使用 scanf 读取字符串 [英] Reading a string with scanf

问题描述

我对某事有点困惑.我的印象是，使用 scanf() 读取 C 字符串的正确方法遵循

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

(不要介意可能的缓冲区溢出，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，以下似乎也有效，

However, the following seems to work too,

scanf( "%s" , &string );

这只是我的编译器(gcc)，纯粹的运气，还是其他什么?

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

推荐答案

一个数组衰减"成指向它的第一个元素的指针，所以 scanf("%s", string) 等价于scanf("%s", &string[0]).另一方面，scanf("%s", &string) 传递了一个指向 char[256] 的指针，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，在处理其参数列表的尾部时，会尝试拉出一个char *.当您传入 string 或 &string[0] 时，这是正确的事情，但是当您传入 &string 时你依赖于语言标准不能保证的东西，即指针 &string 和 &string[0] -- 指向不同对象的指针从相同位置开始的类型和大小 - 以相同的方式表示.

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.

我相信我从未遇到过无法运行的系统，实际上您可能是安全的.尽管如此，这是错误的，并且在某些平台上可能会失败.(假设示例:一个调试"实现，其中包含每个指针的类型信息.我认为 SymbolicsLisp Machines"上的 C 实现做了类似的事情.)

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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