string : "<< stringword << endl;
ofstream myfile;
myfile.open ("writtentext");
myfile << stringword;
myfile.close();
The output of the first program :
Original vector : 10 200 3000 40000
Vector in string : 10200300040000
Writing to File .....
second program will do the following tasks :
read the file
convert the array of string back into original vector
----------------- EDIT -----------------------
Now the writing and reading is fine, thanks to Shark and Jrok,I am using a comma as a separator. The output of first program :
Vector in string : 10,200,3000,40000,
Then I wrote the rest of 2nd program :
string stringword;
ifstream myfile;
myfile.open ("writtentext");
getline (myfile,stringword);
cout << "Read From File = " << stringword << endl;
cout << "Convert back to vector = " ;
for (int i=0;i<stringword.length();i++)
{
if (stringword.find(','))
{
int value;
istringstream (stringword) >> value;
v.push_back(value);
stringword.erase(0, stringword.find(','));
}
}
for (int j=0;j<v.size();i++)
{
cout << v.at(i) << " " ;
}
But it can only convert and push back the first element, the rest is erased. Here is the output :
Read From File = 10,200,3000,40000,
Convert back to vector = 10
What did I do wrong? Thanks
The easiest thing would be to insert a space character as a separator when you're writing, as that's the default separator for operator>>
sw << v[i] << ' ';
Now you can read back into an int variable directly, formatted stream input will do the conversion for you automatically. Use vector's push_back method to add values to it as you go.
Yes, this question is over a year old, and probably completely irrelevant to the original asker, but Google led me here so it might lead others here too.
When posting, please post a complete minimal working example, having to add #include and main and stuff is time better spent helping. It's also important because of your very problem.
Why your second code isn't working is all in this block
for (int i=0;i<stringword.length();i++)
{
if (stringword.find(','))
{
int value;
istringstream (stringword) >> value;
v.push_back(value);
stringword.erase(0, stringword.find(','));
}
}
istringstream (stringword) >> value interprets the data up to the comma as an integer, the first value, which is then stored.
stringword.find(',') gets you the 0-indexed position of the comma. A return value of 0 means that the character is the first character in the string, it does not tell you whether there is a comma in the string. In that case, the return value would be string::npos.
stringword.erase deletes that many characters from the start of the string. In this case, it deletes 10, making stringword ,200,3000,40000. This means that in the next iteration stringword.find(',') returns 0.
if (stringword.find(',')) does not behave as wished. if(0) casts the integer to a bool, where 0 is false and everything else is true. Therefore, it never enters the if-block again, as the next iterations will keep checking against this unchanged string.
And besides all that there's this:
for (int j=0;j<v.size();i++)
{
cout << v.at(i) << " " ;
}
it uses i. That was declared in a for loop, in a different scope.
The code you gave simply doesn't compile, even with the added main and includes. Heck, v isn't even defined in the second program.
It is however not enough, as the for condition stringword.length() is recalculated every loop. In this specific instance it works, because your integers get an extra digit each time, but let's say your input file is 1,2,3,4,:
The loop executes normally three times
The fourth time, stringword is 4, stringword.length() returns 2, but i is already valued 3, so i<stringword.length() is invalid, and the loop exits.
If you want to use the string's length as a condition, but edit the string during processing, store the value before editing. Even if you don't edit the string, this means less calls to length().
If you save length beforehand, in this new scenario that would be 8. However, after 4 loops string is already empty, and it executes the for loop some more times with no effect.
Instead, as we are editing the string to become empty, check for that.
All this together makes for radically different code altogether to make this work:
while (!stringword.empty())
{
int value;
istringstream (stringword) >> value;
v.push_back(value);
stringword.erase(0, stringword.find(',')+1);
}
for (int i = 0; i < v.size(); i++)
{
cout << v.at(i) << " " ;
}
A different way to solve this would have been to not try to find from the start, but from index i onwards, leaving a string of commas. But why stick to messy stuff if you can just do this.
And that's about it.