Syntax error with basic Python 3 issue

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Syntax error on print with Python 3

I know Python is supposed to be the god of interpreted languages... its simplicity, its lack of redundancy, blah blah blah. But since I'm too used to C, C++, Java, Javascript and Php, I must admit this is annoying. Here's my code:

#!/usr/bin/python3.1
def shit(texture, weight):
    if textura is "green":
        texturaTxt = "Shit is green"
    elif textura is "blue":
        texturaTxt = "Shit is blue"
    elif textura is "purple":
        texturaTxt = "Shit is purple"
    else:
        print "Incorrect color"
        break
    if weight < 20:
        pesoTxt = " and weights so few"
    elif weight >= 20 and peso <=50:
        pesoTxt = " and weights mid"
    elif weight > 50:
        pesoTxt = " and weights a damn lot"
    else:
        print "Incorrect weight"
    return texturaTxt + pesoTxt

c = input("Introduce crappy color: ")
e = input("Introduce stupid weight: ")
r = shit(c, w)
print r

I'm trying to learn Python and what I was trying to achieve was:

...
function shit(texture, weight) {
    string textureTxt = "Shit is ", pesoTxt = " and weights ";
    switch(texture) {
        case "green": textureTxt .= "green as an applee"; break;
        case "blue": textureTxt .= "blue as the sky"; break;
        case "purple": textureTxt .= "purple as Barny"; break;
        default: cout<<"Incorrect texture, try again later" <<endl; exit;
    }
    //etc
}
string color = "";
int weight = 0;
cout<<"Introduce color: ";
cin>>color;
//etc 
cout<<shit(color, weight);
...

But I give up, I can't make it work, it throws my all kind of errors. Hope there is some C++ or php or C to python converter out there.

Thanks for your help

unknown user · Answer 1 · 2012-10-20T14:02:13.147

Python3 no longer supports print as a special form with the arguments following the print keyword (like print x in Python 2.x). Instead print is now a function and requires an argument list like in: print(x). See "Print is a function" in http://docs.python.org/release/3.0.1/whatsnew/3.0.html

In addition the break statement cannot occur outside of a loop. Other than the C switch statement, if does not support break. As there is no fall-through logic for if-statements it is not requred. In order to stop execution of the function and return to the caller use return.

Instead of the is operator, use the equality operator ==. is tests if objects are identical, this is a stricter test than equality. More details can be found here.

In addition you are getting the weight as a string. You might want to convert the string to an integer for comparison with other integer values, using the function int(weight).

Also there are a number of other small errors:

You are assigning the user input for the weight to e while you try to use the undefined variable name w in the function call
The first parameter in the function is called texture but you are using textura in the function body.
You are using peso instead of weight in one instance

Here is a (less offensive) rewrite with those errors eliminated:

def stuff(color, weight):

    color_txt = "Your stuff is "
    if color == "green":
        color_txt += "green as an apple"
    elif color == "blue":
        color_txt += "blue as the sky"
    elif color == "purple":
        color_txt += "purple as an eggplant"
    else:
        print("Invalid color!")
        return

    w = 0
    # converting a string to an integer might
    # throw an exception if the input is invalid
    try:
        w = int(weight)
    except ValueError:
        print("Invalid weight!")
        return

    weight_txt = ""
    if w<20:
        weight_txt = " and weighs so few"
    elif 20 <= w <= 50:
        weight_txt = " and weighs mid"
    elif w > 50:
        weight_txt = " and weighs a lot"
    else:
        print("Invalid weight!")

    return color_txt + weight_txt

c = input("Input color: ")
w = input("Input weight: ")
r = stuff(c, w)
print(r)

Syntax error with basic Python 3 issue

1 Answers1