Boolalpha function

Hi guys. I'm new to programming. I have a question. I've seen a book teaching C++ and here's the code they gave:

Code:

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// This programme shows an example of bool data type
#include <iostream>
using namespace std;

int main()
{
    bool excellent = true, good = false;
   
    cout<< "The value of excellent is "<< boolalpha << excellent<< endl;
    cout<< "The value of good is "<< boolalpha<< good;
    cin.get();
}

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It seems that when I didn't add the boolalpha in

Code:

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cout<< "The value of good is "<< boolalpha<< good;

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, it still display the true and false. Why is that so? Can anyone help me?

The boolalpha flag, once set, has effect as long as it is not unset with "noboolalpha".

boolalpha can display true ou false
nonboolalpha display 1 or 0

if you didn't add boolalpha it display true ou false

Quote:

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Originally Posted by anon

The boolalpha flag, once set, has effect as long as it is not unset with "noboolalpha".

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Is that true? So is it just applicable to just a function? For example, if I put boolalpha in main function, can it last until another function prototype?

Assuming cout is a global object, once you set a permanent format flag somewhere, it will take effect from then on everywhere until you reset it.

Quote:

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Originally Posted by anon

Assuming cout is a global object, once you set a permanent format flag somewhere, it will take effect from then on everywhere until you reset it.

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lol...I don't know what you mean...because cout will never be a global object? I'm new to programming, sorry for my newbiness

Quote:

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Originally Posted by junkeat90

lol...I don't know what you mean...because cout will never be a global object? I'm new to programming, sorry for my newbiness

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No, normally, cout _IS_ a global object. You can of course have a local cout, if you do something like:

Code:

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void somefunc(ostream &cout)
{
    cout << "Some text" << endl;
}

int main()
{
  ofstream myoutput("somefile.txt");
  somefunc(myoutput);
  return 0;
}

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[That is not a complete piece of code, but should be sufficient to illustrate the case - and if it makes you feel like "???!", then ignore the whole part of the discussion of cout being a local object and see cout as "always global"]
--
Mats