I never learned about cin.ignore()...

I never really did learn much of anything about buffers and cin.ignore() can anyone enlighten me as to how these work and all? It's my understanding that if you don't use endl to clear the buffer or something like that....a lot of stuff can remain there and be printed out in subsequent output statements?

....Something along those lines?

Hi,

Quote:

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It's my understanding that if you don't use endl to clear the buffer or something like that....a lot of stuff can remain there and be printed out in subsequent output statements?

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'\n' puts a newline at the end of the output, so you can do this:

cout<<"some text\n";

endl tacks on a '\n' to the output and calls flush() if the stream is buffered(is cout buffered?) It doesn't seem like cout is buffered:

Code:

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#include <iostream>
using namespace std;

int main()
{
        cout<<"bye";
        return 0;
}

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displays the text with my compiler. Maybe the compiler flushes the stream before ending the program.

cin.ignore() is the same thing as cin.ignore(1) because 1 is the default value. It ignores one character in the input stream. ignore() actually has two parameters and the second one is the delimiter--ignore will skip the number of characters specified or until it hits the delimiter.

cout may or may not be buffered, cerr is not buffered. The buffer is flushed when the destructor is called(this applies to all streams) cout is also tieed to cin so calling cin >> i calls flush on cout. Thus it's hard to find a situation where you would notice that the buffer has not been flushed.
The only time you might care is if your program dies unexpectedly

Code:

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    std::cout << "aproaching death";
    *(reinterpret_cast<int *>(0)) = 5;

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this may or may not print anything. Adding endl requres that the user will see the output before you do anything else. Most of the time it just wastes cycles.