Index
- Guessing Game
- Common Programming Concepts
- Understanding Ownership
- Using Structs
- Enums and Pattern Matching
- Managing Growing Projects with Packages, Crates, and Modules
- Defining Modules to Control Scope and Privacy
- Paths for Referring to an Item in the Module Tree
- Bringing Paths into Scope with the use Keyword
- Separating Modules into Different Files
- Common Collections
- Error Handling
- Generic Types, Traits, and Lifetimes
- Writing Automated Tests
- Object Oriented Programming
- Adding dependancies
- Option Take
- RefCell
- mem
- Data Structure
- Recipe
- Semi colon
- Calling rust from python
- Default
- Crytocurrency With rust
- Function chaining
- Question Mark Operator
- Tests with println
- lib and bin
- Append vector to hash map
- Random Number
- uuid4
- uwrap and option
- Blockchain with Rust
- Near Protocol
- Actix-web
Semi colon
https://stackoverflow.com/questions/26665471/are-semicolons-optional-in-rustSemicolon is not optional
They're not optional. Semicolons modify the behaviour of an expression statement so it should be a conscious decision whether you use them or not for a line of code.
Almost everything in Rust is an expression. An expression is something that returns a value. If you put a semicolon you are suppressing the result of this expression, which in most cases is what you want.
On the other hand, this means that if you end your function with an expression without a semicolon, the result of this last expression will be returned. The same can be applied for a block in a match statement.
You can use expressions without semicolons anywhere else a value is expected.
For example:
let a = {
let inner = 2;
inner * inner
};
let inner = 2;
inner * inner
};
Here the expression inner * inner does not end with a semicolon, so its value is not suppressed. Since it is the last expression in the block, its value will be returned and assigned to a. If you put a semicolon on this same line, the value of inner * inner won't be returned.
In your specific case, not suppressing the value of your let statement doesn't make sense, and the compiler is rightly giving you an error for it. In fact, let is not an expression.