How to Fix: TypeError: unsupported operand type(s) for +: ‘int’ and ‘str’

This error is the opposite of the common string concatenation error. It means you tried to do math (+), but one of your items was a string. If you see a TypeError unsupported operand type message, it’s likely because the operands involved aren’t of compatible types for the operation you’ve tried.

Problem Code:

age = 25
age_in_five_years = age + "5" # CRASH!
# TypeError: unsupported operand type(s) for +: 'int' and 'str'

Python says: “I know how to add int + int (math). I know how to add str + str (joining). I don’t know how to add int + str.”

# WRONG — "5" is a string, you can't add it to an integer
age = 25
age_in_five_years = age + "5"     # TypeError: unsupported operand type(s) for +: 'int' and 'str'

# WRONG — input() always returns a string, even when the user types a number
age = input("Enter your age: ")   # user types 25 — age is still the string "25"
age_in_five_years = age + 5       # TypeError fires here

# RIGHT — convert the string to int before doing math
age_string = input("Enter your age: ")
age_int = int(age_string)
age_in_five_years = age_int + 5
print(age_in_five_years)          # Output: 30

# RIGHT — production pattern: wrap int() in try/except for unpredictable input
try:
    age = int(input("Enter your age: "))
    print(f"In 5 years you'll be {age + 5}")
except ValueError:
    print("Please enter a whole number.")

The Cause (User Input)

This happens 99% of the time with the input() function. input() always returns a string, even if the user types a number.

age = input("Enter your age: ") # User types '25'
age_in_five_years = age + 5    # CRASH! (age is the string "25")

The Fix: Convert to a Number

You must explicitly convert the string to an integer (int) or float (float) before you can do math with it.

age_string = input("Enter your age: ")
age_int = int(age_string) # Convert the string "25" to the number 25

# Now it works!
age_in_five_years = age_int + 5
print(age_in_five_years)

Pro Tip: Wrap your int() conversion in a try/except block to catch cases where the user types “hello” instead of “25”.

TypeError: unsupported operand type(s) for +: ‘int’ and ‘str’ — The Type Conversion Rule Every Python Developer Needs

TypeError: unsupported operand type(s) for +: ‘int’ and ‘str’ enforces one rule: every value touching a math operator must be a number. Python never converts types silently — you do it explicitly or it crashes.

input() is the single biggest source of this error. It returns a string every time, regardless of what the user types. Wrap every input() that feeds a math operation with int() or float() at the point of assignment — not later in the code where the crash lands.

Three conversion functions cover every case.

int() converts whole number strings: int(“25”) gives 25. It crashes on decimals — int(“25.5”) raises ValueError. Use it for ages, counts, menu choices, and any value that must be a whole number.

float() converts decimal strings: float(“25.5”) gives 25.5. It also handles whole numbers: float(“25”) gives 25.0. Use it for prices, measurements, and any value that might carry a decimal.

int(float()) handles strings that arrive as decimals but need to be integers: int(float(“25.5”)) gives 25. This pattern covers CSV data, API responses, and scraped values where the format isn’t guaranteed.

Wrap the conversion in try/except ValueError on any input you don’t fully control. A user who types “hello”, a CSV cell with “N/A”, or an API field returning null all crash bare int() or float() calls. The try/except pattern costs three lines and eliminates the entire class of conversion crashes permanently.

try:
value = int(float(input(“Enter a number: “).strip()))
except ValueError:
print(“Invalid input — enter a number.”)