Invalid Conversion Problem in C++ - c++

I have the following snippet:
string base= tag1[j];
That gives the invalid conversion error.
What's wrong with my code below? How can I overcome it.
Full code is here:
#include <iostream>
#include <vector>
#include <fstream>
#include <sstream>
#include <time.h>
using namespace std;
int main ( int arg_count, char *arg_vec[] ) {
if (arg_count < 3 ) {
cerr << "expected one argument" << endl;
return EXIT_FAILURE;
}
// Initialize Random Seed
srand (time(NULL));
string line;
string tag1 = arg_vec[1];
string tag2 = arg_vec[2];
double SubsRate = 0.003;
double nofTag = static_cast<double>(atoi(arg_vec[3]));
vector <string> DNA;
DNA.push_back("A");
DNA.push_back("C");
DNA.push_back("G");
DNA.push_back("T");
for (unsigned i=0; i < nofTag ; i++) {
int toSub = rand() % 1000 + 1;
if (toSub <= (SubsRate * 1000)) {
// Mutate
cout << toSub << " Sub" << endl;
int mutateNo = 0;
for (int j=0; j < tag1.size(); j++) {
mutateNo++;
string base = tag1[j]; // This fail
int dnaNo = rand() % 4;
if (mutateNo <= 3) {
// Mutation happen at most at 3 position
base = DNA[dnaNo];
}
cout << tag1[j] << " " << dnaNo << " " << base << endl;
//cout << base;
}
cout << endl;
}
else {
// Don't mutate
//cout << tag1 << endl;
}
}
return 0;
}
Why do I get an Invalid conversion from char to const char* when looping over a string?

The std::string operator [] returns a single char. string cannot be instantiated with a single char.
Use:
string base = string( 1, tag1[j] ) instead

Change it to
char base = tag1[j];

string tag1 = arg_vec[1];
tag1 is a string literal.
string base = tag1[j]; is initialized with a char instead of char *.
Try, char base = tag1[j];

There is no constructor for string that takes just a char (which is what tag1[j] is). You have a couple options:
string base; // construct a default string
base = tag1[j]; // set it to a char (there is an
// assignment from char to string,
// even if there's no constructor
or
string base( 1, tag1[j]); // create a string with a single char
Or as Josh mentioned, you can define base as a char since you're not performing any string operations on it anyway. If you decide to do this you'll need to change DNA to be a vector<char> (and change the initialization of DNA to using chars instead of strings).

One problem is that the error message says the program expects one argument when it actually requires two. You should probably follow the Unix conventions and show the required usage too (or instead):
if (arg_count != 3) {
cerr << "Usage: " << arg_vec[0] << " tag1 tag2";
return EXIT_FAILURE;
}
The names 'argc' and 'argv' are very conventional (and the only major alternative I've seen is 'ac' and 'av'). It might be worth sticking with that.

Related

Converting string to a int32. Is there a better way than this? [duplicate]

I want to convert a string to an int and I don't mean ASCII codes.
For a quick run-down, we are passed in an equation as a string. We are to break it down, format it correctly and solve the linear equations. Now, in saying that, I'm not able to convert a string to an int.
I know that the string will be in either the format (-5) or (25) etc. so it's definitely an int. But how do we extract that from a string?
One way I was thinking is running a for/while loop through the string, check for a digit, extract all the digits after that and then look to see if there was a leading '-', if there is, multiply the int by -1.
It seems a bit over complicated for such a small problem though. Any ideas?
In C++11 there are some nice new convert functions from std::string to a number type.
So instead of
atoi( str.c_str() )
you can use
std::stoi( str )
where str is your number as std::string.
There are version for all flavours of numbers:
long stol(string), float stof(string), double stod(string),...
see http://en.cppreference.com/w/cpp/string/basic_string/stol
The possible options are described below:
1. sscanf()
#include <cstdio>
#include <string>
int i;
float f;
double d;
std::string str;
// string -> integer
if(sscanf(str.c_str(), "%d", &i) != 1)
// error management
// string -> float
if(sscanf(str.c_str(), "%f", &f) != 1)
// error management
// string -> double
if(sscanf(str.c_str(), "%lf", &d) != 1)
// error management
This is an error (also shown by cppcheck) because "scanf without field width limits can crash with huge input data on some versions of libc" (see here, and here).
2. std::sto()*
#include <iostream>
#include <string>
int i;
float f;
double d;
std::string str;
try {
// string -> integer
int i = std::stoi(str);
// string -> float
float f = std::stof(str);
// string -> double
double d = std::stod(str);
} catch (...) {
// error management
}
This solution is short and elegant, but it is available only on on C++11 compliant compilers.
3. sstreams
#include <string>
#include <sstream>
int i;
float f;
double d;
std::string str;
// string -> integer
std::istringstream ( str ) >> i;
// string -> float
std::istringstream ( str ) >> f;
// string -> double
std::istringstream ( str ) >> d;
// error management ??
However, with this solution is hard to distinguish between bad input (see here).
4. Boost's lexical_cast
#include <boost/lexical_cast.hpp>
#include <string>
std::string str;
try {
int i = boost::lexical_cast<int>( str.c_str());
float f = boost::lexical_cast<int>( str.c_str());
double d = boost::lexical_cast<int>( str.c_str());
} catch( boost::bad_lexical_cast const& ) {
// Error management
}
However, this is just a wrapper of sstream, and the documentation suggests to use sstream for better error management (see here).
5. strto()*
This solution is very long, due to error management, and it is described here. Since no function returns a plain int, a conversion is needed in case of integer (see here for how this conversion can be achieved).
6. Qt
#include <QString>
#include <string>
bool ok;
std::string;
int i = QString::fromStdString(str).toInt(&ok);
if (!ok)
// Error management
float f = QString::fromStdString(str).toFloat(&ok);
if (!ok)
// Error management
double d = QString::fromStdString(str).toDouble(&ok);
if (!ok)
// Error management
Conclusions
Summing up, the best solution is C++11 std::stoi() or, as a second option, the use of Qt libraries. All other solutions are discouraged or buggy.
std::istringstream ss(thestring);
ss >> thevalue;
To be fully correct you'll want to check the error flags.
use the atoi function to convert the string to an integer:
string a = "25";
int b = atoi(a.c_str());
http://www.cplusplus.com/reference/clibrary/cstdlib/atoi/
To be more exhaustive (and as it has been requested in comments), I add the solution given by C++17 using std::from_chars.
std::string str = "10";
int number;
std::from_chars(str.data(), str.data()+str.size(), number);
If you want to check whether the conversion was successful:
std::string str = "10";
int number;
auto [ptr, ec] = std::from_chars(str.data(), str.data()+str.size(), number);
assert(ec == std::errc{});
// ptr points to chars after read number
Moreover, to compare the performance of all these solutions, see the following quick-bench link: https://quick-bench.com/q/GBzK53Gc-YSWpEA9XskSZLU963Y
(std::from_chars is the fastest and std::istringstream is the slowest)
1. std::stoi
std::string str = "10";
int number = std::stoi(str);
2. string streams
std::string str = "10";
int number;
std::istringstream(str) >> number
3. boost::lexical_cast
#include <boost/lexical_cast.hpp>
std::string str = "10";
int number;
try
{
number = boost::lexical_cast<int>(str);
std::cout << number << std::endl;
}
catch (boost::bad_lexical_cast const &e) // bad input
{
std::cout << "error" << std::endl;
}
4. std::atoi
std::string str = "10";
int number = std::atoi(str.c_str());
5. sscanf()
std::string str = "10";
int number;
if (sscanf(str .c_str(), "%d", &number) == 1)
{
std::cout << number << '\n';
}
else
{
std::cout << "Bad Input";
}
What about Boost.Lexical_cast?
Here is their example:
The following example treats command line arguments as a sequence of numeric data:
int main(int argc, char * argv[])
{
using boost::lexical_cast;
using boost::bad_lexical_cast;
std::vector<short> args;
while(*++argv)
{
try
{
args.push_back(lexical_cast<short>(*argv));
}
catch(bad_lexical_cast &)
{
args.push_back(0);
}
}
...
}
Admittedly, my solution wouldn't work for negative integers, but it will extract all positive integers from input text containing integers. It makes use of numeric_only locale:
int main() {
int num;
std::cin.imbue(std::locale(std::locale(), new numeric_only()));
while ( std::cin >> num)
std::cout << num << std::endl;
return 0;
}
Input text:
the format (-5) or (25) etc... some text.. and then.. 7987...78hjh.hhjg9878
Output integers:
5
25
7987
78
9878
The class numeric_only is defined as:
struct numeric_only: std::ctype<char>
{
numeric_only(): std::ctype<char>(get_table()) {}
static std::ctype_base::mask const* get_table()
{
static std::vector<std::ctype_base::mask>
rc(std::ctype<char>::table_size,std::ctype_base::space);
std::fill(&rc['0'], &rc[':'], std::ctype_base::digit);
return &rc[0];
}
};
Complete online demo : http://ideone.com/dRWSj
In C++11 we can use "stoi" function to convert string into a int
#include <iostream>
#include <string>
using namespace std;
int main()
{
string s1 = "16";
string s2 = "9.49";
string s3 = "1226";
int num1 = stoi(s1);
int num2 = stoi(s2);
int num3 = stoi(s3);
cout << "stoi(\"" << s1 << "\") is " << num1 << '\n';
cout << "stoi(\"" << s2 << "\") is " << num2 << '\n';
cout << "stoi(\"" << s3 << "\") is " << num3 << '\n';
return 0;
}
It's probably a bit of overkill, but
boost::lexical_cast<int>( theString ) should to the job
quite well.
Well, lot of answers, lot of possibilities. What I am missing here is some universal method that converts a string to different C++ integral types (short, int, long, bool, ...).
I came up with following solution:
#include<sstream>
#include<exception>
#include<string>
#include<type_traits>
using namespace std;
template<typename T>
T toIntegralType(const string &str) {
static_assert(is_integral<T>::value, "Integral type required.");
T ret;
stringstream ss(str);
ss >> ret;
if ( to_string(ret) != str)
throw invalid_argument("Can't convert " + str);
return ret;
}
Here are examples of usage:
string str = "123";
int x = toIntegralType<int>(str); // x = 123
str = "123a";
x = toIntegralType<int>(str); // throws exception, because "123a" is not int
str = "1";
bool y = toIntegralType<bool>(str); // y is true
str = "0";
y = toIntegralType<bool>(str); // y is false
str = "00";
y = toIntegralType<bool>(str); // throws exception
Why not just use stringstream output operator to convert a string into an integral type?
Here is the answer:
Let's say a string contains a value that exceeds the limit for intended integral type. For examle, on Wndows 64 max int is 2147483647.
Let's assign to a string a value max int + 1: string str = "2147483648".
Now, when converting the string to an int:
stringstream ss(str);
int x;
ss >> x;
x becomes 2147483647, what is definitely an error: string "2147483648" was not supposed to be converted to the int 2147483647. The provided function toIntegralType spots such errors and throws exception.
In Windows, you could use:
const std::wstring hex = L"0x13";
const std::wstring dec = L"19";
int ret;
if (StrToIntEx(hex.c_str(), STIF_SUPPORT_HEX, &ret)) {
std::cout << ret << "\n";
}
if (StrToIntEx(dec.c_str(), STIF_SUPPORT_HEX, &ret)) {
std::cout << ret << "\n";
}
strtol,stringstream need to specify the base if you need to interpret hexdecimal.
I know this question is really old but I think there's a better way of doing this
#include <string>
#include <sstream>
bool string_to_int(std::string value, int * result) {
std::stringstream stream1, stream2;
std::string stringednumber;
int tempnumber;
stream1 << value;
stream1 >> tempnumber;
stream2 << tempnumber;
stream2 >> stringednumber;
if (!value.compare(stringednumber)) {
*result = tempnumber;
return true;
}
else return false;
}
If I wrote the code right, this will return a boolean value that tells you if the string was a valid number, if false, it wasn't a number, if true it was a number and that number is now result, you would call this this way:
std::string input;
std::cin >> input;
bool worked = string_to_int(input, &result);
You can use std::stringstream, here's an example:
#include <iostream>
#include <sstream>
using namespace std;
string r;
int main() {
cin >> r;
stringstream tmp(r);
int s;
tmp >> s;
cout << s;
return 0;
}
atoi is a built-in function that converts a string to an integer, assuming that the string begins with an integer representation.
http://www.cplusplus.com/reference/clibrary/cstdlib/atoi/
From http://www.cplusplus.com/reference/string/stoi/
// stoi example
#include <iostream> // std::cout
#include <string> // std::string, std::stoi
int main ()
{
std::string str_dec = "2001, A Space Odyssey";
std::string str_hex = "40c3";
std::string str_bin = "-10010110001";
std::string str_auto = "0x7f";
std::string::size_type sz; // alias of size_t
int i_dec = std::stoi (str_dec,&sz);
int i_hex = std::stoi (str_hex,nullptr,16);
int i_bin = std::stoi (str_bin,nullptr,2);
int i_auto = std::stoi (str_auto,nullptr,0);
std::cout << str_dec << ": " << i_dec << " and [" << str_dec.substr(sz) << "]\n";
std::cout << str_hex << ": " << i_hex << '\n';
std::cout << str_bin << ": " << i_bin << '\n';
std::cout << str_auto << ": " << i_auto << '\n';
return 0;
}
Output:
2001, A Space Odyssey: 2001 and [, A Space Odyssey]
40c3: 16579
-10010110001: -1201
0x7f: 127
My Code:
#include <iostream>
using namespace std;
int main()
{
string s="32"; //String
int n=stoi(s); //Convert to int
cout << n + 1 << endl;
return 0;
}
ll toll(string a){
ll ret=0;
bool minus=false;
for(auto i:a){
if(i=='-'){ minus=true; continue; }
ret*=10;
ret+=(i-'0');
} if(minus) ret*=-1;
return ret;
# ll is defined as, #define ll long long int
# usage: ll a = toll(string("-1234"));
}
To convert from string representation to integer value, we can use std::stringstream.
if the value converted is out of range for integer data type, it returns INT_MIN or INT_MAX.
Also if the string value can’t be represented as an valid int data type, then 0 is returned.
#include
#include
#include
int main() {
std::string x = "50";
int y;
std::istringstream(x) >> y;
std::cout << y << '\n';
return 0;
}
Output:
50
As per the above output, we can see it converted from string numbers to integer number.
Source and more at string to int c++
int stringToInt(std::string value) {
if(value.length() == 0 ) return 0; //tu zmiana..
if (value.find( std::string("NULL") ) != std::string::npos) {
return 0;
}
if (value.find( std::string("null") ) != std::string::npos) {
return 0;
}
int i;
std::stringstream stream1;
stream1.clear();
stream1.str(value);
stream1 >> i;
return i;
};
there is another easy way : suppose you have a character like c='4' therefore you can do one of these steps :
1st : int q
q=(int) c ; (q is now 52 in ascii table ) .
q=q-48; remember that adding 48 to digits is their ascii code .
the second way :
q=c-'0'; the same , character '0' means 48
One line version: long n = strtol(s.c_str(), NULL, base); .
(s is the string, and base is an int such as 2, 8, 10, 16.)
You can refer to this link for more details of strtol.
The core idea is to use strtol function, which is included in cstdlib.
Since strtol only handles with char array, we need to convert string to char array. You can refer to this link.
An example:
#include <iostream>
#include <string> // string type
#include <bitset> // bitset type used in the output
int main(){
s = "1111000001011010";
long t = strtol(s.c_str(), NULL, 2); // 2 is the base which parse the string
cout << s << endl;
cout << t << endl;
cout << hex << t << endl;
cout << bitset<16> (t) << endl;
return 0;
}
which will output:
1111000001011010
61530
f05a
1111000001011010
I think that converting from int to std::string or vice versa needs some special functions like std::stoi()
but if you need to convert a double into a string use to_string() (NOT C#. C# is .ToString() not to_string())
If you wot hard code :)
bool strCanBeInt(std::string string){
for (char n : string) {
if (n != '0' && n != '1' && n != '2' && n != '3' && n != '4' && n != '5'
&& n != '6' && n != '7' && n != '8' && n != '9') {
return false;
}
}
return true;
}
int strToInt(std::string string) {
int integer = 0;
int numInt;
for (char n : string) {
if(n == '0') numInt = 0;
if(n == '1') numInt = 1;
if(n == '2') numInt = 2;
if(n == '3') numInt = 3;
if(n == '4') numInt = 4;
if(n == '5') numInt = 5;
if(n == '6') numInt = 6;
if(n == '7') numInt = 7;
if(n == '8') numInt = 8;
if(n == '9') numInt = 9;
if (integer){
integer *= 10;
}
integer += numInt;
}
return integer;
}

How do I remove repeated words from a string and only show it once with their wordcount

Basically, I have to show each word with their count but repeated words show up again in my program.
How do I remove them by using loops or should I use 2d arrays to store both the word and count?
#include <iostream>
#include <stdio.h>
#include <iomanip>
#include <cstring>
#include <conio.h>
#include <time.h>
using namespace std;
char* getstring();
void xyz(char*);
void tokenizing(char*);
int main()
{
char* pa = getstring();
xyz(pa);
tokenizing(pa);
_getch();
}
char* getstring()
{
static char pa[100];
cout << "Enter a paragraph: " << endl;
cin.getline(pa, 1000, '#');
return pa;
}
void xyz(char* pa)
{
cout << pa << endl;
}
void tokenizing(char* pa)
{
char sepa[] = " ,.\n\t";
char* token;
char* nexttoken;
int size = strlen(pa);
token = strtok_s(pa, sepa, &nexttoken);
while (token != NULL) {
int wordcount = 0;
if (token != NULL) {
int sizex = strlen(token);
//char** fin;
int j;
for (int i = 0; i <= size; i++) {
for (j = 0; j < sizex; j++) {
if (pa[i + j] != token[j]) {
break;
}
}
if (j == sizex) {
wordcount++;
}
}
//for (int w = 0; w < size; w++)
//fin[w] = token;
//cout << fin[w];
cout << token;
cout << " " << wordcount << "\n";
}
token = strtok_s(NULL, sepa, &nexttoken);
}
}
This is the output I get:
I want to show, for example, the word "i" once with its count of 5, and then not show it again.
First of all, since you are using c++, I would recommend you to split text in c++ way(some examples are here), and store every word in map or unordered_map. Example of my realization you can find here
But if you don't want to rewrite your code, you can simply add a variable that will indicate whether a copy of the word was found before or after the word position. If a copy was not found in front, then print your word
This post gives an example to save each word from your 'strtok' function into a vector of string. Then, use string.compare to have each word compared with word[0]. Those indexes match with word[0] are marked in an int array 'used'. The count of match equals to the number marks in the array used ('nused'). Those words of marked are then removed from the vector, and the remaining carries on to the next comparing process. The program ends when no word remained.
You may write a word comparing function to replace 'str.compare(str2)', if you prefer not to use std::vector and std::string.
#include <iostream>
#include <string>
#include <vector>
#include<iomanip>
#include<cstring>
using namespace std;
char* getstring();
void xyz(char*);
void tokenizing(char*);
int main()
{
char* pa = getstring();
xyz(pa);
tokenizing(pa);
}
char* getstring()
{
static char pa[100] = "this is a test and is a test and is test.";
return pa;
}
void xyz(char* pa)
{
cout << pa << endl;
}
void tokenizing(char* pa)
{
char sepa[] = " ,.\n\t";
char* token;
char* nexttoken;
std::vector<std::string> word;
int used[64];
std::string tok;
int nword = 0, nsize, nused;
int size = strlen(pa);
token = strtok_s(pa, sepa, &nexttoken);
while (token)
{
word.push_back(token);
++nword;
token = strtok_s(NULL, sepa, &nexttoken);
}
for (int i = 0; i<nword; i++) std::cout << word[i] << std::endl;
std::cout << "total " << nword << " words.\n" << std::endl;
nsize = nword;
while (nsize > 0)
{
nused = 0;
tok = word[0] ;
used[nused++] = 0;
for (int i=1; i<nsize; i++)
{
if ( tok.compare(word[i]) == 0 )
{
used[nused++] = i; }
}
std::cout << tok << " : " << nused << std::endl;
for (int i=nused-1; i>=0; --i)
{
for (int j=used[i]; j<(nsize+i-nused); j++) word[j] = word[j+1];
}
nsize -= nused;
}
}
Notice that the removal of used words has to do in backward order. If you do it in sequential order, the marked indexes in the 'used' array will need to be changed. A running test:
$ ./a.out
this is a test and is a test and is test.
this
is
a
test
and
is
a
test
and
is
test
total 11 words.
this : 1
is : 3
a : 2
test : 3
and : 2
I read your last comment.
But I am very sorry, I do not know C. So, I will answer in C++.
But anyway, I will answer with the C++ standard approach. That is usually only 10 lines of code . . .
#include <iostream>
#include <algorithm>
#include <map>
#include <string>
#include <regex>
// Regex Helpers
// Regex to find a word
static const std::regex reWord{ R"(\w+)" };
// Result of search for one word in the string
static std::smatch smWord;
int main() {
std::cout << "\nPlease enter text: \n";
if (std::string line; std::getline(std::cin, line)) {
// Words and its appearance count
std::map<std::string, int> words{};
// Count the words
for (std::string s{ line }; std::regex_search(s, smWord, reWord); s = smWord.suffix())
words[smWord[0]]++;
// Show result
for (const auto& [word, count] : words) std::cout << word << "\t\t--> " << count << '\n';
}
return 0;
}

Extract integer from a string

I have string like "y.x-name', where y and x are number ranging from 0 to 100. From this string, what would be the best method to extract 'x' into an integer variable in C++.
You could split the string by . and convert it to integer type directly. The second number in while loop is the one you want, see sample code:
template<typename T>
T stringToDecimal(const string& s)
{
T t = T();
std::stringstream ss(s);
ss >> t;
return t;
}
int func()
{
string s("100.3-name");
std::vector<int> v;
std::stringstream ss(s);
string line;
while(std::getline(ss, line, '.'))
{
v.push_back(stringToDecimal<int>(line));
}
std::cout << v.back() << std::endl;
}
It will output: 3
It seem that this thread has a problem similar to you, it might help ;)
Simple string parsing with C++
You can achieve it with boost::lexical_cast, which utilizes streams like in billz' answer:
Pseudo code would look like this (indices might be wrong in that example):
std::string yxString = "56.74-name";
size_t xStart = yxString.find(".") + 1;
size_t xLength = yxString.find("-") - xStart;
int x = boost::lexical_cast<int>( yxString + xStart, xLength );
Parsing errors can be handled via exceptions that are thrown by lexical_cast.
For more flexible / powerful text matching I suggest boost::regex.
Use two calls to unsigned long strtoul( const char *str, char **str_end, int base ), e.g:
#include <cstdlib>
#include <iostream>
using namespace std;
int main(){
char const * s = "1.99-name";
char *endp;
unsigned long l1 = strtoul(s,&endp,10);
if (endp == s || *endp != '.') {
cerr << "Bad parse" << endl;
return EXIT_FAILURE;
}
s = endp + 1;
unsigned long l2 = strtoul(s,&endp,10);
if (endp == s || *endp != '-') {
cerr << "Bad parse" << endl;
return EXIT_FAILURE;
}
cout << "num 1 = " << l1 << "; num 2 = " << l2 << endl;
return EXIT_FAILURE;
}

Parse a string into two double

I have a string that is in the format #########s###.##
where #### is just a few numbers, and the second piece is usually a decimal, but not always.
I need to break the two number pieces apart, and set them as two doubles(or some other valid number type.
I can only use standard methods for this, as the server it's being run on only has standard modules.
I can currently grab the second piece using find and substr, but can't figure out how to get the first piece. I still haven't done anything that changes the second piece into a numerical type, but hopefully that is much easier.
here's what I have:
string symbol,pieces;
fin >> pieces; //pieces is a string of the type i mentioned #####s###.##
unsigned pos;
pos = pieces.find("s");
string capitals = pieces.substr(pos+1);
cout << "Price of stock " << symbol << " is " << capitals << endl;
istringstream makes it easy.
#include <iostream>
#include <sstream>
#include <string>
int main(int argc, char* argv[]) {
std::string input("123456789s123.45");
std::istringstream output(input);
double part1;
double part2;
output >> part1;
char c;
// Throw away the "s"
output >> c;
output >> part2;
std::cout << part1 << ", " << part2 << std::endl;
return 0;
}
You can specify a count along with an offset when calling substr:
string first = pieces.substr(0, pos);
string second = pieces.substr(pos + 1);
You can do the same thing as you did for the second part:
unsigned pos;
pos = pieces.find("s");
string firstPart = pieces.substr(0,pos);
This code will split the string as you desire and convert them to double, it could easily be changed to convert to float as well:
#include <iostream>
#include <sstream>
#include <string>
#include <stdexcept>
class BadConversion : public std::runtime_error {
public:
BadConversion(std::string const& s)
: std::runtime_error(s)
{ }
};
inline double convertToDouble(std::string const& s,
bool failIfLeftoverChars = true)
{
std::istringstream i(s);
double x;
char c;
if (!(i >> x) || (failIfLeftoverChars && i.get(c)))
throw BadConversion("convertToDouble(\"" + s + "\")");
return x;
}
int main()
{
std::string symbol,pieces;
std::cin >> pieces; //pieces is a string of the type i mentioned #####s###.##
unsigned pos;
pos = pieces.find("s");
std::string first = pieces.substr(0, pos);
std::string second = pieces.substr(pos + 1);
std::cout << "first: " << first << " second " << second << std::endl;
double d1 = convertToDouble(first), d2 = convertToDouble(second) ;
std::cout << d1 << " " << d2 << std::endl ;
}
Just for reference, I took the conversion code from one of my previous answers.
Grabbing the first piece is easy:
string firstpiece = pieces.substr(0, pos);
As for converting to numeric types, I find sscanf() to be particularly useful for that:
#include <cstdio>
std::string pieces;
fin >> pieces; //pieces is a string of the type i mentioned #####s###.##
double firstpiece = 0.0, capitals = 0.0;
std::sscanf(pieces.c_str() "%lfs%lf", &firstpiece, &capitals);
...
Some poeple will complain that this is not C++-y but this is valid C++
char * in = "1234s23.93";
char * endptr;
double d1 = strtod(in,&endptr);
in = endptr + 1;
double d2 = strtod(in, &endptr);

How can I convert a std::string to int?

I want to convert a string to an int and I don't mean ASCII codes.
For a quick run-down, we are passed in an equation as a string. We are to break it down, format it correctly and solve the linear equations. Now, in saying that, I'm not able to convert a string to an int.
I know that the string will be in either the format (-5) or (25) etc. so it's definitely an int. But how do we extract that from a string?
One way I was thinking is running a for/while loop through the string, check for a digit, extract all the digits after that and then look to see if there was a leading '-', if there is, multiply the int by -1.
It seems a bit over complicated for such a small problem though. Any ideas?
In C++11 there are some nice new convert functions from std::string to a number type.
So instead of
atoi( str.c_str() )
you can use
std::stoi( str )
where str is your number as std::string.
There are version for all flavours of numbers:
long stol(string), float stof(string), double stod(string),...
see http://en.cppreference.com/w/cpp/string/basic_string/stol
The possible options are described below:
1. sscanf()
#include <cstdio>
#include <string>
int i;
float f;
double d;
std::string str;
// string -> integer
if(sscanf(str.c_str(), "%d", &i) != 1)
// error management
// string -> float
if(sscanf(str.c_str(), "%f", &f) != 1)
// error management
// string -> double
if(sscanf(str.c_str(), "%lf", &d) != 1)
// error management
This is an error (also shown by cppcheck) because "scanf without field width limits can crash with huge input data on some versions of libc" (see here, and here).
2. std::sto()*
#include <iostream>
#include <string>
int i;
float f;
double d;
std::string str;
try {
// string -> integer
int i = std::stoi(str);
// string -> float
float f = std::stof(str);
// string -> double
double d = std::stod(str);
} catch (...) {
// error management
}
This solution is short and elegant, but it is available only on on C++11 compliant compilers.
3. sstreams
#include <string>
#include <sstream>
int i;
float f;
double d;
std::string str;
// string -> integer
std::istringstream ( str ) >> i;
// string -> float
std::istringstream ( str ) >> f;
// string -> double
std::istringstream ( str ) >> d;
// error management ??
However, with this solution is hard to distinguish between bad input (see here).
4. Boost's lexical_cast
#include <boost/lexical_cast.hpp>
#include <string>
std::string str;
try {
int i = boost::lexical_cast<int>( str.c_str());
float f = boost::lexical_cast<int>( str.c_str());
double d = boost::lexical_cast<int>( str.c_str());
} catch( boost::bad_lexical_cast const& ) {
// Error management
}
However, this is just a wrapper of sstream, and the documentation suggests to use sstream for better error management (see here).
5. strto()*
This solution is very long, due to error management, and it is described here. Since no function returns a plain int, a conversion is needed in case of integer (see here for how this conversion can be achieved).
6. Qt
#include <QString>
#include <string>
bool ok;
std::string;
int i = QString::fromStdString(str).toInt(&ok);
if (!ok)
// Error management
float f = QString::fromStdString(str).toFloat(&ok);
if (!ok)
// Error management
double d = QString::fromStdString(str).toDouble(&ok);
if (!ok)
// Error management
Conclusions
Summing up, the best solution is C++11 std::stoi() or, as a second option, the use of Qt libraries. All other solutions are discouraged or buggy.
std::istringstream ss(thestring);
ss >> thevalue;
To be fully correct you'll want to check the error flags.
use the atoi function to convert the string to an integer:
string a = "25";
int b = atoi(a.c_str());
http://www.cplusplus.com/reference/clibrary/cstdlib/atoi/
To be more exhaustive (and as it has been requested in comments), I add the solution given by C++17 using std::from_chars.
std::string str = "10";
int number;
std::from_chars(str.data(), str.data()+str.size(), number);
If you want to check whether the conversion was successful:
std::string str = "10";
int number;
auto [ptr, ec] = std::from_chars(str.data(), str.data()+str.size(), number);
assert(ec == std::errc{});
// ptr points to chars after read number
Moreover, to compare the performance of all these solutions, see the following quick-bench link: https://quick-bench.com/q/GBzK53Gc-YSWpEA9XskSZLU963Y
(std::from_chars is the fastest and std::istringstream is the slowest)
1. std::stoi
std::string str = "10";
int number = std::stoi(str);
2. string streams
std::string str = "10";
int number;
std::istringstream(str) >> number
3. boost::lexical_cast
#include <boost/lexical_cast.hpp>
std::string str = "10";
int number;
try
{
number = boost::lexical_cast<int>(str);
std::cout << number << std::endl;
}
catch (boost::bad_lexical_cast const &e) // bad input
{
std::cout << "error" << std::endl;
}
4. std::atoi
std::string str = "10";
int number = std::atoi(str.c_str());
5. sscanf()
std::string str = "10";
int number;
if (sscanf(str .c_str(), "%d", &number) == 1)
{
std::cout << number << '\n';
}
else
{
std::cout << "Bad Input";
}
What about Boost.Lexical_cast?
Here is their example:
The following example treats command line arguments as a sequence of numeric data:
int main(int argc, char * argv[])
{
using boost::lexical_cast;
using boost::bad_lexical_cast;
std::vector<short> args;
while(*++argv)
{
try
{
args.push_back(lexical_cast<short>(*argv));
}
catch(bad_lexical_cast &)
{
args.push_back(0);
}
}
...
}
Admittedly, my solution wouldn't work for negative integers, but it will extract all positive integers from input text containing integers. It makes use of numeric_only locale:
int main() {
int num;
std::cin.imbue(std::locale(std::locale(), new numeric_only()));
while ( std::cin >> num)
std::cout << num << std::endl;
return 0;
}
Input text:
the format (-5) or (25) etc... some text.. and then.. 7987...78hjh.hhjg9878
Output integers:
5
25
7987
78
9878
The class numeric_only is defined as:
struct numeric_only: std::ctype<char>
{
numeric_only(): std::ctype<char>(get_table()) {}
static std::ctype_base::mask const* get_table()
{
static std::vector<std::ctype_base::mask>
rc(std::ctype<char>::table_size,std::ctype_base::space);
std::fill(&rc['0'], &rc[':'], std::ctype_base::digit);
return &rc[0];
}
};
Complete online demo : http://ideone.com/dRWSj
In C++11 we can use "stoi" function to convert string into a int
#include <iostream>
#include <string>
using namespace std;
int main()
{
string s1 = "16";
string s2 = "9.49";
string s3 = "1226";
int num1 = stoi(s1);
int num2 = stoi(s2);
int num3 = stoi(s3);
cout << "stoi(\"" << s1 << "\") is " << num1 << '\n';
cout << "stoi(\"" << s2 << "\") is " << num2 << '\n';
cout << "stoi(\"" << s3 << "\") is " << num3 << '\n';
return 0;
}
It's probably a bit of overkill, but
boost::lexical_cast<int>( theString ) should to the job
quite well.
Well, lot of answers, lot of possibilities. What I am missing here is some universal method that converts a string to different C++ integral types (short, int, long, bool, ...).
I came up with following solution:
#include<sstream>
#include<exception>
#include<string>
#include<type_traits>
using namespace std;
template<typename T>
T toIntegralType(const string &str) {
static_assert(is_integral<T>::value, "Integral type required.");
T ret;
stringstream ss(str);
ss >> ret;
if ( to_string(ret) != str)
throw invalid_argument("Can't convert " + str);
return ret;
}
Here are examples of usage:
string str = "123";
int x = toIntegralType<int>(str); // x = 123
str = "123a";
x = toIntegralType<int>(str); // throws exception, because "123a" is not int
str = "1";
bool y = toIntegralType<bool>(str); // y is true
str = "0";
y = toIntegralType<bool>(str); // y is false
str = "00";
y = toIntegralType<bool>(str); // throws exception
Why not just use stringstream output operator to convert a string into an integral type?
Here is the answer:
Let's say a string contains a value that exceeds the limit for intended integral type. For examle, on Wndows 64 max int is 2147483647.
Let's assign to a string a value max int + 1: string str = "2147483648".
Now, when converting the string to an int:
stringstream ss(str);
int x;
ss >> x;
x becomes 2147483647, what is definitely an error: string "2147483648" was not supposed to be converted to the int 2147483647. The provided function toIntegralType spots such errors and throws exception.
In Windows, you could use:
const std::wstring hex = L"0x13";
const std::wstring dec = L"19";
int ret;
if (StrToIntEx(hex.c_str(), STIF_SUPPORT_HEX, &ret)) {
std::cout << ret << "\n";
}
if (StrToIntEx(dec.c_str(), STIF_SUPPORT_HEX, &ret)) {
std::cout << ret << "\n";
}
strtol,stringstream need to specify the base if you need to interpret hexdecimal.
I know this question is really old but I think there's a better way of doing this
#include <string>
#include <sstream>
bool string_to_int(std::string value, int * result) {
std::stringstream stream1, stream2;
std::string stringednumber;
int tempnumber;
stream1 << value;
stream1 >> tempnumber;
stream2 << tempnumber;
stream2 >> stringednumber;
if (!value.compare(stringednumber)) {
*result = tempnumber;
return true;
}
else return false;
}
If I wrote the code right, this will return a boolean value that tells you if the string was a valid number, if false, it wasn't a number, if true it was a number and that number is now result, you would call this this way:
std::string input;
std::cin >> input;
bool worked = string_to_int(input, &result);
You can use std::stringstream, here's an example:
#include <iostream>
#include <sstream>
using namespace std;
string r;
int main() {
cin >> r;
stringstream tmp(r);
int s;
tmp >> s;
cout << s;
return 0;
}
atoi is a built-in function that converts a string to an integer, assuming that the string begins with an integer representation.
http://www.cplusplus.com/reference/clibrary/cstdlib/atoi/
From http://www.cplusplus.com/reference/string/stoi/
// stoi example
#include <iostream> // std::cout
#include <string> // std::string, std::stoi
int main ()
{
std::string str_dec = "2001, A Space Odyssey";
std::string str_hex = "40c3";
std::string str_bin = "-10010110001";
std::string str_auto = "0x7f";
std::string::size_type sz; // alias of size_t
int i_dec = std::stoi (str_dec,&sz);
int i_hex = std::stoi (str_hex,nullptr,16);
int i_bin = std::stoi (str_bin,nullptr,2);
int i_auto = std::stoi (str_auto,nullptr,0);
std::cout << str_dec << ": " << i_dec << " and [" << str_dec.substr(sz) << "]\n";
std::cout << str_hex << ": " << i_hex << '\n';
std::cout << str_bin << ": " << i_bin << '\n';
std::cout << str_auto << ": " << i_auto << '\n';
return 0;
}
Output:
2001, A Space Odyssey: 2001 and [, A Space Odyssey]
40c3: 16579
-10010110001: -1201
0x7f: 127
My Code:
#include <iostream>
using namespace std;
int main()
{
string s="32"; //String
int n=stoi(s); //Convert to int
cout << n + 1 << endl;
return 0;
}
ll toll(string a){
ll ret=0;
bool minus=false;
for(auto i:a){
if(i=='-'){ minus=true; continue; }
ret*=10;
ret+=(i-'0');
} if(minus) ret*=-1;
return ret;
# ll is defined as, #define ll long long int
# usage: ll a = toll(string("-1234"));
}
To convert from string representation to integer value, we can use std::stringstream.
if the value converted is out of range for integer data type, it returns INT_MIN or INT_MAX.
Also if the string value can’t be represented as an valid int data type, then 0 is returned.
#include
#include
#include
int main() {
std::string x = "50";
int y;
std::istringstream(x) >> y;
std::cout << y << '\n';
return 0;
}
Output:
50
As per the above output, we can see it converted from string numbers to integer number.
Source and more at string to int c++
int stringToInt(std::string value) {
if(value.length() == 0 ) return 0; //tu zmiana..
if (value.find( std::string("NULL") ) != std::string::npos) {
return 0;
}
if (value.find( std::string("null") ) != std::string::npos) {
return 0;
}
int i;
std::stringstream stream1;
stream1.clear();
stream1.str(value);
stream1 >> i;
return i;
};
error handling not done
int myatoti(string ip)
{
int ret = 0;
int sign = 1;
if (ip[0] == '-')
{
ip.erase(0, 1);
sign = -1;
}
int p = 0;
for (auto it = ip.rbegin(); it != ip.rend(); it++)
{
int val = *it - 48;
int hun = 1;
for (int k = 0; k < p; k++)
{
hun *= 10;
}
ret += val * hun;
p++;
}
return ret * sign;
}
there is another easy way : suppose you have a character like c='4' therefore you can do one of these steps :
1st : int q
q=(int) c ; (q is now 52 in ascii table ) .
q=q-48; remember that adding 48 to digits is their ascii code .
the second way :
q=c-'0'; the same , character '0' means 48
One line version: long n = strtol(s.c_str(), NULL, base); .
(s is the string, and base is an int such as 2, 8, 10, 16.)
You can refer to this link for more details of strtol.
The core idea is to use strtol function, which is included in cstdlib.
Since strtol only handles with char array, we need to convert string to char array. You can refer to this link.
An example:
#include <iostream>
#include <string> // string type
#include <bitset> // bitset type used in the output
int main(){
s = "1111000001011010";
long t = strtol(s.c_str(), NULL, 2); // 2 is the base which parse the string
cout << s << endl;
cout << t << endl;
cout << hex << t << endl;
cout << bitset<16> (t) << endl;
return 0;
}
which will output:
1111000001011010
61530
f05a
1111000001011010
I think that converting from int to std::string or vice versa needs some special functions like std::stoi()
but if you need to convert a double into a string use to_string() (NOT C#. C# is .ToString() not to_string())
If you wot hard code :)
bool strCanBeInt(std::string string){
for (char n : string) {
if (n != '0' && n != '1' && n != '2' && n != '3' && n != '4' && n != '5'
&& n != '6' && n != '7' && n != '8' && n != '9') {
return false;
}
}
return true;
}
int strToInt(std::string string) {
int integer = 0;
int numInt;
for (char n : string) {
if(n == '0') numInt = 0;
if(n == '1') numInt = 1;
if(n == '2') numInt = 2;
if(n == '3') numInt = 3;
if(n == '4') numInt = 4;
if(n == '5') numInt = 5;
if(n == '6') numInt = 6;
if(n == '7') numInt = 7;
if(n == '8') numInt = 8;
if(n == '9') numInt = 9;
if (integer){
integer *= 10;
}
integer += numInt;
}
return integer;
}