Is there a way to get the function parameters in the callee as the caller put it?

user_1235183 0
Viewed 85

I want to achieve that calling foo(2*3) prints 2*3.

The reason is that I try to create a test framework for querying data files and I want to print the query statement with the assertion result.

I tried to get it work via the inspect module but I could not make it work.

code-inspection python
2 Answers

In general, the answer is "no", since the value received by the function is the result of the expression 2*3, not the expression itself. However, in Python almost anything is possible if you really want it ;-)

For simple cases you could achieve this using the inspect module like this:

#!/usr/bin/env python3

import inspect


def foo(x):
    context = inspect.stack()[1].code_context[0]
    print(context)


def main():
    foo(2 * 3)


if __name__ == '__main__':
    main()

This will print:

    foo(2 * 3)

It would be trivial to get the part 2 * 3 from this string using a regular expression. However, if the function call is not on a single line, or if the line contains multiple statements, this simple trick will not work properly. It might be possible with more advanced coding, but I guess your use case is to simply print the expression in a test report or something like that? If so, this solution might be just good enough.

user_106698750

Because the expression is evaluated before it is passed to the function, it is not possible to print out the un-evaluated expression.

However, there is a possible workaround. You can instead pass the expression as a string and evaluate it inside the function using eval(). As a simple example:

def foo(expr):
    print(expr)
    return(eval(expr))

Please note however that using eval is considered bad practice.

A better solution is to simply pass a string as well as the expression, such as foo(2*3, "2*3").

user_20653890
Related