Im trying to make a number that is positive (already converted into a string) look like "+number" instead of "number" but i can't define it in an if
#include <iostream>
#include <string>
int main()
{
std::string x3s;
int number = 145;
if (number >= 0)
{
x3s = "+" + number;
}
std::cout << x3s << std::endl;
}
Firstly, there is an I/O manipulator std::showpos.
#include <iostream>
int main()
{
int number = 145;
std::cout << std::showpos << number << std::endl;
}
Secondly, you are using the verb "define" incorrectly.
You can use x3s = std::string("+") + std::to_string(number);
Related
I'm starting learning c++ and stepped on this problem, trying to make the following calculation: place + (place / 10)² which if place = 90 it should be 171.
#include <iostream>
#include <iomanip>
#include <cmath>
#include "TestFunction.h"
using namespace std;
int main() {
TestFunction test1 ("John", 90);
test1.getInfo();
}
here is the TestFunction header
#include <iostream>
#include <string>
#include <iomanip>
class TestFunction {
public:
TestFunction(std::string userName, int userPlace) {
name = userName;
place = userPlace;
}
int getPlace() {
return place;
}
int getResult() {
return num1;
}
void getInfo() {
std::cout << "Using getPlace():" << getPlace() << std::endl;
std::cout << "Using getResult(): " << getResult() << std::endl;
std::cout << "Using num1: " << num1 << std::endl;
std::cout << "calculate here: " << getPlace() + pow(getPlace() / 10, 2) << std::endl;
}
private:
std::string name;
int place;
int num1 = place + pow(place / 10, 2);
};
and get this result:
Using getPlace():90
Using getResult(): -2147483648
Using num1: -2147483648
calculate here: 171
I really don't know what I am missing when trying to use getResult() or num1, any advice or simple explanation will be welcome, thanks.
You need to keep track of when your calculations are done.
The init of num1 is done earlier than the initialisation of place, which is something to avoid at all cost.
You could move that calculation into the constructor:
TestFunction(std::string userName, int userPlace) {
name = userName;
place = userPlace;
num1 = place + pow(place / 10, 2);
}
There are other ways, but this is probably most accessable to you.
lexical_cast throws an exception in the following case. Is there a way to use lexical_cast and convert the string to integer.
#include <iostream>
#include "boost/lexical_cast.hpp"
#include <string>
int main()
{
std::string src = "124is";
int iNumber = boost::lexical_cast<int>(src);
std::cout << "After conversion " << iNumber << std::endl;
}
I understand, I can use atoi instead of boost::lexical_cast.
If I'm understanding your requirements correctly it seems as though removing the non-numeric elements from the string first before the lexical_cast will solve your problem. The approach I outline here makes use of the isdigit function which will return true if the given char is a digit from 0 to 9.
#include <iostream>
#include "boost/lexical_cast.hpp"
#include <string>
#include <algorithm>
#include <cctype> //for isdigit
struct is_not_digit{
bool operator()(char a) { return !isdigit(a); }
};
int main()
{
std::string src = "124is";
src.erase(std::remove_if(src.begin(),src.end(),is_not_digit()),src.end());
int iNumber = boost::lexical_cast<int>(src);
std::cout << "After conversion " << iNumber << std::endl;
}
The boost/lexical_cast uses stringstream to convert from string to other types,so you must make sure the string can be converted completely！ or， it will throw the bad_lexical_cast exception,This is an example:
#include <boost/lexical_cast.hpp>
#include <iostream>
#include <string>
#define ERROR_LEXICAL_CAST 1
int main()
{
using boost::lexical_cast;
int a = 0;
double b = 0.0;
std::string s = "";
int e = 0;
try
{
// ----- string --> int
a = lexical_cast<int>("123");//good
b = lexical_cast<double>("123.12");//good
// -----double to string good
s = lexical_cast<std::string>("123456.7");
// ----- bad
e = lexical_cast<int>("abc");
}
catch(boost::bad_lexical_cast& e)
{
// bad lexical cast: source type value could not be interpreted as target
std::cout << e.what() << std::endl;
return ERROR_LEXICAL_CAST;
}
std::cout << a << std::endl; // cout：123
std::cout << b << std::endl; //cout：123.12
std::cout << s << std::endl; //cout：123456.7
return 0;
}
The following code converts an std::string to int and the problem lies with the fact that it cannot discern from a true integer or just a random string. Is there a systematic method for dealing with such a problem?
#include <cstring>
#include <iostream>
#include <sstream>
int main()
{
std::string str = "H";
int int_value;
std::istringstream ss(str);
ss >> int_value;
std::cout<<int_value<<std::endl;
return 0;
}
EDIT: This is the solution that I liked because it is very minimal and elegant! It doesn't work for negative numbers but I only needed positive ones anyways.
#include <cstring>
#include <iostream>
#include <sstream>
int main()
{
std::string str = "2147483647";
int int_value;
std::istringstream ss(str);
if (ss >> int_value)
std::cout << "Hooray!" << std::endl;
std::cout<<int_value<<std::endl;
str = "-2147483648";
std::istringstream negative_ss(str);
if (ss >> int_value)
std::cout << "Hooray!" << std::endl;
std::cout<<int_value<<std::endl;
return 0;
}
You may try to use Boost lexical_cast, it will throw an exception if the cast failed.
int number;
try
{
number = boost::lexical_cast<int>(str);
}
catch(boost::bad_lexical_cast& e)
{
std::cout << str << "isn't an integer number" << std::endl;
}
EDIT
Accorinding to #chris, You may also try to use std::stoi since C++11. It will throw std::invalid_argument exception if no conversion could be performed. You may find more information here: std::stoi
WhozCraig's approach is much nicer and I wanted to expand on it using the approach that the C++ FAQ uses which is as follows:
#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 int convertToInt(std::string const& s,
bool failIfLeftoverChars = true)
{
std::istringstream i(s);
int x;
char c;
if (!(i >> x) || (failIfLeftoverChars && i.get(c)))
throw BadConversion("convertToInt(\"" + s + "\")");
return x;
}
int main()
{
std::cout << convertToInt( "100" ) << std::endl ;
std::cout << convertToInt( "-100" ) << std::endl ;
std::cout << convertToInt( " -100" ) << std::endl ;
std::cout << convertToInt( " -100 ", false ) << std::endl ;
// The next two will fail
std::cout << convertToInt( " -100 ", true ) << std::endl ;
std::cout << convertToInt( "H" ) << std::endl ;
}
This is robust and will know if the conversion fails, you also can optionally choose to fail on left over characters.
/* isdigit example */
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
int main ()
{
char str[]="1776ad";
int year;
if (isdigit(str[0]))
{
year = atoi (str);
printf ("The year that followed %d was %d.\n",year,year+1);
}
return 0;
}
Hello
I know it was asked many times but I hadn't found answer to my specific question.
I want to convert only string that contains only decimal numbers:
For example 256 is OK but 256a is not.
Could it be done without checking the string?
Thanks
The simplest way that makes error checking optional that I can think of is this:
char *endptr;
int x = strtol(str, &endptr, 0);
int error = (*endptr != '\0');
In C++ way, use stringstream:
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
int main()
{
stringstream sstr;
int a = -1;
sstr << 256 << 'a';
sstr >> a;
if (sstr.failbit)
{
cout << "Either no character was extracted, or the character can't represent a proper value." << endl;
}
if (sstr.badbit)
{
cout << "Error on stream.\n";
}
cout << "Extracted number " << a << endl;
return 0;
}
An other way using c++ style : We check the number of digits to know if the string was valid or not :
#include <iostream>
#include <sstream>
#include <string>
#include <cmath>
int main(int argc,char* argv[]) {
std::string a("256");
std::istringstream buffer(a);
int number;
buffer >> number; // OK conversion is done !
// Let's now check if the string was valid !
// Quick way to compute number of digits
size_t num_of_digits = (size_t)floor( log10( abs( number ) ) ) + 1;
if (num_of_digits!=a.length()) {
std::cout << "Not a valid string !" << std::endl;
}
else {
std::cout << "Valid conversion to " << number << std::endl;
}
}
I would like to know what is the easiest way to convert an int to C++ style string and from C++ style string to int.
edit
Thank you very much. When converting form string to int what happens if I pass a char string ? (ex: "abce").
Thanks & Regards,
Mousey
Probably the easiest is to use operator<< and operator>> with a stringstream (you can initialize a stringstream from a string, and use the stream's .str() member to retrieve a string after writing to it.
Boost has a lexical_cast that makes this particularly easy (though hardly a paragon of efficiency). Normal use would be something like int x = lexical_cast<int>(your_string);
You can change "%x" specifier to "%d" or any other format supported by sprintf. Ensure to appropriately adjust the buffer size 'buf'
int main(){
char buf[sizeof(int)*2 + 1];
int x = 0x12345678;
sprintf(buf, "%x", x);
string str(buf);
int y = atoi(str.c_str());
}
EDIT 2:
int main(){
char buf[sizeof(int)*2 + 1];
int x = 42;
sprintf(buf, "%x", x);
string str(buf);
//int y = atoi(str.c_str());
int y = static_cast<int>(strtol(str.c_str(), NULL, 16));
}
This is to convert string to number.
#include "stdafx.h"
#include <iostream>
#include <string>
#include <sstream>
int convert_string_to_number(const std::string& st)
{
std::istringstream stringinfo(st);
int num = 0;
stringinfo >> num;
return num;
}
int main()
{
int number = 0;
std::string number_as_string("425");
number = convert_string_to_number(number_as_string);
std::cout << "The number is " << number << std::endl;
std::cout << "Number of digits are " << number_as_string.length() << std::endl;
}
Like wise, the following is to convert number to string.
#include "stdafx.h"
#include <iostream>
#include <string>
#include <sstream>
std::string convert_number_to_string(const int& number_to_convert)
{
std::ostringstream os;
os << number_to_convert;
return (os.str());
}
int main()
{
int number = 425;
std::string stringafterconversion;
stringafterconversion = convert_number_to_string(number);
std::cout << "After conversion " << stringafterconversion << std::endl;
std::cout << "Number of digits are " << stringafterconversion.length() << std::endl;
}
Use atoi to convert a string to an int. Use a stringstream to convert the other way.