Why the fault is :invalid literal for int () with base 10

Question

print(int("3*(6+4)")+int("2+(4/5)"))

Answer 1

You are trying to parse "3*(6+4)" as an integer. This is actually a string literal which contains chars such as *, (, +, and hence makes the literal invalid to be parsed to integer of base 10.

You might be trying to do:

print(eval("3*(6+4)")+eval("2+(4/5)"))

Answer 2

You can't evaluate the expression with int() because you are trying to convert operators to integers - i.e. while int("6") is 6, int("6 + 4") is not 10.

You can either convert each string number to an integer and then do the math, e.g.
print(int("3") * (int("6") + int("4")) + int("2") + int("4") / int("5"))

or, you can use eval() to evaluate the string expressions, i.e. like this:
print(eval("3*(6+4)")+eval("2+(4/5)"))

Just a word of caution - eval() can open the door to malicious code in some circumstances - see, e.g.:

http://www.geeksforgeeks.org/eval-in-JUMP_LINK__&&__python__&&__JUMP_LINK/

but it does make a fun one-line calculator:
print(eval(input()))

Answer 3

Thank you two