Related
I want to reverse a string without the use of a loop. My code with the loop looks like:
#include <iostream>
#include <string>
using namespace std;
string reverseString(string str) {
string changedString;
int strLength = int(str.length() - 1);
for(int i {strLength}; i >= 0; i--) {
changedString.push_back(str.at(i));
}
return changedString;
}
int main() {
string str;
cout << "Enter a string to reverse it:\n" << flush;
cin >> str;
cout << reverseString(str) << flush;
}
Now I need to write a function without the loop. Only the methods of String should be used. Can you help me solving this problem?
It is very simple to write such a function
std::string reverse( const std::string &s )
{
return { s.rbegin(), s.rend() };
}
Here is a demonstrative program
#include <iostream>
#include <string>
std::string reverse( const std::string &s )
{
return { s.rbegin(), s.rend() };
}
int main()
{
std::string s( "Hello World" );
std::cout << s << '\n';
std::cout << reverse( s ) << '\n';
return 0;
}
Its output is
Hello World
dlroW olleH
Well, you can do that using recursion. Here are some links if you aren't aware what recursion is : link1 and link2.
Technically it won't be a loop.
string reverseString(string str, int index, string ans) {
if (index == -1) return ans;
ans += str[index];
return reverseString(str, index - 1, ans);
}
Parameters for this function will be str as it was by default, index = size(str) - 1 and ans ans = "";
reverseString(str, size(str) - 1, "") for example.
If you want your function to take exactly one argument, then you can write wrapper function and the one I wrote will have different name - reverseStringWrapper for example and in reverseString there will be only one line - return reverseStringWrapper(str, size(str) - 1, "");
string reverseStringWrapper(string str, int index, string ans) {
if (index == -1) return ans;
ans += str[index];
return reverseString(str, index - 1, ans);
}
string reverseString(string str) {
return reverseStringWrapper(str, size(str) - 1, "");
}
How was this?
In c, You can use strrev() function to reverse the string(char*)
In c++, you can either use std::reverse() or StringBuilder.reverse()
method to reverse a string.
.
This way you can reverse the char array(char*).
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
// Function to reverse a given character array using std::reverse
void reverse(char *str)
{
std::reverse(str, str + strlen(str));
}
// main function
int main()
{
/* using C string */
char s[] = "Hello World";
reverse(s);
cout << "Reverse of the given string is : " << s;
return 0;
}
This way you can reverse the string.
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
// Function to reverse a given character array using std::reverse
void reverse(char* str)
{
std::reverse(str, str + strlen(str));
}
// main function
int main()
{
/* using C string */
// char s[] = "Techie Delight";
string s = "hello world";
int n = s.length();
// declaring character array
char char_array[n + 1];
// copying the contents of the
// string to char array
strcpy(char_array, s.c_str());
reverse(char_array);
s = char_array;
cout << "Reverse of the given string is : " << s;
return 0;
}
Hope this might Helps:)
I thought this would be really simple, but it's presenting some difficulties. If I have
std::string name = "John";
int age = 21;
How do I combine them to get a single string "John21"?
In alphabetical order:
std::string name = "John";
int age = 21;
std::string result;
// 1. with Boost
result = name + boost::lexical_cast<std::string>(age);
// 2. with C++11
result = name + std::to_string(age);
// 3. with FastFormat.Format
fastformat::fmt(result, "{0}{1}", name, age);
// 4. with FastFormat.Write
fastformat::write(result, name, age);
// 5. with the {fmt} library
result = fmt::format("{}{}", name, age);
// 6. with IOStreams
std::stringstream sstm;
sstm << name << age;
result = sstm.str();
// 7. with itoa
char numstr[21]; // enough to hold all numbers up to 64-bits
result = name + itoa(age, numstr, 10);
// 8. with sprintf
char numstr[21]; // enough to hold all numbers up to 64-bits
sprintf(numstr, "%d", age);
result = name + numstr;
// 9. with STLSoft's integer_to_string
char numstr[21]; // enough to hold all numbers up to 64-bits
result = name + stlsoft::integer_to_string(numstr, 21, age);
// 10. with STLSoft's winstl::int_to_string()
result = name + winstl::int_to_string(age);
// 11. With Poco NumberFormatter
result = name + Poco::NumberFormatter().format(age);
is safe, but slow; requires Boost (header-only); most/all platforms
is safe, requires C++11 (to_string() is already included in #include <string>)
is safe, and fast; requires FastFormat, which must be compiled; most/all platforms
(ditto)
is safe, and fast; requires the {fmt} library, which can either be compiled or used in a header-only mode; most/all platforms
safe, slow, and verbose; requires #include <sstream> (from standard C++)
is brittle (you must supply a large enough buffer), fast, and verbose; itoa() is a non-standard extension, and not guaranteed to be available for all platforms
is brittle (you must supply a large enough buffer), fast, and verbose; requires nothing (is standard C++); all platforms
is brittle (you must supply a large enough buffer), probably the fastest-possible conversion, verbose; requires STLSoft (header-only); most/all platforms
safe-ish (you don't use more than one int_to_string() call in a single statement), fast; requires STLSoft (header-only); Windows-only
is safe, but slow; requires Poco C++ ; most/all platforms
In C++11, you can use std::to_string, e.g.:
auto result = name + std::to_string( age );
If you have Boost, you can convert the integer to a string using boost::lexical_cast<std::string>(age).
Another way is to use stringstreams:
std::stringstream ss;
ss << age;
std::cout << name << ss.str() << std::endl;
A third approach would be to use sprintf or snprintf from the C library.
char buffer[128];
snprintf(buffer, sizeof(buffer), "%s%d", name.c_str(), age);
std::cout << buffer << std::endl;
Other posters suggested using itoa. This is NOT a standard function, so your code will not be portable if you use it. There are compilers that don't support it.
#include <iostream>
#include <sstream>
std::ostringstream o;
o << name << age;
std::cout << o.str();
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
string itos(int i) // convert int to string
{
stringstream s;
s << i;
return s.str();
}
Shamelessly stolen from http://www.research.att.com/~bs/bs_faq2.html.
This is the easiest way:
string s = name + std::to_string(age);
If you have C++11, you can use std::to_string.
Example:
std::string name = "John";
int age = 21;
name += std::to_string(age);
std::cout << name;
Output:
John21
It seems to me that the simplest answer is to use the sprintf function:
sprintf(outString,"%s%d",name,age);
#include <string>
#include <sstream>
using namespace std;
string concatenate(std::string const& name, int i)
{
stringstream s;
s << name << i;
return s.str();
}
#include <sstream>
template <class T>
inline std::string to_string (const T& t)
{
std::stringstream ss;
ss << t;
return ss.str();
}
Then your usage would look something like this
std::string szName = "John";
int numAge = 23;
szName += to_string<int>(numAge);
cout << szName << endl;
Googled [and tested :p ]
This problem can be done in many ways. I will show it in two ways:
Convert the number to string using to_string(i).
Using string streams.
Code:
#include <string>
#include <sstream>
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
int main() {
string name = "John";
int age = 21;
string answer1 = "";
// Method 1). string s1 = to_string(age).
string s1=to_string(age); // Know the integer get converted into string
// where as we know that concatenation can easily be done using '+' in C++
answer1 = name + s1;
cout << answer1 << endl;
// Method 2). Using string streams
ostringstream s2;
s2 << age;
string s3 = s2.str(); // The str() function will convert a number into a string
string answer2 = ""; // For concatenation of strings.
answer2 = name + s3;
cout << answer2 << endl;
return 0;
}
In C++20 you'll be able to do:
auto result = std::format("{}{}", name, age);
In the meantime you can use the {fmt} library, std::format is based on:
auto result = fmt::format("{}{}", name, age);
Disclaimer: I'm the author of the {fmt} library and C++20 std::format.
If you'd like to use + for concatenation of anything which has an output operator, you can provide a template version of operator+:
template <typename L, typename R> std::string operator+(L left, R right) {
std::ostringstream os;
os << left << right;
return os.str();
}
Then you can write your concatenations in a straightforward way:
std::string foo("the answer is ");
int i = 42;
std::string bar(foo + i);
std::cout << bar << std::endl;
Output:
the answer is 42
This isn't the most efficient way, but you don't need the most efficient way unless you're doing a lot of concatenation inside a loop.
If you are using MFC, you can use a CString
CString nameAge = "";
nameAge.Format("%s%d", "John", 21);
Managed C++ also has a
string formatter.
As a one liner: name += std::to_string(age);
The std::ostringstream is a good method, but sometimes this additional trick might get handy transforming the formatting to a one-liner:
#include <sstream>
#define MAKE_STRING(tokens) /****************/ \
static_cast<std::ostringstream&>( \
std::ostringstream().flush() << tokens \
).str() \
/**/
Now you can format strings like this:
int main() {
int i = 123;
std::string message = MAKE_STRING("i = " << i);
std::cout << message << std::endl; // prints: "i = 123"
}
As a Qt-related question was closed in favour of this one, here's how to do it using Qt:
QString string = QString("Some string %1 with an int somewhere").arg(someIntVariable);
string.append(someOtherIntVariable);
The string variable now has someIntVariable's value in place of %1 and someOtherIntVariable's value at the end.
There are more options possible to use to concatenate integer (or other numerric object) with string. It is Boost.Format
#include <boost/format.hpp>
#include <string>
int main()
{
using boost::format;
int age = 22;
std::string str_age = str(format("age is %1%") % age);
}
and Karma from Boost.Spirit (v2)
#include <boost/spirit/include/karma.hpp>
#include <iterator>
#include <string>
int main()
{
using namespace boost::spirit;
int age = 22;
std::string str_age("age is ");
std::back_insert_iterator<std::string> sink(str_age);
karma::generate(sink, int_, age);
return 0;
}
Boost.Spirit Karma claims to be one of the fastest option for integer to string conversion.
std::ostringstream
#include <sstream>
std::ostringstream s;
s << "John " << age;
std::string query(s.str());
std::to_string (C++11)
std::string query("John " + std::to_string(age));
boost::lexical_cast
#include <boost/lexical_cast.hpp>
std::string query("John " + boost::lexical_cast<std::string>(age));
Here is an implementation of how to append an int to a string using the parsing and formatting facets from the IOStreams library.
#include <iostream>
#include <locale>
#include <string>
template <class Facet>
struct erasable_facet : Facet
{
erasable_facet() : Facet(1) { }
~erasable_facet() { }
};
void append_int(std::string& s, int n)
{
erasable_facet<std::num_put<char,
std::back_insert_iterator<std::string>>> facet;
std::ios str(nullptr);
facet.put(std::back_inserter(s), str,
str.fill(), static_cast<unsigned long>(n));
}
int main()
{
std::string str = "ID: ";
int id = 123;
append_int(str, id);
std::cout << str; // ID: 123
}
Common Answer: itoa()
This is bad. itoa is non-standard, as pointed out here.
You can concatenate int to string by using the given below simple trick, but note that this only works when integer is of single digit. Otherwise, add integer digit by digit to that string.
string name = "John";
int age = 5;
char temp = 5 + '0';
name = name + temp;
cout << name << endl;
Output: John5
There is a function I wrote, which takes the int number as the parameter, and convert it to a string literal. This function is dependent on another function that converts a single digit to its char equivalent:
char intToChar(int num)
{
if (num < 10 && num >= 0)
{
return num + 48;
//48 is the number that we add to an integer number to have its character equivalent (see the unsigned ASCII table)
}
else
{
return '*';
}
}
string intToString(int num)
{
int digits = 0, process, single;
string numString;
process = num;
// The following process the number of digits in num
while (process != 0)
{
single = process % 10; // 'single' now holds the rightmost portion of the int
process = (process - single)/10;
// Take out the rightmost number of the int (it's a zero in this portion of the int), then divide it by 10
// The above combination eliminates the rightmost portion of the int
digits ++;
}
process = num;
// Fill the numString with '*' times digits
for (int i = 0; i < digits; i++)
{
numString += '*';
}
for (int i = digits-1; i >= 0; i--)
{
single = process % 10;
numString[i] = intToChar ( single);
process = (process - single) / 10;
}
return numString;
}
In C++ 20 you can have a variadic lambda that does concatenate arbitrary streamable types to a string in a few lines:
auto make_string=[os=std::ostringstream{}](auto&& ...p) mutable
{
(os << ... << std::forward<decltype(p)>(p) );
return std::move(os).str();
};
int main() {
std::cout << make_string("Hello world: ",4,2, " is ", 42.0);
}
see https://godbolt.org/z/dEe9h75eb
using move(os).str() guarantees that the ostringstream object's stringbuffer is empty next time the lambda is called.
You can use the C function itoa() like this:
char buf[3];
itoa(age, buf, 10);
name += buf;
I am brand new to programming so none of this may be right. I was just messing around trying to get the effect that I (finally) achieved. More for practice than anything else, but I wondered if there was another way to do it.
#include "stdafx.h"
#include <conio.h>
#include <iostream>
#include <string>
using namespace std;
int main()
{
string str1="Don't even think about it!";
string str2;
string str3;
string str4 = "";
for (int i = 0; i < str1.length(); i++){
str2 = str1.substr (0, str1.length()-i);
cout << str2 << str4;
for (int x = str2.length() - 1; x >= 0; x--){
str3 = str1[x];
cout << str3;
}
str4 = str4 + " ";
cout << "\n";
}
getch();
main();
return 0;
}
The question I have is this: Is there a way to make the str3, after it is backwards, its own string that I could then justify right instead of adding spaces with str4?
In order to get a reversed string, just pass the reverse-iterators to the constructor of a new string:
#include <iostream>
#include <string>
int main() {
std::string s = "this is a test";
std::string s_reversed(s.rbegin(), s.rend());
std::cout << s << "\n" << s_reversed << "\n";
}
The output is:
$ g++ test.cc && ./a.out
this is a test
tset a si siht
Applied to your problem:
#include <iostream>
#include <string>
int main() {
std::string s = "this is a test";
for (auto i = s.length(); i > 0; i--) {
std::cout << s;
std::cout << std::string(s.rbegin(), s.rend());
std::cout << '\n';
s[i-1] = ' ';
}
}
Note that there also is an algorithm in the standard library to reverse a container, e.g., a string: std::reverse().
For reference:
http://en.cppreference.com/w/cpp/string/basic_string/basic_string
http://en.cppreference.com/w/cpp/string/basic_string/rbegin
http://en.cppreference.com/w/cpp/string/basic_string/rend
http://en.cppreference.com/w/cpp/algorithm/reverse
You could try this:
replacing letters for whitespace and then reversing the string.
Concat the modified string with the reverse of it and print it.
std::string reverse(std::string str)
{
std::reverse(str.begin(), str.end());
return str;
}
int main()
{
std::string str = "Don't even think about it!";
size_t N = str.length();
for (size_t i = 1; i < N; i++)
{
std::cout << str << reverse(str) << std::endl;
str.replace(N-i,1, " ");
}
return 0;
}
First I would rename the variables so it is more clear what you are trying to achieve. Believe me, in two months you will not have a clue what this code actually does. Suggestion:
str1 -> baseString
str2 -> forwardsClippedString
str3 -> backwardsClippedString
str4 -> gapString
Maybe these names are not even good but I think they are better than str1, ..., str4.
Then I would make a separate method of the inner loop. After that make a separate method of the outer for loop - yes, main() shouldn't be complicated at all.
Finally instead of recursively calling main I suggest to do a while-loop. An infinite one if you please (but it would be nice if one key quits the loop then).
I have to concatenate char with int.
Here's my code:
int count = 100;
char* name = NULL;
sprintf((char *)name, "test_%d", count);
printf("%s\n", name);
Nothing printed. What's the problem?
You didn't allocate any memory into which sprintf could copy its result. You might try:
int count = 100;
char name[20];
sprintf(name, "test_%d", count);
printf("%s\n", name);
Or even:
int count = 100;
char *name = malloc(20);
sprintf(name, "test_%d", count);
printf("%s\n", name);
Of course, if your only goal is the print the combined string, you can just do this:
printf("test_%d\n", 100);
If you programm C++ use sstream instead:
stringstream oss;
string str;
int count =100
oss << count;
str=oss.str();
cout << str;
You have to allocate memory for name first. In C, library functions like sprintf won't make it for you.
In fact, I am very surprised that you didn't get a segmentation fault.
A simple workaround would be using char name[5+11+1] for the case of 32-bit int.
I use boost::format for this.
#include <boost/format.hpp>
int count = 100;
std::string name = boost::str( boost::format("test_%1%") % count );
Since the answer is tagged C++, this is probably how you should do it there:
The C++11 way: std::string str = "Hello " + std::to_string(5);
The Boost way: std::string str = "Hello " + boost::lexical_cast<std::string>(5);
#include <iostream>
#include <string>
#include <sstream>
int count = 100;
std::stringstream ss;
ss << "Helloworld";
ss << " ";
ss << count ;
ss << std::endl;
std::string str = ss.str();
std::cout << str;
const char * mystring = str.c_str();
I thought this would be really simple, but it's presenting some difficulties. If I have
std::string name = "John";
int age = 21;
How do I combine them to get a single string "John21"?
In alphabetical order:
std::string name = "John";
int age = 21;
std::string result;
// 1. with Boost
result = name + boost::lexical_cast<std::string>(age);
// 2. with C++11
result = name + std::to_string(age);
// 3. with FastFormat.Format
fastformat::fmt(result, "{0}{1}", name, age);
// 4. with FastFormat.Write
fastformat::write(result, name, age);
// 5. with the {fmt} library
result = fmt::format("{}{}", name, age);
// 6. with IOStreams
std::stringstream sstm;
sstm << name << age;
result = sstm.str();
// 7. with itoa
char numstr[21]; // enough to hold all numbers up to 64-bits
result = name + itoa(age, numstr, 10);
// 8. with sprintf
char numstr[21]; // enough to hold all numbers up to 64-bits
sprintf(numstr, "%d", age);
result = name + numstr;
// 9. with STLSoft's integer_to_string
char numstr[21]; // enough to hold all numbers up to 64-bits
result = name + stlsoft::integer_to_string(numstr, 21, age);
// 10. with STLSoft's winstl::int_to_string()
result = name + winstl::int_to_string(age);
// 11. With Poco NumberFormatter
result = name + Poco::NumberFormatter().format(age);
is safe, but slow; requires Boost (header-only); most/all platforms
is safe, requires C++11 (to_string() is already included in #include <string>)
is safe, and fast; requires FastFormat, which must be compiled; most/all platforms
(ditto)
is safe, and fast; requires the {fmt} library, which can either be compiled or used in a header-only mode; most/all platforms
safe, slow, and verbose; requires #include <sstream> (from standard C++)
is brittle (you must supply a large enough buffer), fast, and verbose; itoa() is a non-standard extension, and not guaranteed to be available for all platforms
is brittle (you must supply a large enough buffer), fast, and verbose; requires nothing (is standard C++); all platforms
is brittle (you must supply a large enough buffer), probably the fastest-possible conversion, verbose; requires STLSoft (header-only); most/all platforms
safe-ish (you don't use more than one int_to_string() call in a single statement), fast; requires STLSoft (header-only); Windows-only
is safe, but slow; requires Poco C++ ; most/all platforms
In C++11, you can use std::to_string, e.g.:
auto result = name + std::to_string( age );
If you have Boost, you can convert the integer to a string using boost::lexical_cast<std::string>(age).
Another way is to use stringstreams:
std::stringstream ss;
ss << age;
std::cout << name << ss.str() << std::endl;
A third approach would be to use sprintf or snprintf from the C library.
char buffer[128];
snprintf(buffer, sizeof(buffer), "%s%d", name.c_str(), age);
std::cout << buffer << std::endl;
Other posters suggested using itoa. This is NOT a standard function, so your code will not be portable if you use it. There are compilers that don't support it.
#include <iostream>
#include <sstream>
std::ostringstream o;
o << name << age;
std::cout << o.str();
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
string itos(int i) // convert int to string
{
stringstream s;
s << i;
return s.str();
}
Shamelessly stolen from http://www.research.att.com/~bs/bs_faq2.html.
This is the easiest way:
string s = name + std::to_string(age);
If you have C++11, you can use std::to_string.
Example:
std::string name = "John";
int age = 21;
name += std::to_string(age);
std::cout << name;
Output:
John21
It seems to me that the simplest answer is to use the sprintf function:
sprintf(outString,"%s%d",name,age);
#include <string>
#include <sstream>
using namespace std;
string concatenate(std::string const& name, int i)
{
stringstream s;
s << name << i;
return s.str();
}
#include <sstream>
template <class T>
inline std::string to_string (const T& t)
{
std::stringstream ss;
ss << t;
return ss.str();
}
Then your usage would look something like this
std::string szName = "John";
int numAge = 23;
szName += to_string<int>(numAge);
cout << szName << endl;
Googled [and tested :p ]
This problem can be done in many ways. I will show it in two ways:
Convert the number to string using to_string(i).
Using string streams.
Code:
#include <string>
#include <sstream>
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
int main() {
string name = "John";
int age = 21;
string answer1 = "";
// Method 1). string s1 = to_string(age).
string s1=to_string(age); // Know the integer get converted into string
// where as we know that concatenation can easily be done using '+' in C++
answer1 = name + s1;
cout << answer1 << endl;
// Method 2). Using string streams
ostringstream s2;
s2 << age;
string s3 = s2.str(); // The str() function will convert a number into a string
string answer2 = ""; // For concatenation of strings.
answer2 = name + s3;
cout << answer2 << endl;
return 0;
}
In C++20 you'll be able to do:
auto result = std::format("{}{}", name, age);
In the meantime you can use the {fmt} library, std::format is based on:
auto result = fmt::format("{}{}", name, age);
Disclaimer: I'm the author of the {fmt} library and C++20 std::format.
If you'd like to use + for concatenation of anything which has an output operator, you can provide a template version of operator+:
template <typename L, typename R> std::string operator+(L left, R right) {
std::ostringstream os;
os << left << right;
return os.str();
}
Then you can write your concatenations in a straightforward way:
std::string foo("the answer is ");
int i = 42;
std::string bar(foo + i);
std::cout << bar << std::endl;
Output:
the answer is 42
This isn't the most efficient way, but you don't need the most efficient way unless you're doing a lot of concatenation inside a loop.
If you are using MFC, you can use a CString
CString nameAge = "";
nameAge.Format("%s%d", "John", 21);
Managed C++ also has a
string formatter.
As a one liner: name += std::to_string(age);
The std::ostringstream is a good method, but sometimes this additional trick might get handy transforming the formatting to a one-liner:
#include <sstream>
#define MAKE_STRING(tokens) /****************/ \
static_cast<std::ostringstream&>( \
std::ostringstream().flush() << tokens \
).str() \
/**/
Now you can format strings like this:
int main() {
int i = 123;
std::string message = MAKE_STRING("i = " << i);
std::cout << message << std::endl; // prints: "i = 123"
}
As a Qt-related question was closed in favour of this one, here's how to do it using Qt:
QString string = QString("Some string %1 with an int somewhere").arg(someIntVariable);
string.append(someOtherIntVariable);
The string variable now has someIntVariable's value in place of %1 and someOtherIntVariable's value at the end.
There are more options possible to use to concatenate integer (or other numerric object) with string. It is Boost.Format
#include <boost/format.hpp>
#include <string>
int main()
{
using boost::format;
int age = 22;
std::string str_age = str(format("age is %1%") % age);
}
and Karma from Boost.Spirit (v2)
#include <boost/spirit/include/karma.hpp>
#include <iterator>
#include <string>
int main()
{
using namespace boost::spirit;
int age = 22;
std::string str_age("age is ");
std::back_insert_iterator<std::string> sink(str_age);
karma::generate(sink, int_, age);
return 0;
}
Boost.Spirit Karma claims to be one of the fastest option for integer to string conversion.
std::ostringstream
#include <sstream>
std::ostringstream s;
s << "John " << age;
std::string query(s.str());
std::to_string (C++11)
std::string query("John " + std::to_string(age));
boost::lexical_cast
#include <boost/lexical_cast.hpp>
std::string query("John " + boost::lexical_cast<std::string>(age));
Here is an implementation of how to append an int to a string using the parsing and formatting facets from the IOStreams library.
#include <iostream>
#include <locale>
#include <string>
template <class Facet>
struct erasable_facet : Facet
{
erasable_facet() : Facet(1) { }
~erasable_facet() { }
};
void append_int(std::string& s, int n)
{
erasable_facet<std::num_put<char,
std::back_insert_iterator<std::string>>> facet;
std::ios str(nullptr);
facet.put(std::back_inserter(s), str,
str.fill(), static_cast<unsigned long>(n));
}
int main()
{
std::string str = "ID: ";
int id = 123;
append_int(str, id);
std::cout << str; // ID: 123
}
Common Answer: itoa()
This is bad. itoa is non-standard, as pointed out here.
You can concatenate int to string by using the given below simple trick, but note that this only works when integer is of single digit. Otherwise, add integer digit by digit to that string.
string name = "John";
int age = 5;
char temp = 5 + '0';
name = name + temp;
cout << name << endl;
Output: John5
There is a function I wrote, which takes the int number as the parameter, and convert it to a string literal. This function is dependent on another function that converts a single digit to its char equivalent:
char intToChar(int num)
{
if (num < 10 && num >= 0)
{
return num + 48;
//48 is the number that we add to an integer number to have its character equivalent (see the unsigned ASCII table)
}
else
{
return '*';
}
}
string intToString(int num)
{
int digits = 0, process, single;
string numString;
process = num;
// The following process the number of digits in num
while (process != 0)
{
single = process % 10; // 'single' now holds the rightmost portion of the int
process = (process - single)/10;
// Take out the rightmost number of the int (it's a zero in this portion of the int), then divide it by 10
// The above combination eliminates the rightmost portion of the int
digits ++;
}
process = num;
// Fill the numString with '*' times digits
for (int i = 0; i < digits; i++)
{
numString += '*';
}
for (int i = digits-1; i >= 0; i--)
{
single = process % 10;
numString[i] = intToChar ( single);
process = (process - single) / 10;
}
return numString;
}
In C++ 20 you can have a variadic lambda that does concatenate arbitrary streamable types to a string in a few lines:
auto make_string=[os=std::ostringstream{}](auto&& ...p) mutable
{
(os << ... << std::forward<decltype(p)>(p) );
return std::move(os).str();
};
int main() {
std::cout << make_string("Hello world: ",4,2, " is ", 42.0);
}
see https://godbolt.org/z/dEe9h75eb
using move(os).str() guarantees that the ostringstream object's stringbuffer is empty next time the lambda is called.
You can use the C function itoa() like this:
char buf[3];
itoa(age, buf, 10);
name += buf;