How to get function name?

[hanlei0000000009]

Thanks all!

void fun()
{
string str;

str = ???; //How to get the function("call") name which is call this function?

cout << str << endl; // I want output "call"
}

void call()
{
fun();
}

modified on Wednesday, February 25, 2009 2:19 AM

[_Superman_]

The __FUNCTION__ macro will give you the function name.

«_Superman_» I love work. It gives me something to do between weekends.

[Iain Clarke Warrior Programmer]

I've just learned something today - and it's before 9am! I can now close my mind to further information, as I have to ration these things. Iain.

Codeproject MVP for C++, I can't believe it's for my lounge posts...

[CPallini]

I suppose __FUNCTION__ would expand to "foo" while he's looking for "call" (i.e. the caller function name). :)

If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
[My articles]

[Hamid Taebi]

You can see more info about macro's(LINE,..) on the www.functionx.com and www.cplusplus.com. ;)

Of one Essence is the human race thus has Creation put the base One Limb impacted is sufficient For all Others to feel the Mace (Saadi )

[Code o mat]

Check this[^] out, but note that -as far as i know- these only work with debug information present, but i might be wrong.

> The problem with computers is that they do what you tell them to do and not what you want them to do. < > Life: great graphics, but the gameplay sux. <

[Stuart Dootson]

The easiest way is to require the calling function to pass in its name, which it can get with __FUNCTION__:

void fun(const char* callerName)
{
string str;

str = callerName;

cout << str << endl; // I want output "call"
}

void call()
{
fun(__FUNCTION__);
}

Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p