How to use String(format: "%.2f") - Using Swift - Swift Forums
What is format string
What's the difference between String.format() and str. ...
What are String Format's and how would I use them
Hiiii guys, I am a beginner for python programming; I recently learnt format string and is really confused about what is it, and what is the use of it? Thanks for your time for answering my questions
Make sure you use a good IDE so that you have easy access to browse into JDK source code. In Eclipse say, use F3 to open to any declaration. IntelliJ IDEA has similar feature.
If you view the source code for both methods, you can see these calls are identical except that variables this is interchanged with format when comparing the instance vs static method:
public String formatted(Object... args) {
return new Formatter().format(this, args).toString();
}
public static String format(String format, Object... args) {
return new Formatter().format(format, args).toString();
}
So as you've observed: String.format(str, args) is same as str.formatted(args)
Text Blocks were finalized and permanently added in JDK 15 and some additional methods added to support text blocks. One of this methods is:
String::formatted(Object... args)
And I know the functions of two codes below are the same.
As you mentioned in your question both methods do the same job and return same results. The goal of introducing such a method is:
simplify value substitution in the Text Block.
Based on JEP (JDK Enhancement Proposals) 378:
Text blocks do not directly support string interpolation. Interpolation may be considered in a future JEP. In the meantime, the new instance method
String::formattedaids in situations where interpolation might be desired.
As an example you consider this code segment:
String code = String.format("""
public void print(%s o) {
System.out.println(Objects.toString(o));
}
""", type);
We can change it using formatted method as:
String source = """
public void print(%s object) {
System.out.println(Objects.toString(object));
}
""".formatted(type);
Which is cleaner.
Also consider these minor differences between them when using the methods:
public static String format(String format, Object... args)
- Returns a formatted string using the format string and arguments.
- It's a static method of
Stringclass. - It's introduced in Java SE 5 [since 2004].
public String formatted(Object... args)
- Formats using this string as the format string, and the supplied arguments.
- It's an instance method of
Stringclass. - It's introduced in Java SE 15 (JDK 15) [since 2020].
- This method is equivalent to
String.format(this, args)
Use sprintf. (This is NOT safe, but OP asked for an ANSI C answer. See the comments for a safe version.)
int sprintf ( char * str, const char * format, ... );
Write formatted data to string Composes a string with the same text that would be printed if format was used on printf, but instead of being printed, the content is stored as a C string in the buffer pointed by str.
The size of the buffer should be large enough to contain the entire resulting string (see snprintf for a safer version).
A terminating null character is automatically appended after the content.
After the format parameter, the function expects at least as many additional arguments as needed for format.
Parameters:
str
Pointer to a buffer where the resulting C-string is stored. The buffer should be large enough to contain the resulting string.
format
C string that contains a format string that follows the same specifications as format in printf (see printf for details).
... (additional arguments)
Depending on the format string, the function may expect a sequence of additional arguments, each containing a value to be used to replace a format specifier in the format string (or a pointer to a storage location, for n). There should be at least as many of these arguments as the number of values specified in the format specifiers. Additional arguments are ignored by the function.
Example:
// Allocates storage
char *hello_world = (char*)malloc(13 * sizeof(char));
// Prints "Hello world!" on hello_world
sprintf(hello_world, "%s %s!", "Hello", "world");
If you have a POSIX-2008 compliant system (any modern Linux), you can use the safe and convenient asprintf() function: It will malloc() enough memory for you, you don't need to worry about the maximum string size. Use it like this:
char* string;
if(0 > asprintf(&string, "Formatting a number: %d\n", 42)) return error;
log_out(string);
free(string);
This is the minimum effort you can get to construct the string in a secure fashion. The sprintf() code you gave in the question is deeply flawed:
There is no allocated memory behind the pointer. You are writing the string to a random location in memory!
Even if you had written
char s[42];you would be in deep trouble, because you can't know what number to put into the brackets.
Even if you had used the "safe" variant
snprintf(), you would still run the danger that your strings gets truncated. When writing to a log file, that is a relatively minor concern, but it has the potential to cut off precisely the information that would have been useful. Also, it'll cut off the trailing endline character, gluing the next log line to the end of your unsuccessfully written line.If you try to use a combination of
malloc()andsnprintf()to produce correct behavior in all cases, you end up with roughly twice as much code than I have given forasprintf(), and basically reprogram the functionality ofasprintf().
If you are looking at providing a wrapper of log_out() that can take a printf() style parameter list itself, you can use the variant vasprintf() which takes a va_list as an argument. Here is a perfectly safe implementation of such a wrapper:
//Tell gcc that we are defining a printf-style function so that it can do type checking.
//Obviously, this should go into a header.
void log_out_wrapper(const char *format, ...) __attribute__ ((format (printf, 1, 2)));
void log_out_wrapper(const char *format, ...) {
char* string;
va_list args;
va_start(args, format);
if(0 > vasprintf(&string, format, args)) string = NULL; //this is for logging, so failed allocation is not fatal
va_end(args);
if(string) {
log_out(string);
free(string);
} else {
log_out("Error while logging a message: Memory allocation failed.\n");
}
}
Modern C++ makes this super simple.
C++20
C++20 introduces std::format, which allows you to do exactly that. It uses replacement fields similar to those in python:
#include <iostream>
#include <format>
int main() {
std::cout << std::format("Hello {}!\n", "world");
}
Code from cppreference.com, CC BY-SA and GFDL
Check out the compiler support page to see if it's available in your standard library implementation.
As of 2024-08-24, support is available starting from:
- Visual Studio 2019 16.10, released on 2021-05-25 (experimental)
- Visual Studio 2022 17.0, released on 2021-11-08
- Clang 14, with
-fexperimental-libraryflag - Clang 17, full support
- GCC 13.1, released on 2023-04-26 though support is officially "experimental" (and I couldn't find proof this was changed with versions 14 and 15)
- XCode 15.3, released on 2024-03-05
In all other cases, you can resort to the C++11 solution below, or use the {fmt} library, which has the same semantics as std::format.
C++11
With C++11s std::snprintf, this already became a pretty easy and safe task.
#include <memory>
#include <string>
#include <stdexcept>
template<typename ... Args>
std::string string_format( const std::string& format, Args ... args )
{
int size_s = std::snprintf( nullptr, 0, format.c_str(), args ... ) + 1; // Extra space for '\0'
if( size_s <= 0 ){ throw std::runtime_error( "Error during formatting." ); }
auto size = static_cast<size_t>( size_s );
std::unique_ptr<char[]> buf( new char[ size ] );
std::snprintf( buf.get(), size, format.c_str(), args ... );
return std::string( buf.get(), buf.get() + size - 1 ); // We don't want the '\0' inside
}
The code snippet above is licensed under CC0 1.0.
Line by line explanation:
Aim: Write to a char* by using std::snprintf and then convert that to a std::string.
First, we determine the desired length of the char array using a special condition in snprintf. From cppreference.com:
Return value
[...] If the resulting string gets truncated due to buf_size limit, function returns the total number of characters (not including the terminating null-byte) which would have been written, if the limit was not imposed.
This means that the desired size is the number of characters plus one, so that the null-terminator will sit after all other characters and that it can be cut off by the string constructor again. This issue was explained by @alexk7 in the comments.
int size_s = std::snprintf( nullptr, 0, format.c_str(), args ... ) + 1;
snprintf will return a negative number if an error occurred, so we then check whether the formatting worked as desired. Not doing this could lead to silent errors or the allocation of a huge buffer, as pointed out by @ead in the comments.
if( size_s <= 0 ){ throw std::runtime_error( "Error during formatting." ); }
Because we know that size_s can't be negative, we use a static cast to convert it from a signed int to an unsigned size_t. This way, even the most pedantic compiler won't complain about the conversions that would otherwise happen on the next lines.
size_t size = static_cast<size_t>( size_s );
Next, we allocate a new character array and assign it to a std::unique_ptr. This is generally advised, as you won't have to manually delete it again.
Note that this is not a safe way to allocate a unique_ptr with user-defined types as you can not deallocate the memory if the constructor throws an exception!
std::unique_ptr<char[]> buf( new char[ size ] );
In C++14, you could instead use make_unique, which is safe for user-defined types.
auto buf = std::make_unique<char[]>( size );
After that, we can of course just use snprintf for its intended use and write the formatted string to the char[].
std::snprintf( buf.get(), size, format.c_str(), args ... );
Finally, we create and return a new std::string from that, making sure to omit the null-terminator at the end.
return std::string( buf.get(), buf.get() + size - 1 );
You can see an example in action here.
If you also want to use std::string in the argument list, take a look at this gist.
Additional information for Visual Studio users:
As explained in this answer, Microsoft renamed std::snprintf to _snprintf (yes, without std::). MS further set it as deprecated and advises to use _snprintf_s instead, however _snprintf_s won't accept the buffer to be zero or smaller than the formatted output and will not calculate the outputs length if that occurs.
So in order to get rid of the deprecation warnings during compilation, you can insert the following line at the top of the file which contains the use of _snprintf:
#pragma warning(disable : 4996)
Final thoughts
A lot of answers to this question were written before the time of C++11 and use fixed buffer lengths or vargs. Unless you're stuck with old versions of C++, I wouldn't recommend using those solutions. Ideally, go the C++20 way.
Because the C++11 solution in this answer uses templates, it can generate quite a bit of code if it is used a lot. However, unless you're developing for an environment with very limited space for binaries, this won't be a problem and is still a vast improvement over the other solutions in both clarity and security.
If space efficiency is super important, these two solution with vargs and vsnprintf can be useful. DO NOT USE any solutions with fixed buffer lengths, that is just asking for trouble.
You can't do it directly, because you don't have write access to the underlying buffer (until C++11; see Dietrich Epp's comment). You'll have to do it first in a c-string, then copy it into a std::string:
char buff[100];
snprintf(buff, sizeof(buff), "%s", "Hello");
std::string buffAsStdStr = buff;
But I'm not sure why you wouldn't just use a string stream? I'm assuming you have specific reasons to not just do this:
std::ostringstream stringStream;
stringStream << "Hello";
std::string copyOfStr = stringStream.str();