c++ - 修改和非修改 putback() 之间的区别

Question

问题来自https://en.cppreference.com/w/cpp/io/basic_istream/putback , 示例代码。

#include <sstream>
#include <iostream>

int main()
{
    std::istringstream s2("Hello, world"); // input-only stream
    s2.get();
    if (s2.putback('Y')) // cannot modify input-only buffer
        std::cout << s2.rdbuf() << '\n';
    else
        std::cout << "putback failed\n";

    s2.clear();
    if (s2.putback('H')) // non-modifying putback
        std::cout << s2.rdbuf() << '\n';
    else
        std::cout << "putback failed\n";
}

为什么 s2.putback('Y') 失败但 s2.putback('H') 成功？后者不也是修改input-only stream buffer的操作吗？

---

此外，我在做一些实验时发现了一些令人困惑的事情。与上面的示例相比，我添加了 1 行代码，但第二个结果失败了。这是为什么？

#include <sstream>
#include <iostream>

int main()
{
    std::istringstream s2("Hello, world"); // input-only stream
    s2.get();
    if (s2.putback('Y')) // cannot modify input-only buffer
        std::cout << s2.rdbuf() << '\n';
    else
        std::cout << "putback failed\n";

    std::cout << s2.rdbuf() << '\n';   //1 line code added

    s2.clear();
    if (s2.putback('H')) // non-modifying putback
        std::cout << s2.rdbuf() << '\n';
    else
        std::cout << "putback failed\n";

}

Answer 1

Why s2.putback('Y') fails but s2.putback('H') succeed? Isn't the latter also an operation to modify the input-only stream buffer?

调用 s2.putback('H') 可能 修改缓冲区，但在这种情况下，它不会，因为数据已经以 'H'.

你可以举例说明这样的行为:

s2.clear();

assert(s2.putback('H')); // Ok, replacing 'H' with 'H' doesn't change anything
assert(!s2.putback('Z')); // Can't modify.