Aprenda Imutabilidade de String

In Python, strings are immutable: once created, the characters they contain cannot be changed in place. Any "modification" you do actually creates a new string. This matters for correctness (no accidental in-place edits) and for performance (how you build larger strings).

No In-Place Edits

You can read characters by index, but you cannot assign to them.


              123456
            
s = "hello"

t = "H" + s[1:]   # Creates a new string: "Hello"
print(t)

s[0] = "H"      # TypeError: strings don't support item assignment

Most string methods return a new string and leave the original unchanged.


              12345678
            
# Cleaning up user input from a registration form
user_name = "  Alice  "

user_name.strip()              # returns "Alice", but the variable still has spaces
print(user_name)               # "  Alice  "

user_name = user_name.strip()  # assign the cleaned value back
print(user_name)               # "Alice" → cleaned and ready to store

Chaining is fine, but remember you're getting a new object each step.


              123456
            
# Normalizing a user's chat message before saving it
user_message = "  hello\n"

clean_message = user_message.strip().upper()   
print(user_message)   # original remains "  hello\n"
print(clean_message)  # "HELLO" → cleaned and ready for processing

"Modifying" By Creating a New String

Use slicing, replace, or concatenation to produce a new value.


              1234567
            
s = "data"

s = s.replace("t", "T")       # "daTa"
print(s)

s = s[:1] + "A" + s[2:]       # "dATa"
print(s)

Efficient Building

Repeated + in large loops can be slow (many intermediate strings). A common pattern is to collect pieces and join once:


              1234
            
# Combining message parts received from a device
response_parts = ["Status:", " ", "200", "\n", "Success"]
response_message = "".join(response_parts)  # "Status: 200\nSuccess"
print(response_message)

Note

You'll learn more joining/formatting patterns in the next chapter.

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Obrigado pelo seu feedback!

Seção 3. Capítulo 4

Pergunte à IA

Pergunte o que quiser ou experimente uma das perguntas sugeridas para iniciar nosso bate-papo

Deslize para mostrar o menu

No In-Place Edits

You can read characters by index, but you cannot assign to them.


              123456
            
s = "hello"

t = "H" + s[1:]   # Creates a new string: "Hello"
print(t)

s[0] = "H"      # TypeError: strings don't support item assignment

Most string methods return a new string and leave the original unchanged.


              12345678
            
# Cleaning up user input from a registration form
user_name = "  Alice  "

user_name.strip()              # returns "Alice", but the variable still has spaces
print(user_name)               # "  Alice  "

user_name = user_name.strip()  # assign the cleaned value back
print(user_name)               # "Alice" → cleaned and ready to store

Chaining is fine, but remember you're getting a new object each step.


              123456
            
# Normalizing a user's chat message before saving it
user_message = "  hello\n"

clean_message = user_message.strip().upper()   
print(user_message)   # original remains "  hello\n"
print(clean_message)  # "HELLO" → cleaned and ready for processing

"Modifying" By Creating a New String

Use slicing, replace, or concatenation to produce a new value.


              1234567
            
s = "data"

s = s.replace("t", "T")       # "daTa"
print(s)

s = s[:1] + "A" + s[2:]       # "dATa"
print(s)

Efficient Building

Repeated + in large loops can be slow (many intermediate strings). A common pattern is to collect pieces and join once:


              1234
            
# Combining message parts received from a device
response_parts = ["Status:", " ", "200", "\n", "Success"]
response_message = "".join(response_parts)  # "Status: 200\nSuccess"
print(response_message)

Note

You'll learn more joining/formatting patterns in the next chapter.

Tudo estava claro?

Obrigado pelo seu feedback!

Seção 3. Capítulo 4

Imutabilidade de String

No In-Place Edits

"Modifying" By Creating a New String

Efficient Building

1. Which line attempts to modify a string in place and will raise an error?

2. What will the code output?

3. You need to assemble a long string from many small pieces. What's recommended?

Imutabilidade de String

No In-Place Edits

"Modifying" By Creating a New String

Efficient Building

1. Which line attempts to modify a string in place and will raise an error?

2. What will the code output?

3. You need to assemble a long string from many small pieces. What's recommended?