函数指针奇怪问题，泛型函数可以赋值给变量，不能赋值给结构体的字段

use std::io::{ Result, Error, ErrorKind, Write }; pub fn should_not_call<A, B, R>(_a: A, _b: B) -> Result<R> { Err(Error::new(ErrorKind::Unsupported, "should not be called"))? } pub struct Dump<W: Write> { pub w: W, pub write_u64: fn(&mut Dump<W>, u64) -> Result<usize>, } pub fn write_le<W: Write>(d: &mut Dump<W>, v: u64) -> Result<usize> { d.w.write(u64::to_le_bytes(v).as_slice()) } fn ok<W: Write>(w: W) { let mut write_u64: fn(_,_) -> _ = write_le::<W>; write_u64 = should_not_call; } fn failed<W: Write>(w: W) { let mut d = Dump { w, write_u64: write_le::<W>, }; //d.write_u64 = should_not_call; //d.write_u64 = should_not_call as fn(_,_) -> _; //d.write_u64 = should_not_call as for<'a> fn(&'a mut Dump<W>, u64) -> Result<usize>; } 如上，should_not_call 这个通用泛型函数，可以赋值给 ok 的变量 write_u64，但没法赋值给 failed 的 d.write_u64。 注释掉的第一种写法，报错 expected fn pointer found fn item 注释掉的第二种写法，报错 mismatched types: one type is more general than the other 注释掉的第三种写法，报错 non-primitive cast: invalid cast