Reputation: 198
I want to print the source code of a built-in method. for example, math is a built-in module of python and i want to print the source code of ceil
i know how to print the source code of a custom module using inspect.getsource
need help
I am trying to create a programme where i can call any builtin methods or functions and it will display only the source code of that function or module.
Python has almost everything in builtin library, i want to use these libraries
example:
input: factorial
output:
def factorial(n):
if n == 0:
return 1
else:
return n * factorial(n-1)
import inspect
inspect.getsource(factorial)
doesn't work ... results in Type Error
import inspect
import math
print(inspect.getsource(math.ceil)
TypeError: <built-in function ceil> is not a module, class, method, function, traceback, frame, or code object
thanks in advance :)
Upvotes: 0
Views: 9261
Reputation: 77251
If the source is Python you can do this:
import inspect
import math
try:
print(inspect.getsource(math))
except OSError:
print("Source not available, possibly a C module.")
As other people already commented, many builtin modules are C. If it is the case you will have to dive in the source - fortunately it is not that hard to find, the structure is quite intuitive.
For math.ceil the source code is at line 1073 of Modules/mathmodule.c in cpython:
/*[clinic input]
math.ceil
x as number: object
/
Return the ceiling of x as an Integral.
This is the smallest integer >= x.
[clinic start generated code]*/
static PyObject *
math_ceil(PyObject *module, PyObject *number)
/*[clinic end generated code: output=6c3b8a78bc201c67
input=2725352806399cab]*/
{
_Py_IDENTIFIER(__ceil__);
PyObject *method, *result;
method = _PyObject_LookupSpecial(number, &PyId___ceil__);
if (method == NULL) {
if (PyErr_Occurred())
return NULL;
return math_1_to_int(number, ceil, 0);
}
result = _PyObject_CallNoArg(method);
Py_DECREF(method);
return result;
}
Upvotes: 6