It might sound stupid, but I was just wondering how can I parse a string to an integer in C++?
This is for a school project, and the explanation sheet says :
"Only the use of iostream and string classes and system() function is permitted; The use of any other class or function is prohibited."
I have searched around a bit, but the only suggestions I found are using classes like atoi or atof.
The input string is already checked for error before the parsing, so it will always only contain an integer.
I wouldn't mind doing the parsing manually with conditionals, but I don't think it would be my teacher's (or anyone's) preferred way.
Thanks a bunch if you can help.
So, you can use system(), huh? Behold this masterpiece of engineering:
#include <fstream>
void download_boost() {
system("wget http://downloads.sourceforge.net/"
"project/boost/boost/1.47.0/boost_1_47_0.tar.bz2"
"?r=http%3A%2F%2Fwww.boost.org%2Fusers%2Fhistory%2F"
"version_1_47_0.html&ts=1316116936&use_mirror=kent"
" -O boost_1_47_0.tar.bz2");
}
void unpack_boost() {
system("tar --bzip2 -xf boost_1_47_0.tar.bz2");
}
void write_program() {
std::ofstream os("blah.cpp");
os << "#include \"boost/lexical_cast.hpp\"\n"
"#include <iostream>\n"
"#include <string>\n"
"int main() { std::string s; std::cin >> s;"
"int i = boost::lexical_cast<int>(s);"
"std::cout << i; }";
}
void compile_program() {
system("g++ -Iboost_1_47_0 blah.cpp");
}
void run_program() {
system("./a.out");
}
int main() {
download_boost();
unpack_boost();
write_program();
compile_program();
run_program();
}
(I'm assuming a typical Linux installation with some common tools installed.)
#include <iostream>
#include <string>
int stringToInt(const std::string &text)
{
int number = 0;
int powerIndex = 1;
for (int i = text.length() - 1; i >= 0; i--)
{
number += powerIndex * (text.at(i) - '0');
powerIndex *= 10;
}
return number;
}
std::string intToString (int number)
{
std::string text = "";
int numberHolder = number;
while (numberHolder)
{
char digit = (numberHolder % 10) + '0';
text = digit + text;
numberHolder /= 10;
}
return text;
}
int main ()
{
//Testing...
int number = stringToInt("123");
std::string text = intToString(456);
std::cout << number << "\n" << text << "\n";
return 0;
}
stringstreams come closest to what you want to do, although it may seem a little cumbersome at first.
Example:
#include <string>
#include <sstream>
// (...)
std::string str = "12345";
std::istringstream ss(str);
int num;
ss >> num;
As a function (and optimized):
#include <sstream>
#include <string>
int stringToInt(const std::string &str) {
static std::istringstream ss;
ss.clear();
ss.str(str);
int num;
ss >> num;
return num;
}
Here, I am reusing the std::istringstream by applying the static keyword. I have created a very simple benchmark that demonstrates that this is approximately 2 times faster than not reusing: http://pastebin.com/vLSmCyMF
Warning regarding thread-safety: As рытфолд has noted in the comments, the above implementation of stringToInt is not thread-safe. If you want to call this function safely from multiple threads, you should use the thread_local storage class specifier (available since C++11).
std::string mystring = "12";
std::ostringstream str(mystring);
int someint;
str >> someint;
Basically, given a std::string mystring that contains only an integer:
Start with an int result = 0;.
While mystring has digits
Multiply result by ten
Add biggest digit (first in string)
Remember that the character '0' does not have the value of 0.
Remove the biggest digit from the string
So:
If I have the string "1543", result = 0
we multiply result by ten : result = 0
we add the first digit: result = 1
remove the first digit from the string "543"
we multiply result by ten : result = 10
we add the first digit: result = 15
remove the first digit from the string "43"
we multiply result by ten : result = 150
we add the first digit: result = 154
remove the first digit from the string "3"
we multiply result by ten : result = 1540
we add the first digit: result = 1543
remove the first digit from the string ""
string is empty, so we're done
I wrote code, but then remembered this was a homework problem.
Related
I am faced with a simple yet complex challenge today.
In my program, I wish to insert a - character every three characters of a string. How would this be accomplished? Thank you for your help.
#include <iostream>
int main()
{
std::string s = "thisisateststring";
// Desired output: thi-sis-ate-sts-tri-ng
std::cout << s << std::endl;
return 0;
}
There is no need to "build a new string".
Loop a position iteration, starting at 3, incrementing by 4 with each pass, inserting a - at the position indicated. Stop when the next insertion point would breach the string (which has been growing by one with each pass, thus the need for the 4 slot skip):
#include <iostream>
#include <string>
int main()
{
std::string s = "thisisateststring";
for (std::string::size_type i=3; i<s.size(); i+=4)
s.insert(i, 1, '-');
// Desired output: thi-sis-ate-sts-tri-ng
std::cout << s << std::endl;
return 0;
}
Output
thi-sis-ate-sts-tri-ng
just take an empty string and append "-" at every count divisible by 3
#include <iostream>
int main()
{
std::string s = "thisisateststring";
std::string res="";
int count=0;
for(int i=0;i<s.length();i++){
count++;
res+=s[i];
if(count%3==0){
res+="-";
}
}
std::cout << res << std::endl;
return 0;
}
output
thi-sis-ate-sts-tri-ng
A general (and efficient) approach is to build a new string by iterating character-by-character over the existing one, making any desired changes as you go. In this case, every third character you can insert a hyphen:
std::string result;
result.reserve(s.size() + s.size() / 3);
for (size_t i = 0; i != s.size(); ++i) {
if (i != 0 && i % 3 == 0)
result.push_back('-');
result.push_back(s[i]);
}
Simple. Iterate the string and build a new one
Copy each character from the old string to the new one and every time you've copied 3 characters add an extra '-' to the end of the new string and restart your count of copied characters.
Like 99% problems with text, this one can be solved with a regular expression one-liner:
std::regex_replace(input, std::regex{".{3}"}, "$&-")
However, it brings not one, but two new problems:
it is not a very performant solution
regex library is huge and bloats resulting binary
So think twice.
You could write a simple functor to add the hyphens, like this:
#include <iostream>
struct inserter
{
unsigned n = 0u;
void operator()(char c)
{
std::cout << c;
if (++n%3 == 0) std::cout << '-';
}
};
This can be passed to the standard for_each() algorithm:
#include <algorithm>
int main()
{
const std::string s = "thisisateststring";
std::for_each(s.begin(), s.end(), inserter());
std::cout << std::endl;
}
Exercise: extend this class to work with different intervals, output streams, replacement characters and string types (narrow or wide).
This question already has answers here:
How to check if input is numeric in C++
(8 answers)
Closed 6 years ago.
I'm new to C++,
can I ask is there a way to check if a input string is a number?
If it is a number, change it to integer.
I know we can use either atoi or stoi.
But how can we create it in a function?
Construct a std::istringstream from the std::string.
Extract a number from it. If extraction is successful, use the number. If not, move on to the next thing.
std::string s = "123";
std::istringstream str(s);
int n;
if ( str >> n )
{
// Extraction is successful.
// Use n.
}
Further Enahncements
To make your program more robust, you can add a further check to make sure that:
"123FAST" is not treated as an integer.
"123.56" is not treated as an integer.
"123.56xyz" is not treated as an integer.
std::string s = "123";
std::istringstream str(s);
int n;
if ( str >> n )
{
// Extraction is successful.
// Add another check.
// Get the next character.
// It has to be a whitespace character or EOF for the input
// to be an integer.
// If it is neither, don't accept the input as an integer.
std::istream::int_type c = str.get();
if ( std::isspace(c) || str.eof() )
{
// Use n.
}
}
You just check each character in your input string is a number, I have a simple example for your problem:
#include <stdio.h>
#include <string.h>
bool is_number(const char * s)
{
int length = strlen(s);
for (int i = 0; i < length; i++)
{
if (s[i] < '0' || s[i] > '9')
return false;
}
return true;
}
int main()
{
printf("Enter a string: ");
char myString[10];
gets(myString);
if (is_number(myString))
printf("%s is a number", myString);
else
printf("%s is not a number", myString);
}
I don't think you even need to check. According to the C++ API:
If the first sequence of non-whitespace characters in str is not a
valid integral number, or if no such sequence exists because either
str is empty or it contains only whitespace characters, no conversion
is performed and zero is returned.
SEE:
http://www.cplusplus.com/reference/cstdlib/atoi/
-
So, you should just be able to do:
// Convert the string
int value = atoi( myString.c_str() );
if ( value == 0 )
{
// The string was not a valid integer
}
else
{
// The string was a valid integer
}
'isnum()' function below checks if given string is comprised of numbers only. However it might not work if number is very large.
#include <bits/stdc++.h>
using namespace std;
bool isnum(string str){
for(int i=0;i<str.length();i++){
if((str[i]<'0') || (str[i]>'9'))return false;
}
return true;
}
int main() {
// your code goes here
int n=0,r=1;
string num;
cin>>num;
if(isnum(num)){
for(int i=num.length()-1;i>=0;i--){
n+=(r)*(int)(num[i]-'0');
r*=10;
}
cout<<n;
}
return 0;
}
You can check if there is any non-digit character in the string:
#include <iostream>
#include <string>
#include <algorithm>
#include <cctype>
int main()
{
std::string test;
std::cin >> test;
if(std::find_if(test.begin(), test.end(), [](auto x){return !isdigit(x);}) == test.end())
std::cout << "its a number" << std::endl;
else
std::cout << "not a number" << std::endl;
}
I'm trying to do parsing to some input string reactions read from file at formula :2W+B=8A+10Z, I'm not interested in characters i need only to split and extract the integer values to put them in a vector i.e vector associated with the reaction here is :[2 1 8 10]
i thought about many things: std::strtok(),isdigital(),find_first_of() but they all didn't work for integer values ... can any body help ??
here my try:
int main()
{
std::string input;
std::getline(std::cin, input);
std::stringstream stream(input);
while(1) {
int n;
stream >> n;
char * pch;
pch = strtok (input," ");
while (pch != NULL)
{
printf ("%s\n",pch);
pch = strtok (NULL, " ,.");
}
}
}
This will do what you want in this particular case. However, i suggest that you look into regex to parse your equation better. You may want to consider all possible cases for your input. This includes \,-,* and other operators that you may want to add in your equation. Also, I'm assuming variables in your equation has only one character.
int main()
{
string input;
getline(std::cin, input);
stringstream stream(input);
char tmp[256];
const char *in = input.c_str();
char str[256];
strcpy(str,in);
int x;
tmp[0]='\0';
char c;
vector<int> vec;
//Scan for the digit
//if it is, store the rest of the string back to str
//if it isn't, store the part of the string before a digit to tmp
while (sscanf(str,"%d%s",&x,str) || sscanf(str,"%[^0123456789]%s",tmp,str) > 1)
{
//check if tmp has the form [variable name]+[a string]
//a string can include another variable name and an operator, = in this case
while(sscanf(tmp,"%c+%[^0123456789]",&c,tmp) > 1)
vec.push_back(1);
if (tmp[0]=='\0')
vec.push_back(x);
tmp[0]='\0';
}
//just in case there're more special cases
while(sscanf(str,"%c+%[^0123456789]",&c,str) > 1)
vec.push_back(1);
for(int i = 0; i < vec.size(); i++)
cout << vec[i] << endl;
}
Output:
2
1
8
10
See comments for explanation.
EDIT
Be careful when you have a special case 2W+B=8A+10Z+C+D. Notice the last C D should both have coefficients 1. This could happen in the middle of the equation too.
Here is another solution:
#include <iostream>
#include <vector>
#include <string>
using namespace std;
int main()
{
string equ;
vector<int> digits;
cout << "enter your equation: \n";
cin >> equ;
for (auto i : equ)
{
if (isdigit(i))
{
digits.push_back(stoi(string{i}));
}
}
for (auto& i : digits)
{
cout << i << endl;
}
system("pause");
return 0;
}
You could simply do something like this, for comments see code
#include <iostream>
#include <string>
#include <vector>
std::vector<int> Split(std::string str)
{
std::vector<int> result; // will contain the different ints
// set pointer to first character in the string
char const* pch = str.c_str();
std::string digit; // buffer to keep digits if more than one
int sign = 1; // each number has a sign -1 or 1
for (; *pch; ++pch)
{
if (std::isdigit(*pch)) // if a digit, put in temp buffer
{
digit += *pch;
}
else if (std::isalpha(*pch)) // if not a digit evaluate the ones we have
{
if (digit.empty()) // none so assume 1 before letter e.g. W+2B
{
result.push_back(1*sign);
}
else
{
result.push_back(stoi(digit)*sign);
digit = "";
}
}
else // determine sign of number
{
digit = "";
if (*pch == '+')
{
sign = 1;
}
else if (*pch == '-')
{
sign = -1;
}
}
}
return result;
}
int main()
{
using namespace std;
string expr{"-2W+B=-8A+10Z"};
auto digits = Split(expr);
for (auto& digit : digits)
{
cout << digit << endl;
}
return 0;
}
I don't know whether or not this is possible but I have used this technique in different languages but am struggling to use it in C++. I have 10 images that I am trying to load into an array using a loop as so:
for (int i = 0; i < 10; i++)
{
Sprite[i] = IMG_LoadTexture(renderer, "Graphics/Player" + i + ".png");
}
This however does not seem to work in C++ so I was wondering what I am doing wrong, or what can I do to get the same result without having to load each image individually like so:
Sprite[0] = IMG_LoadTexture(renderer, "Graphics/Player0.png");
My error is: "Expression must have integral or unscoped enum type"
Thanks for any help =)
You cannot do this:
"This is my number: " + (int)4 + "!";
This is illegal. It will give you an error for trying to operator+ a const char* and a const char[SOME_INT_GOES_HERE] or another error for trying to use operator+ to add an int onto a string. Things just don't work that way.
You'd either have to use C (i.e. snprintf()) or a string stream. Here's my test code for isolating your problem:
#include <iostream>
#include <string>
int main()
{
int a = 1;
std::string str = "blah";
std::string end = "!";
//std::string hello = str + a + end;// GIVES AN ERROR for operator+
std::string hello = "blah" + a + "!";
//const char* c_str = "blah" + a + "end";
//std::cout << c_str << std::endl;
std::cout << hello << std::endl;
return 0;
}
Here's an alternative solution using string streams.
#include <iostream>
#include <string>
#include <sstream>
int main()
{
int i = 0;
std::string str;
std::stringstream ss;
while (i < 10)
{
//Send text to string stream.
ss << "text" << i;
//Set string to the text inside string stream
str = ss.str();
//Print out the string
std::cout << str << std::endl;
//ss.clear() doesn't work. Calling a constructor
//for std::string() and setting ss.str(std::string())
//will set the string stream to an empty string.
ss.str(std::string());
//Remember to increment the variable inside of while{}
++i;
}
}
Alternatively, you can also use std::to_string() if you're using C++11 (which just requires -std=c++11) but std::to_string() is broken on some sets of compilers (i.e. regular MinGW). Either switch to another flavor where it works (i.e. MinGW-w64) or just write your own to_string() function using string streams behind the scenes.
snprintf() may be the fastest way of doing such a thing, but for safer C++ and better style, it is recommended you use a non-C way of doing things.
I had a similar problem and I solwed it this way:
#include <iostream>
using namespace std;
int main() {
string line;
for (int i = 0; i < 10; i++) {
line = "Graphics/Player" + inttostr(i) + ".png"; //I wrote inttostr function because built in inttostr functions messed up my program (see below)
char charger[line.length()]; //creating char array
for (int i = 0; i < sizeof(line); i++) {
charger[i] = line[i]; // copying string to char arry
}
Sprite[i] = IMG_LoadTexture(renderer, charger);
}
}
string inttostr(int integer) { //I know it isn't the best way to convert integer to string, but it works
string charakter;
int swap;
bool negativ = false;
if (integer < 0) {
integer = -integer;
negativ = true;
}
if (integer == 0) {
charakter = "0";
}
while (integer >= 1) {
swap = integer % 10;
integer = integer / 10;
charakter = char(swap + 48) + charakter;
}
if (negativ) {
charakter = "-" + charakter;
}
return charakter;
}
I have a string in form "blah-blah..obj_xx..blah-blah" where xx are digits. E.g. the string may be "root75/obj_43.dat".
I want to read "xx" (or 43 from the sample above) as an integer. How do I do it?
I tried to find "obj_" first:
std::string::size_type const cpos = name.find("obj_");
assert(std::string::npos != cpos);
but what's next?
My GCC doesn't support regexes fully, but I think this should work:
#include <iostream>
#include <string>
#include <regex>
#include <iterator>
int main ()
{
std::string input ("blah-blah..obj_42..blah-blah");
std::regex expr ("obj_([0-9]+)");
std::sregex_iterator i = std::sregex_iterator(input.begin(), input.end(), expr);
std::smatch match = *i;
int number = std::stoi(match.str());
std::cout << number << '\n';
}
With something this simple you can do
auto b = name.find_first_of("0123456789", cpos);
auto e = name.find_first_not_of("0123456789", b);
if (b != std::string::npos)
{
auto digits = name.substr(b, e);
int n = std::stoi(digits);
}
else
{
// Error handling
}
For anything more complicated I would use regex.
How about:
#include <iostream>
#include <string>
int main()
{
const std::string test("root75/obj_43.dat");
int number;
// validate input:
const auto index = test.find("obj_");
if(index != std::string::npos)
{
number = std::stoi(test.substr(index+4));
std::cout << "number: " << number << ".\n";
}
else
std::cout << "Input validation failed.\n";
}
Live demo here. Includes (very) basic input validation (e.g. it will fail if the string contains multiple obj_), variable length numbers at the end, or even more stuff following it (adjust the substr call accordingly) and you can add a second argument to std::stoi to make sure it didn't fail for some reason.
Here's another option
//your code:
std::string::size_type const cpos = name.find("obj_");
assert(std::string::npos != cpos);
//my code starts here:
int n;
std::stringstream sin(name.substr(cpos+4));
sin>>n;
Dirt simple method, though probably pretty inefficient, and doesn't take advantage of the STL:
(Note that I didn't try to compile this)
unsigned GetFileNumber(std::string &s)
{
const std::string extension = ".dat";
/// get starting position - first character to the left of the file extension
/// in a real implementation, you'd want to verify that the string actually contains
/// the correct extension.
int i = (int)(s.size() - extension.size() - 1);
unsigned sum = 0;
int tensMultiplier = 1;
while (i >= 0)
{
/// get the integer value of this digit - subtract (int)'0' rather than
/// using the ASCII code of `0` directly for clarity. Optimizer converts
/// it to a literal immediate at compile time, anyway.
int digit = s[i] - (int)'0';
/// if this is a valid numeric character
if (digit >= 0 && digit <= 9)
{
/// add the digit's value, adjusted for it's place within the numeric
/// substring, to the accumulator
sum += digit * tensMultiplier;
/// set the tens place multiplier for the next digit to the left.
tensMultiplier *= 10;
}
else
{
break;
}
i--;
}
return sum;
}
If you need it as a string, just append the found digits to a result string rather than accumulating their values in sum.
This also assumes that .dat is the last part of your string. If not, I'd start at the end, count left until you find a numeric character, and then start the above loop. This is nice because it's O(n), but may not be as clear as the regex or find approaches.