How to invoke a Perl script that need args from Python

Question

I'm writing my script in Python, and I want to invoke a Perl script from it.

This is the line I want to call:

perl conv_intel_hex_2_verilog.pl arg1 arg2

while arg1 is the input from the Python script and arg2 is a new file that I'm creating.

So far I have this:

def main(argv):
    file  = open("test.hex", "w")
    input = argv[0];
    subprocess.call(["perl", "conv_intel_hex_2_verilog.pl", input]);
    file.close();

This runs and does nothing.

And when I'm changing the

subprocess.call(["perl", "conv_intel_hex_2_verilog.pl", input]);

to

subprocess.call(["perl", "conv_intel_hex_2_verilog.pl", input ,file]);

it doesn't compile...

Since the perl script just takes an output file name as an argument, you don't need to pre-create the file; just pass "test.hex" as the second argument , not file. — chepner
– chepner, Commented Jun 3, 2015 at 15:50

Rob Kennedy · Accepted Answer · 2015-06-03 15:50:07Z

4

As you've described it, the program you're running expects its second argument to be the name of a file, so that's what you need to give it. In your current code, you're giving it a Python file object, which is not the same as the string name of a file.

subprocess.call(["perl", "conv_intel_hex_2_verilog.pl", input, "test.hex"]);

There is no need to create the file object prior to running that code, either. The other program will surely create it for you, and might even balk if it finds that the file already exists.

answered Jun 3, 2015 at 15:50

Rob Kennedy

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simbabque · Accepted Answer · 2015-06-03 15:49:51Z

3

You cannot just give it a filehandle (or whatever that is called in Python). You are constructing a call to another program, and that program expects a filename, so pass the filename as a string.

subprocess.call(["perl", "conv_intel_hex_2_verilog.pl", input ,"test.hex"]);

You also don't need to open the file first.

answered Jun 3, 2015 at 15:49

simbabque

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