This is my code.
#include <iostream>
using namespace std;
typedef struct
{
int polski;
int wf;
int matma;
}oceny;
int funkcja_liczaca(int suma, int ile_liczb, int ktory_przedmiot, oceny &temporary);
int main()
{
int suma = 0;
int temp[3];
int ile_liczb_zostalo_wprowadzonych = 0;
oceny database;
string teksty[3] = {"polski: ", "wf: ", "matma: "};
for (int i=0; i!=3; i++)
{
cout << teksty[i] << endl;
while(temp[i]!=0)
{
cin >> temp[i];
if(cin.good()) //floating point exception here. the code don't even step into this one.
{
{
suma = temp[i] + suma;
ile_liczb_zostalo_wprowadzonych++;
if(temp[i]==0){ile_liczb_zostalo_wprowadzonych--;}
}
}else cout << "error";
};
funkcja_liczaca(suma, ile_liczb_zostalo_wprowadzonych, i, database);
suma = 0;
ile_liczb_zostalo_wprowadzonych = 0;
}
cout << "output of struct members in main() \n";
cout << database.polski << endl;
cout << database.wf << endl;
cout << database.matma << endl;
return 0;
}
int funkcja_liczaca(int suma, int ile_liczb, int ktory_przedmiot, oceny &temporary)
{
if(ktory_przedmiot==0){temporary.polski=suma/ile_liczb;cout << temporary.polski << endl;}
if(ktory_przedmiot==1){temporary.wf=suma/ile_liczb;cout << temporary.wf << endl;}
if(ktory_przedmiot==2){temporary.matma=suma/ile_liczb;cout << temporary.matma << endl;}
}
It counts arithmetic average of inputed numbers untill user input 0 which ends loop. then the arithmetic average of thoose numbers is counted in the funkcja_liczaca() and it's saved into the members of struct oceny.
everything works fine but i want to implement something like "stream" check while inputing from keyboard to prevent inputing bad variables into integer type variable.
so inputing 'g' into temp[i] is causing floating point exception. the question is why? cin.good() and cin.fail() is not working.
When you want to deal with errors in the input stream, it's better to read the input line by line as a string and then attempt to extract your data from the string. If extraction of the data from the string is successful, proceed to process the data. Otherwise, attempt to read the next line of text. Here's the core logic for that.
while ( true )
{
cout << teksty[i] << endl;
std::string line;
if ( !getline(cin, line) )
{
// Problem reading a line of text.
// Exit.
exit(EXIT_FAILURE);
}
// Construct a istringstream object to extract the data.
std::istringstream istr(line);
if ( istr >> temp[i] )
{
// Extracting the number was successful.
// Add any additional checks as necessary.
// Break out of the while loop.
break.
}
// Bad input. Continue to the next iteration of the loop
// and read the next line of text.
}
Related
I have a text file with the following contents:
2
S 8
push 2 push 3 push 5 push 7 pop print push 6 print
S 4
pop print push 1 print
An assignment gives:
The first line of the input file means the number of test cases. For each test case, the first character means which container adapter (Stack or Queue) that you need to use Linked lists to implement. The correct output should be:
The values in the stack : 2 3 5
The values in the stack : 2 3 5 6
The values in the stack :
The values in the stack : 1
I've written some working functions for stack and queue struct, though I am working on the input of the stack function first.
#include <iostream>
#include <fstream>
#include <string>
#include <vector>
using namespace std;
void push(int);
int pop();
void printStack();
struct Stack {
int s[100];
int top;
Stack() { top = -1; }
};
Stack st;
void push(int n) {
if (st.top == 99) {
cout << "Stack full" << endl;
}
else {
st.s[++st.top] = n;
}
}
int pop() {
if (st.top == -1) {
cout << "Stack is empty" << endl;
}
else {
return st.s[st.top--];
}
}
void printStack() {
cout << "Elements";
for (int i = 0;i <= st.top;i++) {
cout << st.s[i] << ' ';
}
cout << endl;
}
void clearStack() {
st.top = -1;
}
The main part of the code is giving me trouble. I want to read every token of the text file while keeping the line structure; e.g. being able to parse by order of line. However, I do not know the length of each line, only the number of lines. How may I read the file correctly and finish this assignment?
int main() {
std::ifstream file("input1.txt");
std::string item_name;
int numTestCases;
vector<string> file_content{};
while (std::getline(file, item_name))
{
//cout << item_name << "\n";
char str[252];
strcpy(str, item_name.c_str());
char* pch;
//cout << "str0:" << str[0] << "\n";
file_content.push_back(str);
}
cout << "printing file content:" << endl;
for (int i = 0;i < file_content.size(); i++) {
cout << file_content[i] << endl;
}
}
Okay, you basically have two distinct problems.
Read your input file so you know what actions to perform.
Implement both Stack and Queue.
So start by breaking them up. I looked at your stack code. Your problem says to use linked lists, which isn't what you're doing. Maybe that's just because you haven't gotten that far yet.
Code like this would give you the number of test cases.
std::string numCasesStr;
if (!getline(file, numCasesStr)) {
cout << "early end of file detected\n";
return;
}
int numCases = std::stoi(numCasesStr);
At this point, you can now do this:
for (int testCase = 0; testCase < numCases; ++testCase) {
std::string typeAndCountStr;
if (!getline(file, typeAndCountStr)) {
cout << "early end of file detected\n";
return;
}
char typeC = typeAndCountStr.at(0);
if (typeC == 'S') {
...
}
else if (typeC == 'Q') {
...
}
}
The harder part is parsing the next line. You're going to get input in a similar fashion, but then you have to break it into pieces. This is called tokenizing. Basically you split it at each space. What's useful is the find method on a string.
do {
size_t lastPos = 0;
size_t pos = str.find(' ', lastPos);
string thisArg;
if (pos != string::npos) {
thisArg = str.substr(lastPos, pos);
lastPos = pos + 1;
}
else {
thisArg = str.substr(lastPos);
}
// At this point, thisArg contains one argument. You still have more
// to do, but this is one way to split your string into pieces.
} while (lastPos != string::npos);
What I do with that is stuff the individual pieces into a std::vector<std::string> and now it's a lot easier to deal with. You can traverse the vector, looking at each string, and depending upon what it is, you know if you have to grab the next item in the list (like push 8 -- you get push and then you get the 8) or just use the current item.
Overall -- break the problem down into smaller pieces. For main:
Get the number of test cases
Loop from 0..testCaseCnt
Get the type of tests (stack or queue)
Get the next input
Split it into tokens broken at each space
Traverse the tokens and Do The Right Thing (tm).
Code for main to read inputs:
int main() {
int numTestCases;
vector<string> file_content{};
fstream ifs;
ifs.open("input2.txt");
if (!ifs.is_open()) {
cout << "Failed to open file.\n";
}
else {
ifs >> numTestCases;
char type;
int numberOps;
//ifs >> type;
cout << numTestCases;
for (int j = 0;j < numTestCases;j++) {
ifs >> type;
ifs >> numberOps;
if (type == 'S') {
Stack st;
clearStack();
for (int i = 0;i < numberOps;i++) {
string operation;
ifs >> operation;
if (operation == "push") {
int pushed;
ifs >> pushed;
push(pushed);
}
if (operation == "pop") {
pop();
}
if (operation == "print") {
printStack();
}
}
}
}
ifs.close();
}
}
I input a number in char type variable. like 12 or 22. but, console show me a 1 or 2.
How i get a whole number 12 ,22 in console?
#include <iostream>
int main()
{
using namespace std;
char a = 0;
cin >> a;
cout << a << endl;
return 0;
}
Here is console result.
12
1
C:\Users\kdwyh\source\repos\MyFirstProject\Debug\MyFirstProject.exe(프로세스 18464개)이(가) 종료되었습니다(코드: 0개).
이 창을 닫으려면 아무 키나 누르세요...
The reason I don't use int, string and something is because I want to get both number and Character in one variable.
So I want to see the results of combined numbers and character at the same time.
in that process i can't get a whole number.
#include <iostream>
using namespace std;
int index = 0;
constexpr int pagenum = 10;
void chapterlist(void);
void nextlist(void);
void beforelist(void);
void movechapter(char a);
int main(void)
{
char userin = 0;
bool toggle = 0;
cout << "결과를 볼 챕터를 고르시오." << endl;
chapterlist();
cout << "다음 페이지로 이동: n" << endl;
cin >> userin;
if (userin == 'n')
{
backflash:
while(toggle == 0)
{
nextlist();
cin >> userin;
if (userin == 'b')
{
toggle = 1;
goto backflash;
}
else if (userin == 'n')
continue;
else
{
system("cls");
movechapter(userin);
break;
}
}
while(toggle == 1)
{
beforelist();
cin >> userin;
if (userin == 'n')
{
toggle = 0;
goto backflash;
}
else if (userin == 'b')
continue;
else
{
system("cls");
movechapter(userin);
break;
}
}
}
else
{
system("cls");
movechapter(userin);
}
return 0;
}
void chapterlist(void)
{
int x = 0;
for (x = index + 1; x <= index + 10; x++)
cout << "Chapter." << x << endl;
}
void nextlist(void)
{
system("cls");
cout << "결과를 볼 챕터를 고르시오." << endl;
index = index + pagenum;
chapterlist();
cout << "다음 페이지로 이동: n" << endl;
cout << "이전 페이지로 이동: b" << endl;
}
void beforelist(void)
{
system("cls");
cout << "결과를 볼 챕터를 고르시오." << endl;
index = index - pagenum;
chapterlist();
cout << "다음 페이지로 이동: n" << endl;
cout << "이전 페이지로 이동: b" << endl;
}
void movechapter(char a)
{
cout << "선택한 Chapter." << a << "의 결과입니다." << endl;
}
In movechapter(), console show me a is 1 or 2, not 12, 22.
First, you have to understand what achar type is.
Character types: They can represent a single character, such as 'A' or '$'. The most basic type is char, which is a one-byte character. Other types are also provided for wider characters.
To simplify that, char can only hold one character.
Where as with your code, "12" is actually 2 separate characters, '1' and '2', and that's the reason it would not work.
Instead of declaring a as a char type, you could declare it as an int type, which is a type designed to hold numbers. So you would have:
int a = 0;
However, do note that int often has a maximum value of 2^31.
Or you could use std::string to store character strings. However, do note that if you wish to do any calculations to your string type, you would need to convert them to a number type first:
int myInt = std::stoi(myString);
Edit:
So I have re-checked your code after your update, there is nothing wrong with using std::string in your case. You can still check if user have input n or b by:
if (userin == "n")
Note that you would use double quotation mark, or "letter", around the content that you want to check.
On the other hand, you could use:
if(std::all_of(userin .begin(), userin.end(), ::isdigit))
To check if user have input a number.
Although char is just a number, it's presumed to mean "single character" here for input. Fix this by asking for something else:
int a = 0;
You can always cast that to char as necessary, testing, of course, for overflow.
You should be reading characters into a string, and then converting that string into an int. It would also probably make more sense to use something like getline() to read input, rather than cin >> a.
#include <string>
#include <iostream>
#include <stdexcept>
#include <stdio.h>
int main() {
std::string input_string;
/* note that there is no function that will convert an int string
to a char, only to an int. You can cast this to a char if needed,
or bounds check like I do */
int value;
while(1) {
getline(std::cin, input_string);
/* std::stoi throws std::invalid_argument when given a string
that doesn't start with a number */
try {
value = std::stoi(input_string);
} catch (std::invalid_argument) {
printf("Invalid number!\n");
continue;
}
/* You wanted a char, the max value of a `char` is 255. If
you are happy for any value, this check can be removed */
if (value > 255) {
printf("Too big, input a number between 0-255\n");
continue;
}
break;
}
printf("Number is %hhu\n", value);
}
As I am very new to exceptions, I am having some trouble with the below memory exception code. Basically, I would need to calculate and show sum of 10 positive inputs. During the input, my program should use exception mechanism to display a message that it doesn't allow negative numbers / not-a-numbers (NaN) if so is the input and then exit the program. I am told to use std::cin.fail() to detect whether the value entered fits the variable type(but I am not sure how to implement it). Would appreciate your help thanks!
#include <iostream>
int main() {
int number;
int a=-2;
try {
for(int i=0; i<10;i++) {
std::cin>>number;
}
}
catch (...) {
number==-number?
std::cout << "Its negative"<<std::endl;
number==a?
std::cout << "Its NaN"<<std::endl;
}
}
You say "10 positive inputs", but you have int number;. If your number is going to be an integer, then it can't take fractions like 33.44 for instance, and it also can't be NaN (so you wouldn't need to check for that). If you want to allow fractions, do double number instead. That aside, the check could look like this:
for (int i = 0; i < 10; i++) {
std::cin >> number;
if (std::cin.fail()) {
std::cin.clear();
std::string input;
std::cin >> input;
std::cout << "input failed! Invalid input: " << input << std::endl;
return -1;
}
After every inputted number, you do the std::cin.fail() check. If it failed, you can just return. In this example, I also made it print the invalid input. For that you need to call std::cin.clear(); to reset that error flag, and then you can put that invalid input into an std::string (which should work, unlike with the number). You need to include <string> for that. If you don't need to print anything, you can just return and forget about the clearing, the string and the output. Also, in this example, since the mess made by the invalid input is neatly cleaned up, you could take more (valid) inputs again if you don't return instead.
Next, checking for negative numbers:
if (number < 0) {
std::cout << "Error: Negative number!" << std::endl;
return -1;
}
Finally, the NaN check (for double):
if (isnan(number)) {
std::cout << "Error: number is NaN!" << std::endl;
return -1;
}
You need to include <math.h> for the isnan.
Putting it all together, it could look like this:
#include <iostream>
#include <string> // std::istream >> std::string
#include <math.h> // isnan
int main() {
double number;
for (int i = 0; i < 10; i++) {
std::cin >> number;
if (std::cin.fail()) {
std::cin.clear();
std::string input;
std::cin >> input;
std::cout << "input failed! Invalid input: " << input << std::endl;
return -1;
}
if (number < 0) {
std::cout << "Error: Negative number!" << std::endl;
return -1;
}
if (isnan(number)) {
std::cout << "Error: number is NaN!" << std::endl;
return -1;
}
}
return 0;
}
I was typing this and it asks the user to input two integers which will then become variables. From there it will carry out simple operations.
How do I get the computer to check if what is entered is an integer or not? And if not, ask the user to type an integer in. For example: if someone inputs "a" instead of 2, then it will tell them to reenter a number.
Thanks
#include <iostream>
using namespace std;
int main ()
{
int firstvariable;
int secondvariable;
float float1;
float float2;
cout << "Please enter two integers and then press Enter:" << endl;
cin >> firstvariable;
cin >> secondvariable;
cout << "Time for some simple mathematical operations:\n" << endl;
cout << "The sum:\n " << firstvariable << "+" << secondvariable
<<"="<< firstvariable + secondvariable << "\n " << endl;
}
You can check like this:
int x;
cin >> x;
if (cin.fail()) {
//Not an int.
}
Furthermore, you can continue to get input until you get an int via:
#include <iostream>
int main() {
int x;
std::cin >> x;
while(std::cin.fail()) {
std::cout << "Error" << std::endl;
std::cin.clear();
std::cin.ignore(256,'\n');
std::cin >> x;
}
std::cout << x << std::endl;
return 0;
}
EDIT: To address the comment below regarding input like 10abc, one could modify the loop to accept a string as an input. Then check the string for any character not a number and handle that situation accordingly. One needs not clear/ignore the input stream in that situation. Verifying the string is just numbers, convert the string back to an integer. I mean, this was just off the cuff. There might be a better way. This won't work if you're accepting floats/doubles (would have to add '.' in the search string).
#include <iostream>
#include <string>
int main() {
std::string theInput;
int inputAsInt;
std::getline(std::cin, theInput);
while(std::cin.fail() || std::cin.eof() || theInput.find_first_not_of("0123456789") != std::string::npos) {
std::cout << "Error" << std::endl;
if( theInput.find_first_not_of("0123456789") == std::string::npos) {
std::cin.clear();
std::cin.ignore(256,'\n');
}
std::getline(std::cin, theInput);
}
std::string::size_type st;
inputAsInt = std::stoi(theInput,&st);
std::cout << inputAsInt << std::endl;
return 0;
}
Heh, this is an old question that could use a better answer.
User input should be obtained as a string and then attempt-converted to the data type you desire. Conveniently, this also allows you to answer questions like “what type of data is my input?”
Here is a function I use a lot. Other options exist, such as in Boost, but the basic premise is the same: attempt to perform the string→type conversion and observe the success or failure:
template <typename T>
auto string_to( const std::string & s )
{
T value;
std::istringstream ss( s );
return ((ss >> value) and (ss >> std::ws).eof()) // attempt the conversion
? value // success
: std::optional<T> { }; // failure
}
Using the optional type is just one way. You could also throw an exception or return a default value on failure. Whatever works for your situation.
Here is an example of using it:
int n;
std::cout << "n? ";
{
std::string s;
getline( std::cin, s );
auto x = string_to <int> ( s );
if (!x) return complain();
n = *x;
}
std::cout << "Multiply that by seven to get " << (7 * n) << ".\n";
limitations and type identification
In order for this to work, of course, there must exist a method to unambiguously extract your data type from a stream. This is the natural order of things in C++ — that is, business as usual. So no surprises here.
The next caveat is that some types subsume others. For example, if you are trying to distinguish between int and double, check for int first, since anything that converts to an int is also a double.
There is a function in c called isdigit(). That will suit you just fine. Example:
int var1 = 'h';
int var2 = '2';
if( isdigit(var1) )
{
printf("var1 = |%c| is a digit\n", var1 );
}
else
{
printf("var1 = |%c| is not a digit\n", var1 );
}
if( isdigit(var2) )
{
printf("var2 = |%c| is a digit\n", var2 );
}
else
{
printf("var2 = |%c| is not a digit\n", var2 );
}
From here
If istream fails to insert, it will set the fail bit.
int i = 0;
std::cin >> i; // type a and press enter
if (std::cin.fail())
{
std::cout << "I failed, try again ..." << std::endl
std::cin.clear(); // reset the failed state
}
You can set this up in a do-while loop to get the correct type (int in this case) propertly inserted.
For more information: http://augustcouncil.com/~tgibson/tutorial/iotips.html#directly
You can use the variables name itself to check if a value is an integer.
for example:
#include <iostream>
using namespace std;
int main (){
int firstvariable;
int secondvariable;
float float1;
float float2;
cout << "Please enter two integers and then press Enter:" << endl;
cin >> firstvariable;
cin >> secondvariable;
if(firstvariable && secondvariable){
cout << "Time for some simple mathematical operations:\n" << endl;
cout << "The sum:\n " << firstvariable << "+" << secondvariable
<<"="<< firstvariable + secondvariable << "\n " << endl;
}else{
cout << "\n[ERROR\tINVALID INPUT]\n";
return 1;
}
return 0;
}
I prefer to use <limits> to check for an int until it is passed.
#include <iostream>
#include <limits> //std::numeric_limits
using std::cout, std::endl, std::cin;
int main() {
int num;
while(!(cin >> num)){ //check the Input format for integer the right way
cin.clear();
cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
cout << "Invalid input. Reenter the number: ";
};
cout << "output= " << num << endl;
return 0;
}
Under C++11 and later, I have the found the std::stoi function very useful for this task. stoi throws an invalid_argument exception if conversion cannot be performed. This can be caught and handled as shown in the demo function 'getIntegerValue' below.
The stoi function has a second parameter 'idx' that indicates the position of the first character in the string after the number. We can use the value in idx to check against the string length and ascertain if there are any characters in the input other than the number. This helps eliminate input like 10abc or a decimal value.
The only case where this approach fails is when there is trailing white space after the number in the input, that is, the user enters a lot of spaces after inputting the number. To handle such a case, you could rtrim the input string as described in this post.
#include <iostream>
#include <string>
bool getIntegerValue(int &value);
int main(){
int value{};
bool valid{};
while(!valid){
std::cout << "Enter integer value: ";
valid = getIntegerValue(value);
if (!valid)
std::cout << "Invalid integer value! Please try again.\n" << std::endl;
}
std::cout << "You entered: " << value << std::endl;
return 0;
}
// Returns true if integer is read from standard input
bool getIntegerValue(int &value){
bool isInputValid{};
int valueFromString{};
size_t index{};
std::string userInput;
std::getline(std::cin, userInput);
try {
//stoi throws an invalid_argument exception if userInput cannot be
//converted to an integer.
valueFromString = std::stoi(userInput, &index);
//index being different than length of string implies additional
//characters in input that could not be converted to an integer.
//This is to handle inputs like 10str or decimal values like 10.12
if(index == userInput.length()) isInputValid = true;
}
catch (const std::invalid_argument &arg) {
; //you could show an invalid argument message here.
}
if (isInputValid) value = valueFromString;
return isInputValid;
}
You could use :
int a = 12;
if (a>0 || a<0){
cout << "Your text"<<endl;
}
I'm pretty sure it works.
I'm new to c++ programming and i tried to use struct arrays ( dunno if I say it in the way you understand ). Program does read the variables from the text file but it reads it for 2 times, prob It's not a huge problem as the variables are not changed but just in case
struct menesine_temperatura
{
int diena[10];
int rytas[10];
int pietus[10];
int vakaras[10];
};
int main()
{
menesine_temperatura mt;
int n;
ifstream failas("duomenys.txt");
while(!failas.eof())
{
failas >> n;
cout << n << endl;
for(int i = 0; i < n; i++)
{
failas >> mt.diena[i] >> mt.rytas[i] >> mt.pietus[i] >> mt.vakaras[i];
cout << mt.diena[i] << mt.rytas[i] << mt.pietus[i] << mt.vakaras[i] << endl;
}
}
failas.close();
return 0;
}
as I have mentioned before I' new to this stuff, so help would be appreciated.
With the code you have you, it will appear as if the last set of numbers are being read twice.
Here's a simple example that explains the logic error.
int main()
{
int n;
std::ifstream inf("data.txt");
while ( !inf.eof() )
{
inf >> n;
std::cout << n << std::endl;
}
}
and your input file consists of
10
20
In the first iteration of the loop, 10 is read into n and 10 is printed.
In the second iteration of the loop, 20 is read into n and 20 is printed.
At this point inf.eof() is still false. So you enter the loop the third time. At this time, nothing is read into n but the value of n still 20 and 20 is printed.
The logic error is that just because inf.eof() is false, you assume that there is still some data in the file that can be read.
A better way would be to use:
int main()
{
int n;
std::ifstream inf("data.txt");
while ( inf )
{
if ( inf >> n )
{
std::cout << n << std::endl;
}
}
}
I hope this gives you enough information on how to update your program.