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2 changes: 1 addition & 1 deletion book/zh-cn/03-runtime.md
Original file line number Diff line number Diff line change
Expand Up @@ -333,7 +333,7 @@ void foo() {
```

由于 `int&` 不能引用 `double` 类型的参数,因此必须产生一个临时值来保存 `s` 的值,
从而当 `increase()` 修改这个临时值时,从而调用完成后 `s` 本身并没有被修改。
从而当 `increase()` 修改这个临时值时,调用完成后 `s` 本身并没有被修改。

第二个问题,为什么常量引用允许绑定到非左值?原因很简单,因为 Fortran 需要。

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2 changes: 1 addition & 1 deletion code/3/3.3.rvalue.cpp
Original file line number Diff line number Diff line change
Expand Up @@ -20,7 +20,7 @@ void reference(std::string&& str) {
int main()
{
std::string lv1 = "string,"; // lv1 is a lvalue
// std::string&& r1 = s1; // illegal, rvalue can't ref to lvalue
// std::string&& r1 = lv1; // illegal, rvalue can't ref to lvalue
std::string&& rv1 = std::move(lv1); // legal, std::move can convert lvalue to rvalue
std::cout << rv1 << std::endl; // string,

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