[Bug]The compiler detects the error if a function returns a value, and if it is not assigned to a variable.

[aleo-debugger]

🐛 Bug Report

If a function returns a value, and if it is not assigned to a variable, the compiler detects this as an error “Mismatched types. Expected type(), found type u32 ”
At the same time, if the function does not return a value, then it can be assigned to a variable as well as could be called without being assigned. Looks like an incorrect behaviour.

Steps to Reproduce

Code snippet to reproduce

function excar () {
    let ex = 33u32;
}
function exbaz () -> u32 {
    return 33u32 
}

function main() {
    let excar1 = excar();
    excar();   

    let exbaz1 = exbaz();
    exbaz();   
}

Stack trace & error message

 Compiling Starting...
 Compiling Compiling main program... ("c:\\Users\\test\\Aleo Studio\\function-return\\src/main.leo")
 Compiling     --> "c:\\Users\\test\\Aleo Studio\\function-return\\src/main.leo": 13:5
     |
  13 |      exbaz();
     |  ^^^^
     |
     = Mismatched types. Expected type `()`, found type `u32`.

Error: Crate("leo-compiler", "Program failed due to previous error")

Expected Behavior

(Write what you expected to happen here)

Your Environment

• leo 1.0.7
• rust 1.48.0
• Edition: Windows 10 Pro Version: 20H2
  OS build: 19042.685
  Experience: Windows Feature Experience Pack 120.2212.551.0

[collinc97]

The above code is correct. By default, functions in Leo return the unit type (). https://en.wikipedia.org/wiki/Unit_type

If a function returns a specific type, then that result must be handled - assigned to a variable or used in a statement.

The following code is illegal in Leo

function main() {
    0u8; // Must be assigned to a variable or used in a statement.
}

Shoutout to @gluax for identifying this!