## What's the result of this expression? ```go "\"Hello\\"" + ` \"World\"` ``` 1. "Hello" "World" 2. "Hello" \"World\" *CORRECT* 3. "Hello" `"World"` 4. "\"Hello\" `\"World\"`" > 1. Go doesn't interpret the escape sequences in raw string literals. > 2. That's right. Go interprets `\"` as `"` but it doesn't do so for ` \"World\"`. ## What's the best way to represent the following text in the code? ```xml "Teddy Bear" ``` 1. *CORRECT* ```go ` "Teddy Bear" ` ``` 2. ```go " "Teddy Bear" " ``` 3. ```go " \"Teddy Bear\" " ``` 4. ```go ` \"Teddy Bear\" ` ``` > 2-3. You can't write a string literal like that. It can't be multiple-lines. > 4. You don't need to use escape sequences inside raw string literals. ## What's the result of the following expression? ```go len("lovely") ``` 1. 7 2. 8 3. 6 *CORRECT* 4. 0 > 2. Remember! "a" is 1 char. `a` is also 1 char. ## What's the result of the following expression? ```go len("very") + len(`\"cool\"`) ``` 1. 8 2. 12 *CORRECT* 3. 16 4. 10 > 1. There are also double-quotes, count them as well. > 2. That's right. Go doesn't interpreted \" in raw string literals. > 3. Remember! "very" is 4 characters. `very` is also 4 characters. > 4. Remember! Go doesn't interpreted \" in raw string literals. ## What's the result of the following expression? ```go len("very") + len("\"cool\"") ``` 1. 8 2. 12 3. 16 4. 10 *CORRECT* > 1. There are also double-quotes, count them as well. > 2. Remember! Go interprets escape sequences in string literals. > 4. That's right. Go does interpret \" in a string literal. So, "\"" means ", which is 1 character. ## What's the result of the following expression? ```go len("péripatéticien") ``` **HINT:** é is 2 bytes long. And, the len function counts the bytes not the letters. **USELESS INFORMATION:** "péripatéticien" means "wanderer". 1. 14 2. 16 *CORRECT* 3. 18 4. 20 > 1. Remember! é is 2 bytes long. > 2. An english letter is 1 byte long. However, é is 2 bytes long. So, that makes up 16 bytes. Cool. > 3. You didn't count the double-quotes, did you? ## How can you find the correct length of the characters in this string literal? ```go "péripatéticien" ``` 1. `len(péripatéticien)` 2. `len("péripatéticien")` 3. `utf8.RuneCountInString("péripatéticien")` *CORRECT* 4. `unicode/utf8.RuneCountInString("péripatéticien")` > 1. Where are the double-quotes? > 2. This only finds the bytes in a string value. > 4. You're close. But, the package's name is utf8 not unicode/utf8. ## What's the result of the following expression? ```go utf8.RuneCountInString("péripatéticien") ``` 1. 16 2. 14 *CORRECT* 3. 18 4. 20 > 1. This is its byte count. `RuneCountInString` counts the runes (codepoints) not the bytes. > 2. That's right. `RuneCountInString` returns the number of runes (codepoints) in a string value. ## Which package contains string manipulation functions? 1. string 2. unicode/utf8 3. strings *CORRECT* 4. unicode/strings ## What's the result of this expression? ```go strings.Repeat("*x", 3) + "*" ``` **HINT:** Repeat function repeats the given string. 1. `*x*x*x` 2. `x*x*x` 3. `*x3` 4. `*x*x*x*` *CORRECT* > 1. You're close but you missed the concatenation at the end. > 2. Look closely. > 3. Wow! You should really watch the lectures again. Sorry. > 4. That's right. Repeat function repeats the given string. And, the concatenation operator combines the strings. ## What's the result of this expression? ```go strings.ToUpper("bye bye ") + "see you!" ``` 1. `bye bye see you!` 2. `BYE BYE SEE YOU!` 3. `bye bye + see you!` 4. `BYE BYE see you!` *CORRECT* > 1. You missed the ToUpper? > 2. You're close but look closely. ToUpper only changes the first part of the string there. > 3. Not even close. Sorry. > 4. Perfect! Good catch! ToUpper only changes the first part of the string there.