let mut input_player = String::new();
let mut player_name = Vec::new();
println!("Enter names of the player: ");
for a in 1..(player_number+1){
println!("Enter names of the player # {}: ", a);
io::stdin()
.read_line(&mut input_player)
.expect("failed to read from stdin");
player_name.push(input_player);
}
I am facing problem while initializing Vector through input string in for loop
There are 2 misconceptions that you have here.
First you are trying to read into the string without clearing it first.
Lets say the first name was Foo and second was Bar.
After the first read_line, value of input_player would be Foo indeed. But the second read_line will not overwrite, it will append. So now the value will be FooBar, which probably is not what you want.
The second and more important is that you gave the string away to the vec in the first loop iteration. But you are trying to use it in second iteration. This comes into the domain of ownership, which you can read more about here.
What you need is a new string for each iteration of the loop. So define a new input_player inside the loop.
Related
Now that I'm typing it, this seems like a very convoluted process that could definitely be solved easier. Ignoring that for the moment, I'm trying to take a string (from user input), separate the characters into an array, then call individual characters to make a new string. The issue I'm running into is that the "join" function doesn't like working with the "Vec" function (not sure if function is the right term, sorry). Here is my code so far:
use std::io;
fn main() {
println!("Enter P1:");
let mut mono = String::new();
io::stdin()
.read_line(&mut mono)
.expect("Failed to read line");
let mono: Vec<char> = mono.chars().collect();
let x = [mono[0],mono[1]].join("");
println!("Square 1: {x}");
}
I'm very new to Rust, so any suggestions are extremely helpful. Thank you!
You could always just slice the original string str[a..b].to_string().
If you need to work with char arrays, there's String::from_iter and collecting into a String. Example:
fn main() {
let string = "My String".to_string();
let substr1 = string[0..3].to_string();
let substr2 = string[3..].to_string();
println!("substr1 = {}", substr1);
println!("substr2 = {}", substr2);
let chars: Vec<_> = string.chars().collect();
let collected_substr1: String = chars.iter().take(2).collect();
let collected_substr2: String = chars.iter().skip(3).collect();
println!("collected_substr1 = {}", collected_substr1);
println!("collected_substr2 = {}", collected_substr2);
let from_iter_substr1 = String::from_iter([chars[0], chars[1]].iter());
let from_iter_substr2 = String::from_iter(chars.iter().skip(3));
println!("from_iter_substr1 = {}", from_iter_substr1);
println!("from_iter_substr2 = {}", from_iter_substr2);
}
Vec is a type, FYI.
join only works on string slices (&str), not chars. Assuming you are just trying to join two characters without a separator, you can do
let x: String = mono.chars().take(2).collect();
If the goal is only to extract a substring from the input it can be done much simpler
use std::io;
fn main() {
println!("Enter P1:");
let mut mono = String::new();
io::stdin()
.read_line(&mut mono)
.expect("Failed to read line");
let x = &mono[..2].to_string(); // get rhe slice you need and create a new string from it
println!("Square 1: {x}");
}
Edit
As pointed out in comments, note that in a real life use case you should check the length of your string before slicing in it with arbitrary indexes... this example could easily crash at runtime.
So I have a code which constantly asks for input and then executes your input as a shell command. I understand that the output I am getting from the shell commands is in a buffer of some sort. Now, however, as there are many commands which output lots of lines, I would like to get all of the output into one single string.
extern crate subprocess;
use std::io;
use std::io::{BufRead, BufReader};
use subprocess::Exec;
fn main() {
loop {
let mut mycommand_string: String = String::new();
io::stdin()
.read_line(&mut mycommand_string)
.expect("Failed to read line");
let mycommand: &str = &*mycommand_string;
let x = Exec::shell(mycommand).stream_stdout().unwrap();
let br = BufReader::new(x);
let full: String = " ".to_string();
let string = for (i, line) in br.lines().enumerate() {
let string: String = line.unwrap().to_string();
let full = format!("{}{}", full, string);
println!("{}", full);
};
println!("{}", string);
}
}
This is my code. As you can see, the thing I am aiming for is to somehow iterate over br.lines() and for each line of output it contains, append or concatenate it to a string, so that all the output ends up in one single string, preferably with "\n" in between each line, but not neccesarilly.
Specifically I would like to iterate over the result of the variable br which has a type I dont understand and to concatenate all the strings together.
If you have an iterator of lines, then you can simply collect that into a string:
br.lines().collect();
Of course we should not ignore that there do not seem to be many possible reasons for ever doing that...
I created a empty Mutable String variable using String::new() before a loop start. Then I printed the string value as soon as I entered the loop, changed the type of the same variable to integer u32 by with user input, after trimming all spaces, \n, \r etc..
At the next Iteration of the loop, The value of the variable is back to String and was about to change its type, but when I checked the value of String by printing it, It had some ghost \n and spaces or some ghost characters inherited from the previous integer value.
if the integer is 3 digit, for eg 534 it has 5 characters
if the integer is 1 digit, for eg 3 it has 3 characters
if I give empty value as input, The parsing fails it stays as String, but still in next iteration the String has 2 characters.
I created a function to keep track of the type of variable.
use std::io;
//function to return type of a variable
fn type_of<T>(_: &T) -> String {
return format!("{}", std::any::type_name::<T>());
}
fn main() {
let mut guess = String::new();
loop {
println!(
"At start of loop : {},{}",
type_of(&guess),
guess.chars().count()
);
println!("value : {}", guess);
//emptying string
String::clear(&mut guess);
println!(
"after clearing : {},{}",
type_of(&guess),
guess.chars().count()
);
//getting input for string
println!("Enter value :");
io::stdin()
.read_line(&mut guess)
.expect("Failed to read line");
//converted its values to interger u32 after trimming spaces,\n and \r and stuffs like that
let guess: u32 = match guess.trim().parse() {
Ok(a) => a,
Err(b) => {
println!("{}", b);
println!("after reciving error : {}", type_of(&guess));
continue;
}
};
println!("after type conversion : {}", type_of(&guess));
println!("value: {}", guess);
}
}
the output was :
At start of loop : alloc::string::String,0
value :
after clearing : alloc::string::String,0
Enter value :
111
after type conversion : u32
value: 111
At start of loop : alloc::string::String,5
value : 111
after clearing : alloc::string::String,0
Enter value :
1
after type conversion : u32
value: 1
At start of loop : alloc::string::String,3
value : 1
after clearing : alloc::string::String,0
Enter value :
cannot parse integer from empty string
after reciving error : alloc::string::String
At start of loop : alloc::string::String,2
value :
after clearing : alloc::string::String,0
Enter value :
What causes this?
Is there a way to maintain the value before the loop, at the start of every iteration?
or may be maintain the value of Integer from previous iteration and u32 Type at the same time?
I ran into this problem when I was trying to learn rust using "The Book" from rust docs, to be specific when I was trying to mess around with the code from Chapter 2 (Guess a number project).
There is a misunderstanding on how variables work in Rust. Different variables with the same name can exist, a process called shadowing. In this program, we have two variables called guess.
The following simplification of the previous code shows this pattern.
let guess: mut = String::new(); // <-- guess #1, lives outside loop
loop {
guess.clear();
println!("Enter value :");
io::stdin()
.read_line(&mut guess)
.expect("Failed to read line");
// guess #2, lives inside loop
// vvvvv
let guess: u32 = match guess.trim().parse() {
Ok(a) => a,
Err(b) => {
eprintln!("{}", b);
continue;
}
};
println!("value: {}", guess);
}
The first one is always of type String, and the second one is always of type u32. Variables can never change type. What does change is which one can be seen and used in what scope. Since the second guess is only declared in the middle of the loop, mentioning guess before that declaration will mean the first guess, which is the string.
Combining the two facts that:
read_line reads and includes newline characters into the output string;
trim only returns a string slice, without modifying the underlying String value.
then it makes sense that guess will contain trailing newline characters at the beginning of the loop statement after the first iteration.
Is there a way to maintain the value before the loop, at the start of every iteration? or may be maintain the value of Integer from previous iteration and u32 Type at the same time?
With this last question rephrased to mean "a way to maintain the integer value from the previous iteration", then that is possible by giving it a new name and moving it up. In the example below, guess_num is reassigned on each iteration rather than declared each time.
let guess: mut = String::new();
let mut guess_num: u32 = 0;
loop {
println!("Previous number (or 0 if first iteration): {}", guess_num);
guess.clear();
println!("Enter value :");
io::stdin()
.read_line(&mut guess)
.expect("Failed to read line");
guess_num = match guess.trim().parse() {
Ok(a) => a,
Err(b) => {
eprintln!("{}", b);
continue;
}
};
println!("value: {}", guess);
}
See also:
Why do I need rebinding/shadowing when I can have mutable variable binding?
How to ignore the line break while printing a string read from stdin?
The Rust Programming Language, 3.1 Variables and Mutability
I am learning Rust from The Book and I just finished the first exercise, the guessing game.
I use cargo to build and run my little crate.
$ cargo --version
cargo 1.37.0 (9edd08916 2019-08-02)
$ rustc --version
rustc 1.37.0 (eae3437df 2019-08-13)
Everything run fine, including release mode. Nevertheless, I do not understand a the following behaviour of Rust: I have to redeclare the variable which contains the user input at each iteration of the loop.
Since the exercise is guided step-by-step, my code is the same that the one from the book. Nevertheless, the code from the book is the following:
loop {
// Some code to display instructions.
// Reallocate a new string at each iteration!
let mut guess = String::new();
io::stdin().read_line(&mut guess)
.expect("Failed to read line");
let guess: u32 = match guess.trim().parse() {
Ok(num) => num,
Err(_) => continue,
};
// Some code to check if the player found the secret number.
}
Noticing this systematic reallocation, I moved the string declaration outside the loop:
// Allocate the string once.
let mut guess = String::new();
loop {
// Some code to display instructions.
io::stdin().read_line(&mut guess)
.expect("Failed to read line");
let guess: u32 = match guess.trim().parse() {
Ok(num) => num,
Err(_) => continue,
};
// Some code to check if the player found the secret number.
}
However, Rust did not appreciate this: at the second iteration of the loop, it panics every time.
Why can't I reuse the same mutable variable more than once? Do I not understand something?
EDIT: read_line does not clear the content of the previous input, but appends to it the following one.
Let's say the player enters 1 then 2, the final value of guess will be "1\n2\n".
However, trim() removes the "blank" characters at the beginning and the end of the string, leaving a \n in the middle: parse() panics!
Your code as-is compiles and runs fine on my setup (same version of rust). The panic must happen in the commented-out part of your code. Some comments, though: the scoping in your loop is tricky: guess at the top-half of the loop is the string declared outside the loop, and is the parsed integer in the second half.
More importantly, multiple calls to read_line being passed the same string appends to the string, which probably isn't your intention given the way you're parsing the string. Sprinkling in println!'s of your guess variables should be illuminating. Your code will probably be fixed if you add a guess.clear() on the string after you've parsed the number, but to do that, you'll probably want to rename the u32 guess.
As an aside, you might consider using a BufReader and the for line in reader.lines()) pattern described here.
I am attempting to write a lexer for fun, however something keeps bothering me.
let mut chars: Vec<char> = Vec::new();
let mut contents = String::new();
let mut tokens: Vec<&String> = Vec::new();
let mut append = String::new();
//--snip--
for _char in chars {
append += &_char.to_string();
append = append.trim().to_string();
if append.contains("print") {
println!("print found at: \n{}", append);
append = "".to_string();
}
}
Any time I want to do something as simple as append a &str to a String I have to convert it using .to_string, String::from(), .to_owned, etc.
Is there something I am doing wrong, so that I don't have to constantly do this, or is this the primary way of appending?
If you're trying to do something with a type, check the documentation. From the documentation for String:
push: "Appends the given char to the end of this String."
push_str: "Appends a given string slice onto the end of this String."
It's important to understand the differences between String and &str, and why different methods accept and return each of them.
A &str or &mut str are usually preferred in function arguments and return types. That's because they are just pointers to data so nothing needs to be copied or moved when they are passed around.
A String is returned when a function needs to do some new allocation, while &str and &mut str are slices into an existing String. Even though &mut str is mutable, you can't mutate it in a way that increases its length because that would require additional allocation.
The trim function is able to return a &str slice because that doesn't involve mutating the original string - a trimmed string is just a substring, which a slice perfectly describes. But sometimes that isn't possible; for example, a function that pads a string with an extra character would have to return a String because it would be allocating new memory.
You can reduce the number of type conversions in your code by choosing different methods:
for c in chars {
append.push(c); // append += &_char.to_string();
append = append.trim().to_string();
if append.contains("print") {
println!("print found at: \n{}", append);
append.clear(); // append = "".to_string();
}
}
There isn't anything like a trim_in_place method for String, so the way you have done it is probably the only way.