Java
To convert a String to an int in Java, use Integer.parseInt() for primitive int or Integer.valueOf() for an Integer object. Both methods throw a NumberFormatException if the string is not a valid integer.
Example using
parseInt():String s = "123"; int n = Integer.parseInt(s); System.out.println(n); // Output: 123Handling exceptions: Use try-catch blocks to manage invalid inputs:
try { int n = Integer.parseInt("abc"); } catch (NumberFormatException e) { System.out.println("Invalid number format"); }
C++
In C++, use stoi() (C++11 and later) for std::string, which supports base specification and throws std::invalid_argument or std::out_of_range.
Example:
std::string s = "45"; int n = std::stoi(s); std::cout << n; // Output: 45Supports different bases:
std::stoi("1010", nullptr, 2); // Binary → 10 std::stoi("FF", nullptr, 16); // Hexadecimal → 255
Other methods include atoi() (for C-style strings), stringstream, sscanf(), and manual loop-based conversion.
C#
Use Convert.ToInt32() or int.TryParse() for safe conversion.
Using
Convert.ToInt32():string s = "123"; int n = Convert.ToInt32(s);Using
TryParse()(recommended for error handling):string s = "abc"; if (int.TryParse(s, out int n)) { Console.WriteLine(n); } else { Console.WriteLine("Invalid input"); }
General Notes
Always validate input to avoid runtime exceptions.
For parsing strings with leading/trailing spaces or signs, trim the string first.
Use
try-catchorTryParsevariants in production code to handle invalid inputs gracefully.
String myString = "1234";
int foo = Integer.parseInt(myString);
If you look at the Java documentation you'll notice the "catch" is that this function can throw a NumberFormatException, which you can handle:
int foo;
try {
foo = Integer.parseInt(myString);
}
catch (NumberFormatException e) {
foo = 0;
}
(This treatment defaults a malformed number to 0, but you can do something else if you like.)
Alternatively, you can use an Ints method from the Guava library, which in combination with Java 8's Optional, makes for a powerful and concise way to convert a string into an int:
import com.google.common.primitives.Ints;
int foo = Optional.ofNullable(myString)
.map(Ints::tryParse)
.orElse(0)
Answer from Rob Hruska on Stack OverflowJava - Convert integer to string - Stack Overflow
c - Integer array to string - Code Review Stack Exchange
java - How do I convert from int to String? - Stack Overflow
How does computer converts integer to string?
Videos
String myString = "1234";
int foo = Integer.parseInt(myString);
If you look at the Java documentation you'll notice the "catch" is that this function can throw a NumberFormatException, which you can handle:
int foo;
try {
foo = Integer.parseInt(myString);
}
catch (NumberFormatException e) {
foo = 0;
}
(This treatment defaults a malformed number to 0, but you can do something else if you like.)
Alternatively, you can use an Ints method from the Guava library, which in combination with Java 8's Optional, makes for a powerful and concise way to convert a string into an int:
import com.google.common.primitives.Ints;
int foo = Optional.ofNullable(myString)
.map(Ints::tryParse)
.orElse(0)
For example, here are two ways:
Integer x = Integer.valueOf(str);
// or
int y = Integer.parseInt(str);
There is a slight difference between these methods:
valueOfreturns a new or cached instance ofjava.lang.IntegerparseIntreturns primitiveint.
The same is for all cases: Short.valueOf/parseShort, Long.valueOf/parseLong, etc.
There are multiple ways:
String.valueOf(number)(my preference)"" + number(I don't know how the compiler handles it, perhaps it is as efficient as the above)Integer.toString(number)
Integer class has static method toString() - you can use it:
int i = 1234;
String str = Integer.toString(i);
Returns a String object representing the specified integer. The argument is converted to signed decimal representation and returned as a string, exactly as if the argument and radix 10 were given as arguments to the toString(int, int) method.
Normal ways would be Integer.toString(i) or String.valueOf(i).
The concatenation will work, but it is unconventional and could be a bad smell as it suggests the author doesn't know about the two methods above (what else might they not know?).
Java has special support for the + operator when used with strings (see the documentation) which translates the code you posted into:
StringBuilder sb = new StringBuilder();
sb.append("");
sb.append(i);
String strI = sb.toString();
at compile-time. It's slightly less efficient (sb.append() ends up calling Integer.getChars(), which is what Integer.toString() would've done anyway), but it works.
To answer Grodriguez's comment: ** No, the compiler doesn't optimise out the empty string in this case - look:
simon@lucifer:~$ cat TestClass.java
public class TestClass {
public static void main(String[] args) {
int i = 5;
String strI = "" + i;
}
}
simon@lucifer:~$ javac TestClass.java && javap -c TestClass
Compiled from "TestClass.java"
public class TestClass extends java.lang.Object{
public TestClass();
Code:
0: aload_0
1: invokespecial #1; //Method java/lang/Object."<init>":()V
4: return
public static void main(java.lang.String[]);
Code:
0: iconst_5
1: istore_1
Initialise the StringBuilder:
2: new #2; //class java/lang/StringBuilder
5: dup
6: invokespecial #3; //Method java/lang/StringBuilder."<init>":()V
Append the empty string:
9: ldc #4; //String
11: invokevirtual #5; //Method java/lang/StringBuilder.append:
(Ljava/lang/String;)Ljava/lang/StringBuilder;
Append the integer:
14: iload_1
15: invokevirtual #6; //Method java/lang/StringBuilder.append:
(I)Ljava/lang/StringBuilder;
Extract the final string:
18: invokevirtual #7; //Method java/lang/StringBuilder.toString:
()Ljava/lang/String;
21: astore_2
22: return
}
There's a proposal and ongoing work to change this behaviour, targetted for JDK 9.
It's acceptable, but I've never written anything like that. I'd prefer this:
String strI = Integer.toString(i);
How does computer or programming languages out there converts binary integers into a string? what is the algorithm behind it? Like int 300 to string "300"
Hello, I'm a first year CS student and am learning c++ at my university. I need to build a function that inputs a string from the user, "1234" for example, and then return the integer 1234.
I am not allowed to use the stoi function, and <iostream> and <string> are the only allowed libraries. I'm at a wall, I would appreciate any help! Thanks.
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.