This question already has answers here:
How to create a formatted String out of a literal in Rust?
(2 answers)
Closed 4 months ago.
Is there a way to use String::from with variables? I tried using it similarly to println!(), but doesn't seem to work.
String::from("[INFO]: {}", message)
Expected:
// Returns: "[INFO]: My info message here"
Got:
// Error: argument unexpected
Rust doesn’t really practice variadic functions because we have big metaprogramming potential: modern generic system and about 6 types of macros.
In this case you need the format macro:
format!("[INFO]: {}", message)
Related
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Long Int literal - Invalid Syntax?
(2 answers)
Closed 2 years ago.
I'm trying to convert a small file from Python2 to Python3.
This hex variable assignment is blocking me.
a = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2FL
The error I'm receiving is:
a = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2FL
^
SyntaxError: invalid syntax
Any suggestions?
Is there a problem if you remove the L? Is that not the value you expect?
>>> 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F
115792089237316195423570985008687907853269984665640564039457584007908834671663
Wolfram alpha thinks that's the correct value!
This question already has answers here:
Using functions defined in primitive modules
(1 answer)
no function or associated item named `from_str` found for type `hyper::mime::Mime`
(1 answer)
How to create an `IpAddr` without knowing the specific IP version?
(1 answer)
Closed 3 years ago.
I'm trying to use the Cidr crate, and need to convert strings to CIDRs. But it's not working and I can't figure out why. I think I need to use something, but std::str::FromStr doesn't help. What's worse is that I don't even know where to look next.
error[E0599]: no variant named `from_string` found for type `cidr::AnyIpCidr` in the current scope
--> src/main.rs:45:30
|
45 | let c = cidr::AnyIpCidr::from_string(ipstr);
| -----------------^^^^^^^^^^^
| |
| variant not found in `cidr::AnyIpCidr`
This question already has answers here:
How to disable unused code warnings in Rust?
(11 answers)
How to quiet a warning for a single statement in Rust?
(2 answers)
Closed 4 years ago.
I have lines that return Result. The result does not matter and I don't want unwrap() or any logging. (The frequent failures are expected)
How can I silence the unused warnings for only these lines?
#[allow(unused_must_use)] seems to work when applied to fn level, but this does not work:
#[allow(unused_must_use)]
std::fs::remove_file(&path1);
#[allow(unused_must_use)]
std::fs::remove_file(&path2);
Edit:
while let _ = .. works as a workaround, #[allow(unused..)] specifically exists for this case, and the compiler also suggests to use it. Reading let _ = .. adds another layer of thinking. (assign and then abandon) So I prefer #[allow(..)] although it is more verbose. (If I see let _ = .. frequent enough and I get used to it, I may change my preference)
So I searched around and found some codes that apply macros to the statement level.
This question asks for why #[allow(unused_must_use)] for a line doesn't work-- wrong syntax? bug? just unimplemented yet for unused_must_use?
Edit:
According to this:
How to quiet a warning for a single statement in Rust?
My code should work.
Reading these:
https://github.com/rust-lang/rust/issues/36675
https://github.com/rust-lang/rust/issues/15701
I tried the allow on the scope level and it worked.
#[allow(unused_must_use)] {
std::fs::remove_file(&path1);
std::fs::remove_file(&path2);
}
This question already has answers here:
Get the type of a variable in VBScript
(6 answers)
Closed 7 years ago.
I am passing one object to function which needs to take action according to object type. e.g. If object is of "Scripting.Dictionary" then count keys and if it is "Scripting.FileSystemObject" then close it.
In short as like typeof in c# and variableName.class in java, how we can find which scripting object we are using ?
Use the TypeName function to get the type of a variable.
TypeName Function
Returns a string that provides Variant subtype information about a variable.
TypeName(varname)
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Annoying PHP error: “Strict Standards: Only variables should be passed by reference in”
I have this line of code,
$extension=end(explode(".", $srcName));
when I fun my function I get
PHP Strict Standards: Only variables should be passed by reference in
I am not sure how to solve this
The function end() requires a variable to be passed-by-reference and passing the return-value of a function doesn't acheive this. You'll need to use two lines to accomplish this:
$exploded = explode(".", $srcName);
$extension = end($exploded);
If you're simply trying to get a file-extension, you could also use substr() and strrpos() to do it in one line:
$extension = substr($srcName, strrpos($srcName, '.'));
Or, if you know the number of .'s that appear in the string, say it's only 1, you can use list() (but this won't work if there is a dynamic number of .'s:
list(,$extension) = explode('.', $srcName);