break

Is there a way to have nested for loops in Rust and break the outer one from inside the inner one, the way one could do e.g. in Java? I know Rust supports named breaks in loop but I can't seem to find information about the same regarding for . Lily Ballard Yes. It uses the same syntax as lifetimes. fn main() { 'outer: for x in 0..5 { 'inner: for y in 0..5 { println!("{},{}", x, y); if y == 3 { break 'outer; } } } } See loop labels documentation and the related section of the reference . 来源： https://stackoverflow.com/questions/22905752/named-breaks-in-for-loops-in-rust

I'm currently trying to figure out a way to break out of a for loop from within a function called in that loop. I'm aware of the possibility to just have the function return a value and then check against a particular value and then break, but I'd like to do it from within the function directly. This is because I'm using an in-house library for a specific piece of hardware that mandates the function signature of my function to look like this: void foo (int passV, int aVal, long bVal) I'm aware that not using a return value is very bad practice , but alas circumstances force me to, so please

def funcA(i): if i%3==0: print "Oh! No!", print i break for i in range(100): funcA(i) print "Pass", print i I know script above won't work. So, how can I write if I need put a function with break or continue into a loop? A function cannot cause a break or continue in the code from which it is called. The break/continue has to appear literally inside the loop. Your options are: return a value from funcA and use it to decide whether to break raise an exception in funcA and catch it in the calling code (or somewhere higher up the call chain) write a generator that encapsulates the break logic and

Suppose I need to break out of three or four nested for loops at once at the occurence of some event inside the innermost loop. What is a neat way of doing that? what I do is use flags like this: int i, j, k; int flag1 = 0; int flag2 = 0; for (i = 0; i < 100; i++) { for (j = 0; j < 100; j++) { for (k = 0; k < 100; k++) { if (k == 50) { flag1 = 1; flag2 = 1; break; } } if (flag1 == 1)break; } if (flag2 == 1)break; } I don't think this is particularly neat. How would you accomplish the same thing? (w/o using jumps) use goto. it's clean and simple. Put all the loops in a function and just return

Let's say I have this: while(a){ while(b){ if(b == 10) break; } } Question: Will the break statement take me out of both loops or only from the inner one? Thank you. Abhishekkumar In your example break statement will take you out of while(b) loop while(a) { while(b) { if(b == 10) { break; } } // break will take you here. } Ivan Koblik It will break only the most immediate while loop. Using a label you can break out of both loops: take a look at this example taken from here public class Test { public static void main(String[] args) { outerloop: for (int i=0; i < 5; i++) { for (int j=0; j < 5; j

This is looking like an impossible task. Absolutely nothing I've found works. The question is how to cleanly close a console application started with Process.Start that has been started with no console window and without using shell execute: ( ProcessStartInfo.CreateNoWindow = true; ProcessStartInfo.UseShellExecute = false; ). It is given that the application being started will shut down "cleanly" if it receives a ctrl-c or ctrl-break signal, but there seems to be no way to send it one that works (particularly GenerateConsoleCtrlEvent). Process.Kill doesn't work. It leaves corrupt files behind

I know a break statement jumps out of a loop, but does it jump out of nested loops or just the one its currently in? Without any adornment, break will just break out of the innermost loop. Thus in this code: while (true) { // A while (true) { // B break; } } the break only exits loop B , so the code will loop forever. However, Java has a feature called "named breaks" in which you can name your loops and then specify which one to break out of. For example: A: while (true) { B: while (true) { break A; } } This code will not loop forever, because the break explicitly leaves loop A . Fortunately,