尝试运行插入排序算法时,如下面的Rust 1.15所示。
fn main() {
println!("The sorted set is now: {:?}", insertion_sort(vec![5,2,4,6,1,3]));
}
fn insertion_sort(set: Vec<i32>) -> Vec<i32> {
let mut A = set.to_vec();
for j in 1..set.len() {
let key = A[j];
let mut i = j - 1;
while (i >= 0) && (A[i] > key) {
A[i + 1] = A[i];
i = i - 1;
}
A[i + 1] = key;
}
A
}
我收到错误:
thread 'main' panicked at 'attempt to subtract with overflow', insertion_sort.rs:12
note: Run with `RUST_BACKTRACE=1` for a backtrace
为什么在这里发生溢出以及问题如何得到缓解?
答案 0 :(得分:2)
原因是你试图在0 - 1
类型中计算usize
,这是无符号的(非负)。这可能会导致Rust出错。
为什么usize
?因为Rust期望长度和索引usize
。您可以明确地将它们转换为签名的,例如isize
。
fn main() {
println!("The sorted set is now: {:?}", insertion_sort(vec![5,2,4,6,1,3]));
}
fn insertion_sort(set: Vec<i32>) -> Vec<i32> {
let mut A = set.to_vec();
for j in 1..set.len() as isize {
let key = A[j as usize];
let mut i = j - 1;
while (i >= 0) && (A[i as usize] > key) {
A[(i + 1) as usize] = A[i as usize];
i = i - 1;
}
A[(i + 1) as usize] = key;
}
A
}
我建议的另一个解决方案是避免负指数。在这种情况下,您可以使用i + 1
代替i
,如下所示:
fn main() {
println!("The sorted set is now: {:?}", insertion_sort(vec![5,2,4,6,1,3]));
}
fn insertion_sort(set: Vec<i32>) -> Vec<i32> {
let mut A = set.to_vec();
for j in 1..set.len() {
let key = A[j];
let mut i = j;
while (i > 0) && (A[i - 1] > key) {
A[i] = A[i - 1];
i = i - 1;
}
A[i] = key;
}
A
}