endl

What's the difference between std::endl and '\n'

由 蓝咒 提交于 2021-02-09 11:10:46
问题 I've read the difference between std::endl and '\n' is that std::endl flushes the buffer and '\n' doesn't. However, as far as I know stdout on linux is line-buffered anyway, so does it mean that std::cout << ... << std::endl is the same as std::cout << ... << '\n' ? 回答1: std::ostream os; os << std::endl; // more concise os << '\n' << std::flush; // more explicit about flushing Those two lines have the exact same effect. The manual flushing is often a waste of time: If the output stream is

endl与 \n ; 网罗的自己捋捋~

由 荒凉一梦 提交于 2020-02-28 16:46:34
endl会把一个'\n'写入流。 区别是，endl除了写'\n'进输出流之外，还调用输出流的flush函数，刷新缓冲区，让数据直接写入文件或者屏幕上。 他们都可以用的，不过如果需要立即显示，比如输出到显示器的场合，最用用endl，如果只是写一个'\n'不一定会立即显示，因为数据可能被存在缓冲区里，没有理解写入设备。 输出流是文件考虑效率的时候，可以直接写'\n'，这样不刷新缓冲区，会更快 首先，endl是一个操作符（Manipulators），但我们必须知道endl是一个什么类型的变量。endl是跟在”<<“运算符后面，故endl应该是一个参数。其实endl是一个函数名， 它是一个"<<"运算符重载函数中的参数，参数类型为函数指针。下面我们看下内部函数实现。 1 ostream& ostream::operator << ( ostream& (*op) (ostream&)) 2 { 3 // call the function passed as parameter with this stream as the argument 4 return (*op) (*this); 5 } 1 std::ostream& std::endl (std::ostream& strm) 2 { 3 // write newline 4 strm.put('\n'); 5 //

Overload handling of std::endl?

由 微笑、不失礼 提交于 2019-12-27 12:17:25
问题 I want to define a class MyStream so that: MyStream myStream; myStream << 1 << 2 << 3 << std::endl << 5 << 6 << std::endl << 7 << 8 << std::endl; gives output [blah]123 [blah]56 [blah]78 Basically, I want a "[blah]" inserted at the front, then inserted after every non terminating std::endl ? The difficulty here is NOT the logic management, but detecting and overloading the handling of std::endl . Is there an elegant way to do this? Thanks! EDIT: I don't need advice on logic management. I need

C++ stream as a parameter when overloading operator<<

由 无人久伴 提交于 2019-12-18 00:33:12
问题 I'm trying to write my own logging class and use it as a stream: logger L; L << "whatever" << std::endl; This is the code I started with: #include <iostream> using namespace std; class logger{ public: template <typename T> friend logger& operator <<(logger& log, const T& value); }; template <typename T> logger& operator <<(logger& log, T const & value) { // Here I'd output the values to a file and stdout, etc. cout << value; return log; } int main(int argc, char *argv[]) { logger L; L <<

Using ofstream* wrapper class with overloaded << operator on endl

由 谁说我不能喝 提交于 2019-12-12 02:22:31
问题 C++ This is an attempt to make a class that mimics the output behavior of using the << operator of an ofstream , as far as being able to use std::endl and write string s is concerned. The class has a single data member, the ofstream pointer. The class has two overloaded << operators, one that takes an std::string and another that takes a pointer to a function, whose argument is an ostream reference and returns an ostream reference. That is the signature of std::endl , according to this.

XCode 4 only displaying cout statements with endl

由 谁说我不能喝 提交于 2019-12-11 06:17:59
问题 I have a really weird issue with my cout statements. I've only tried this out with XCode 4. For instance, if I write, cout << "this works" << endl; cout << "this doesnt"; cout << memorySizeLimit << " blocks of memory available." << endl; I see all three output statements in my debugger console. However, if I change the order to, cout << memorySizeLimit << " blocks of memory available." << endl; cout << "this works" << endl; cout << "this doesn't"; I only see the first two couts. Even stranger

C++ style Logger that supports __LINE__ macro and others

由 ≯℡__Kan透↙ 提交于 2019-12-04 17:14:59
I want to make a Logger that can be used like std::cout , but I want to log some extra data like date, time, __LINE__ , __func__ , and __FILE__ which should be saved to the file automatically. Example ToolLogger log; log << "some data" << std::endl; Expected output [14.11.2015 21:10:12.344 (main.cpp) (main,14): some data Inadequate solution To do this I have to put macros like __LINE__ direct in the line where I call my logger, otherwise the macros won't work correct. I found that I can replace std::endl with my macro that will do this black magic like this: #define __FILENAME__ (strrchr(_

std::endl in a string variable?

由 你离开我真会死。 提交于 2019-12-02 00:12:19
问题 Hi i want to save multiply lines in a string. I got a string logstring and i want to save multiplay error logs which i later can print in a txt file or as a console output. Is there a possibility to use endl to format a string variable? I searched the internet but i only find cout << "" << endl; Now my idea was: std::string logstring; logstring = logstring + "Error Message" + "Number" + "Time + Date"; logstring += endl; Is something like this possible or is there no way to format string

std::endl in a string variable?

由 僤鯓⒐⒋嵵緔 提交于 2019-12-01 20:17:56
Hi i want to save multiply lines in a string. I got a string logstring and i want to save multiplay error logs which i later can print in a txt file or as a console output. Is there a possibility to use endl to format a string variable? I searched the internet but i only find cout << "" << endl; Now my idea was: std::string logstring; logstring = logstring + "Error Message" + "Number" + "Time + Date"; logstring += endl; Is something like this possible or is there no way to format string variables? Later i want to print them to a log.txt file? Is it possible to use a string like this? std:

Difference between “endl” and “\\n” [duplicate]

由 一笑奈何 提交于 2019-11-30 07:22:52
Possible Duplicate: C++: “std::endl” vs “\n” I'm wondering if there is any significant difference between these two ways to print newline : cout << endl; //approach1 cout << "\n"; //approach2 Is there any practical difference? Yes, they're different. "\n" is just a string of length 1 that gets appended to stdout. std::endl , instead, is an object that will cause to append the newline character ( "\n" ) AND to flush stdout buffer. For this reason it will take more processing. 来源： https://stackoverflow.com/questions/4512631/difference-between-endl-and-n
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