FFI 函数可以修改未声明为可变的变量吗? [英] Can an FFI function modify a variable that wasn't declared mutable?

问题描述

fn main() {
    let val = 0;
    unsafe { foo(&val) }
}

extern "C" {
    pub fn foo(val: *const u32);
}

在 C 中实现:

void foo(unsigned* val) { *val=1; }

当然，我应该传递val: *mut u32，但是如果我传递了一个不可变的引用，会发生什么?哪些编译器规则适用?即使我传递了一个指向局部变量的指针，val 是否保持不变?

Of course, I should pass val: *mut u32, but what happens in the case that I pass an immutable reference? What compiler rules apply? Does val remain unchanged even though I'm passing a pointer to the local variable?

推荐答案

我想说 未定义行为:

变异非可变数据——即通过共享引用或let绑定拥有的数据到达的数据)，除非该数据包含在UnsafeCell代码>.

Mutating non-mutable data — that is, data reached through a shared reference or data owned by a let binding), unless that data is contained within an UnsafeCell<U>.

这可能包括:

• 如果您在 FFI 调用后使用 val，它可能会忽略您所做的写入(例如将值缓存在寄存器中或由于 不断传播)
• FFI 中的段错误，因为引用的内存可能是只读的
• 来自 FFI 的写入可能会出现在看似无关的位置，因为编译器重用了内存并假设它具有明确定义的值
• 更糟:)

• if you use val after the FFI-call it might ignore the writes you did (e.g. cached the value in a register or due to constant propagation)
• segfault in FFI because the referenced memory might be read-only
• the write from FFI might show up in seemingly unrelated locations because the compiler reused the memory and assumed it had a well-defined value
• and worse :)

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