改变const对象 - 没有警告？此外，在这种情况下，它是UB？ [英] Changing const object - no warning? Also, in which case it is UB?

问题描述

为什么会出现在以下code无警告？

Why is there no warning in the following code?

int deserialize_students(const Student *dest, const int destCapacityMax)
{
    FILE *ptr_file;
    int i=0;


    ptr_file =fopen("output.txt","r");
    if (!ptr_file)
        return -1;

    if(destCapacityMax==0)
        return -2;

    while (!feof (ptr_file))
    {  
        fscanf (ptr_file, "%d", &dest[i].id);    // UB?
        fscanf (ptr_file, "%s",  dest[i].name);     
        fscanf (ptr_file, "%d", &dest[i].gender);      
        i++;

        if(i==destCapacityMax)
            return 0;

    }
    fclose(ptr_file);
    return 0;
 }

这是我怎么把它称为：

Student students[5];
deserialize_students(students,5);

此外我有以下问题：是我做的未定义行为？
注：

Also I have following question: is what I did undefined behaviour? Note:

• 我觉得通过学生是很好，因为函数需要常量学生* ，我可以通过非常量。对吗？


• 但是，当我修改的fscanf 该对象，并触发我UB？或者，这取决于学生是否被声明为const还是不摆在首位（即功能之外）？

• I think passing students is fine because function expects const Student* and I can pass non const. Right?
• But when I modify that object in fscanf, did I trigger UB? Or it depends whether students was declared const or not in the first place(outside that function)?

推荐答案

有在你的code无未定义的行为。

There's no undefined behaviour in your code.

通过来电者是一个什么可变对象。因此，它的优良直接修改或明确的转换：

What the caller passed is a mutable object. So it's fine to modify it either directly or by explicit cast:

int func(const int *p) {
  int *q = (int*)p;
  *q = 5;
}

是好的（可能是不明智的方式，但法律），只要FUNC传递给对象（）是一个可变的。

但是，如果传递的对象是常量合格的话它会被不确定的行为。所以你的情况，它的不的定义。

But if the object passed was const qualified then it would have been undefined behaviour. So in your case, it's not undefined.

限定符常量仅在该函数不应该修改合约 DEST 。它有一个对象的实际的可变性无关。因此，修改常量限定调用UB与否要看对象是否传递到有任何这样的限定词。

The qualifier const is only a contract that the function is not supposed to modify dest. It has no bearing on the actual mutability of an object. So the modifying a const-qualified invokes UB or not depends whether the object passed to has any such qualifier.

至于警告，海湾合作委员会（5.1.1）警告为：

As for the warning, GCC (5.1.1) warns for:

int func(const int *p) {
   fscanf(stdin, "%d", p);
}

与

warning: writing into constant object (argument 3) [-Wformat=]

大概VS不承认的fscanf（）修改对象。但C标准只是说，如果修改一个const限定对象是不确定的：

Probably VS doesn't recognize that fscanf() modifies the object. But the C standard only says that it's undefined if you modify a const-qualified object:

的（C11草案6.7.3，6型预选赛）的

如果试图修改与一个定义的对象
  通过使用一个左的非const限定的const限定型
  类型，行为是不确定的。

If an attempt is made to modify an object defined with a const-qualified type through use of an lvalue with non-const-qualified type, the behavior is undefined.

有没有诊断如果code调用不确定的行为由C标准要求。一般情况下，你是你自己，如果你的code使UB和编译器可能无法帮助你在所有的原因。

There's no diagnostic required by the C standard if the code invokes undefined behaviour. In general, you are on your own if your code causes UB and a compiler may not be able to help you in all causes.

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