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)
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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.
You can use sprintf to do it, or maybe snprintf if you have it:
char str[ENOUGH];
sprintf(str, "%d", 42);
Where the number of characters (plus terminating char) in the str can be calculated using:
(int)((ceil(log10(num))+1)*sizeof(char))
As pointed out in a comment, itoa() is not a standard, so better use the sprintf() approach suggested in the rival answer!
You can use the itoa() function to convert your integer value to a string.
Here is an example:
int num = 321;
char snum[5];
// Convert 123 to string [buf]
itoa(num, snum, 10);
// Print our string
printf("%s\n", snum);
If you want to output your structure into a file there isn't any need to convert any value beforehand. You can just use the printf format specification to indicate how to output your values and use any of the operators from printf family to output your data.
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"