Python SyntaxError

3SyntaxError: explained step by step

A SyntaxError means Python could not understand the grammar of your code. Unlike most errors, it is raised before the program starts: Python reads the entire file first, and if any line breaks the rules, none of the file runs, not even the lines above the mistake.

To show this safely on the page, the examples below pass code as text to compile(), which checks the grammar without running it. In your own file you would simply see the traceback.

source = 'print("Starting...")\nif score > 50\n    print("pass")'
try:
    compile(source, "main.py", "exec")
except SyntaxError as err:
    print("SyntaxError:", err.msg)
    print("line:", err.lineno, "column:", err.offset)
    print("text:", err.text.strip())

"Starting..." was never printed, because the whole file was rejected first.

How to read a SyntaxError and its caret

  File "main.py", line 2
    if score > 50
                 ^
SyntaxError: expected ':'

1. The last line is the message. Recent Python versions are specific: expected ':', '(' was never closed, unterminated string literal.
2. The line above shows the code, and the ^ caret points to where Python gave up, which is often just after the real mistake.
3. If the reported line looks perfectly fine, check the line before it. An unclosed bracket or a missing quote makes Python swallow the next line, so the complaint arrives late.

Why a missing colon causes SyntaxError

if, elif, else, for, while, def, class and try all end with a colon, which tells Python that an indented block follows. See Python if/else and Python functions for the full shapes.

def show(source):
    try:
        compile(source, "main.py", "exec")
        print("OK")
    except SyntaxError as err:
        print(f"line {err.lineno}: {err.msg}")

show('if score > 50\n    print("pass")')
show('def greet(name)\n    return name')
show('if score > 50:\n    print("pass")')

Unclosed or mismatched brackets in Python

Every (, [ and { needs its partner. Python reports an unclosed bracket at the line where it was opened, even if the code that confused it is lower down.

def show(source):
    try:
        compile(source, "main.py", "exec")
        print("OK")
    except SyntaxError as err:
        print(f"line {err.lineno}: {err.msg}")

show('total = (4 +\n    tax\nprint(total)')
show('x = 1 + 2)')
show('values = [1, 2, 3)')
show('total = (4 +\n    5)')

Unterminated string literal errors

A string that opens with a quote must close with the same quote on the same line. An apostrophe inside single quotes ends the string early; use double quotes around the text, or escape the apostrophe. The same rule applies to f-strings, where an empty {} is an error. More about text in Python strings.

def show(source):
    try:
        compile(source, "main.py", "exec")
        print("OK")
    except SyntaxError as err:
        print(f"line {err.lineno}: {err.msg}")

show("print('It's ready')")
show('print("It\'s ready")')
show('name = "Asha')
show('label = f"Total: {}"')

Missing comma between items

In a list, dictionary or call, forgetting a comma gives invalid syntax. Perhaps you forgot a comma?. Beware the one case where Python does not complain: two string literals side by side are silently joined, so ["red" "blue"] is a list with one item, "redblue". See Python lists.

def show(source):
    try:
        compile(source, "main.py", "exec")
        print("OK")
    except SyntaxError as err:
        print(f"line {err.lineno}: {err.msg}")

show("prices = [4, 5 6]")
print(["red" "blue", "green"])

Using = instead of == in a condition

A single = assigns; == compares. Putting = inside an if or while condition is a SyntaxError, and so is assigning to a literal. See Python operators.

def show(source):
    try:
        compile(source, "main.py", "exec")
        print("OK")
    except SyntaxError as err:
        print(f"line {err.lineno}: {err.msg}")

show("if answer = 5:\n    pass")
show("5 = answer")
show("if answer == 5:\n    pass")

Reserved words and Python 2 code

Keywords such as class, for, def, in and True cannot be variable names. Older tutorials use print "hi" without parentheses, which Python 3 rejects.

def show(source):
    try:
        compile(source, "main.py", "exec")
        print("OK")
    except SyntaxError as err:
        print(f"line {err.lineno}: {err.msg}")

show('class = "Maths"')
show('print "hello"')
show('print("hello")')

If the message instead says IndentationError, the grammar was fine but the spacing was not; see Python IndentationError. If the code parses but a name is missing, you get a NameError, and the basic grammar rules are in Python syntax.