抛弃函数参数的常量是未定义的行为吗? [英] Is it Undefined Behaviour to cast away the constness of a function parameter?

问题描述

假设我有这个 C 函数(以及头文件中的相应原型)

Imagine I have this C function (and the corresponding prototype in a header file)

void clearstring(const char *data) {
    char *dst = (char *)data;
    *dst = 0;
}

上述代码中是否存在未定义行为，将 const 丢弃，还是只是一种非常糟糕的编程习惯?

Is there Undefined Behaviour in the above code, casting the const away, or is it just a terribly bad programming practice?

假设没有使用 const 限定的对象

Suppose there are no const-qualified objects used

char name[] = "pmg";
clearstring(name);

推荐答案

尝试写入 *dst 是 UB if 调用者将指针传递给 const对象，或指向字符串文字的指针.

The attempt to write to *dst is UB if the caller passes you a pointer to a const object, or a pointer to a string literal.

但是如果调用者传递给你一个指向实际上是可变数据的指针，那么行为就被定义了.创建指向可修改 char 的 const char* 不会使该 char 不可变.

But if the caller passes you a pointer to data that in fact is mutable, then behavior is defined. Creating a const char* that points to a modifiable char doesn't make that char immutable.

所以:

char c;
clearstring(&c);    // OK, sets c to 0
char *p = malloc(100);
if (p) {
    clearstring(p); // OK, p now points to an empty string
    free(p);
}
const char d = 0;
clearstring(&d);    // UB
clearstring("foo"); // UB

也就是说，你的函数是非常不明智的，因为调用者很容易导致 UB.但实际上可以将其用于已定义的行为.

That is, your function is extremely ill-advised, because it is so easy for a caller to cause UB. But it is in fact possible to use it with defined behavior.

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