I have a problem in python. I want to create an if clause from a string that is given as a parameter to a function. It looks something like this:
def function(foo)
if(foo):
print("bar")
test = "5 > 10"
function(test)
I would like to see nothing printed, because 5 > 10 is False. But actually it is printing "bar". Is there a way to get this right without asking if (foo.split(" ")[2] == ">") do something
You could eval the foo argument:
def function(foo):
if eval(foo):
print("bar")
Use eval()
Ex:
def function(foo):
if(eval(foo)):
print("bar")
test = "5 > 10"
function(test)
An "if" statement takes an expression and evaluate it. If the evaluation is 0 for a numerical expression, 'false' for a boolean expression or NULL for strings , it's considered as false expression and the "then" block is not executed. Otherwise, the expression is true.
In your case, a string's evaluation is always true because it's not NULL. the expression 5 > 10 is 'false', but the string "5 > 10" is 'true'.
You should use eval if you want to evaluate the string:
function(eval('5 >10'))
eval()eval(). Even if it's more verbose, it's safer to do theif foo.split(" ")[1] == ">"thing you mentioned in your question.