读取与scanf函数的字符串 [英] Reading a string with scanf

问题描述

我是一个有点困惑的东西。我是读一个C字符串scanf（）的正确的方式沿

I'm a little bit confused about something. I was under the impression that the correct way of reading a C string with scanf() went along the lines of

（心中永远的可能缓冲区溢出，这只是一个简单的例子）

(never mind the possible buffer overflow, it's just a simple example)

char string[256];
scanf( "%s" , string );

但是，以下似乎太工作，

However, the following seems to work too,

scanf( "%s" , &string );

这只是我的编译器（GCC），纯粹是运气，还是其他什么东西？

Is this just my compiler (gcc), pure luck, or something else?

在此先感谢

推荐答案

这是阵腐朽为指针，以它的第一要素，所以 scanf函数（％S，字符串）等同于 scanf函数（％S，＆安培;串[0]）。在另一方面， scanf函数（％S，＆安培;字符串）将指针传递到 - 的char [256] ，但它指向同一个地方。

An array "decays" into a pointer to its first element, so scanf("%s", string) is equivalent to scanf("%s", &string[0]). On the other hand, scanf("%s", &string) passes a pointer-to-char[256], but it points to the same place.

然后 scanf函数，处理它的参数列表的尾部时，会尽量拉出一个的char * 。这就是当你在字符串已经通过正确的事或＆放大器;字符串[0] ，但是，当你字符串你根据什么说的语言标准并不能保证，即指针＆放大器;字符串＆放过去了$ C>和＆放大器;字符串[0] - 指针到开始在同一地点不同类型和大小的物体 - 重新presented以同样的方式

Then scanf, when processing the tail of its argument list, will try to pull out a char *. That's the Right Thing when you've passed in string or &string[0], but when you've passed in &string you're depending on something that the language standard doesn't guarantee, namely that the pointers &string and &string[0] -- pointers to objects of different types and sizes that start at the same place -- are represented the same way.

我不相信，我曾经遇到过哪些不工作的系统，并在实践中你可能是安全的。尽管如此，这是错误的，它可能会失败在某些平台上。 （假设的例子：。一调试实现，包括与每一个指针类型的信息我的认为的在对符号Lisp机器C实现做了这样的事情）

I don't believe I've ever encountered a system on which that doesn't work, and in practice you're probably safe. None the less, it's wrong, and it could fail on some platforms. (Hypothetical example: a "debugging" implementation that includes type information with every pointer. I think the C implementation on the Symbolics "Lisp Machines" did something like this.)

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