How to Find the Name of the Calling Function

How do I find the name of the calling function?

Here are two options:

You can get a full stacktrace (including the name, module, and offset of the calling function) with recent versions of glibc with the GNU backtrace functions. See my answer here for the details. This is probably the easiest thing.
If that isn't exactly what you're looking for, then you might try libunwind, but it's going to involve more work.

Keep in mind that this isn't something you can know statically (as with PRETTY_FUNCTION); you actually have to walk the stack to figure out what function called you. So this isn't something that's really worth doing in ordinary debug printfs. If you want to do more serious debugging or analysis, though, then this might be useful for you.

How do you find out the caller function in JavaScript?

Note that this solution is deprecated and should no longer be used according to MDN documentation

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Function/caller

function Hello()
{
    alert("caller is " + Hello.caller);
}

Note that this feature is non-standard, from Function.caller:

Non-standard

This feature is non-standard and is not on a standards track. Do not use it on production sites facing the Web: it will not work for every user. There may also be large incompatibilities between implementations and the behavior may change in the future.

The following is the old answer from 2008, which is no longer supported in modern Javascript:

function Hello()
{
    alert("caller is " + arguments.callee.caller.toString());
}

How can we know the caller function's name?

There's nothing you can do only in a.

However, with a simple standard macro trick, you can achieve what you want, IIUC showing the name of the caller.

void a()
{
    /* Your code */
}

void a_special( char const * caller_name )
{
    printf( "a was called from %s", caller_name );
    a();
}

#define a() a_special(__func__)

void b()
{
    a();
}

How to get the name of the calling function inside the called routine?

Thanks @GavinSimpson and @RicardoSporta, but I've figured it out. I'll post an answer in case somebody searches for this in SO.

The name of the function that generated the current call can be retrieved by

deparse(sys.calls()[[sys.nframe()-1]])

This returns a string that contains not only the name of the function, but the entire call object. The name alone can be retrieve by subsetting sys.calls()[[sys.nframe()-1]] before deparsing.

I wanted this because I wrote a function that checks the arguments and halts execution in case of an error. But I wanted this function to (i) dump the environment and (ii) show name of the function one level above in the execution stack. (i) is easy, but I was stuck in (ii).

As for the second question in my post, this is what happens: the expression stop("invalid input") is evaluated in the environment of the test function, but this is not the same thing as if the expression was part of test's body, because the execution stack is different in this 2 scenarios. In the latter case, stop has only test above it, but in the first, it has eval, check and then test upwards. The execution stack, returned by sys.calls() is not the same thing as the enclosing environments. This is what may cause confusion.

How do get the name of a calling function in Javascript?

A function's name is an immutable property of that function, set in the initial function expression.

var notTheName = function thisIsTheName() { ... }

someObj.stillNotTheName = function stillTheName() { ... }

If your function expression does not have a name, there is (unsurprisingly) no way to identify it by name. Assigning a function to a variable does not give it a name; if that were the case, you could not determine the name of an expression assigned to multiple variables.

You should set firstFunction's name property by expressing it as

firstFunction: function firstFunction(){
    this.secondFunction();
}

Also, arguments.callee is deprecated. See Why was the arguments.callee.caller property deprecated in JavaScript? for a very good explanation of the history of arguments.callee.

Getting the name of the calling function in C (without using the preprocessor)

No, there isn't. C isn't a particularly introspective language - things like the name of a function (or pieces of your call stack) simply aren't available at runtime in any sane fashion.

If, for some reason, you are looking for a lot of work for very little benefit, then you can build your programs with debug symbols, and you can write stack-walking and debug symbol lookup code. Then you might be able to find this out on the fly. But be careful, because the symbols you'll see in the debug info will be decorated with type info if you've got any C++ involved.

You've tagged this post gcc, so the relevant details ARE available, however this falls into the 'not recommended' and 'not guaranteed to be the same between compiler versions' territory.

Get the calling function name from the called function

new StackFrame(1, true).GetMethod().Name

Note that in release builds the compiler might inline the method being called, in which case the above code would return the caller of the caller, so to be safe you should decorate your method with:

[MethodImpl(MethodImplOptions.NoInlining)]

How to get name of calling function/method in PHP?

The debug_backtrace() function is the only way to know this, if you're lazy it's one more reason you should code the GetCallingMethodName() yourself. Fight the laziness! :D

Getting the caller function name inside another function in Python?

You can use the inspect module to get the info you want. Its stack method returns a list of frame records.

For Python 2 each frame record is a list. The third element in each record is the caller name. What you want is this:
```
>>> import inspect
>>> def f():
...     print inspect.stack()[1][3]
...
>>> def g():
...     f()
...
>>> g()
g
```

For Python 3.5+, each frame record is a named tuple so you need to replace
```
print inspect.stack()[1][3]
```
with
```
print(inspect.stack()[1].function)
```
on the above code.