这是C中未定义的行为吗?如果没有逻辑上的预测输出 [英] Is this undefined behaviour in C ? If not predict the output logically

问题描述

代码1

#include <stdio.h>
int f(int *a, int b) 
{
  b = b - 1;
  if(b == 0) return 1;
  else {
    *a = *a+1;

    return *a + f(a, b);
  }
}

int main() {
  int X = 5;
  printf("%d\n",f(&X, X));
}

考虑此C代码.这里的问题是预测输出.从逻辑上讲，我的输出是31. (计算机上的输出)

Consider this C code. The question here is to predict the output. Logically, I get 31 as ouput. (Output on machine)

当我将return语句更改为

When I change the return statement to

return f(a, b) + *a;

我在逻辑上是37.(计算机上的输出)

I logically get 37. (Output on machine)

我的一个朋友说，在

return *a + f(a, b);

我们计算树的深度的值，即*首先计算然后调用f(a, b)的值，而在

we compute the value of a while going depth of the tree, i.e. *a first computed then f(a, b) is called, whereas in

return f(a,b) + *a;

返回时解析，即先计算f(a, b)然后调用*a.

It is resolved while returning back, i.e. f(a, b) is computed first then *a is called.

使用这种方法，我尝试自己预测以下代码的输出:

With this approach, I tried predicting the output of the following code myself:

代码2

#include <stdio.h>
int foo(int n) 
{
    static int r;
    if(n <= 1)
        return 1;

    r = n + r;
    return r + foo(n - 2);
} 

int main () {
   printf("value : %d",foo(5));
}

对于return(r+foo(n-2));

我在逻辑上得到14作为输出(计算机上的输出)

Im getting 14 as Output Logically (Output on machine)

对于return(foo(n-2)+r);

我得到17作为输出. (计算机上的输出)

I get 17 as output. (Output on machine)

但是，当我在系统上运行代码时，两种情况下我都得到17.

However, when I run the code on my system I get 17 in both cases.

我的问题:

• 我朋友给的方法正确吗?
• 如果是这样，为什么在计算机上运行时为什么在代码2 中得到相同的输出?
• 否则，解释代码1 和代码2的正确方法是什么?
• 因为C不支持通过引用传递，所以是否存在任何未定义的行为?由于它在代码1 中使用，很难使用指针来实现吗?

• Is the approach given by my friend correct?
• If so, why do I get the same output in Code 2 when I run in a machine?
• If not, what is the correct way to interpret Code 1 and Code 2?
• Is there any undefined behaviour because C does not support pass by Reference? As it is being used in Code 1 tough it can be implemented using pointers?

简而言之，我只是想知道在上述4种情况下预测输出的正确方法.

In a nutshell, I simply wanted to know the correct way to predict the output in the 4 cases mentioned above.

推荐答案

代码1

对于代码1 ，因为标准未指定return *a + f(a, b);(和return f(a, b) + *a;)中术语的评估顺序，并且该函数修改了a为指向，您的代码具有未指定的行为，并且可能有各种答案.

Code 1

For Code 1, because the order of evaluation of the terms in return *a + f(a, b); (and in return f(a, b) + *a;) is not specified by the standard and the function modifies the value that a is pointing at, your code has unspecified behaviour and various answers are possible.

正如您从评论中的愤怒中可以看出的那样，术语未定义的行为"，未指定的行为"等在C标准中具有技术含义，并且该答案的早期版本在应有的地方误用了未定义的行为".使用了未指定".

As you can tell from the furor in the comments, the terms 'undefined behaviour', 'unspecified behaviour' and so on have technical meanings in the C standard, and earlier versions of this answer misused 'undefined behaviour' where it should have used 'unspecified'.

问题的标题是这是C中的未定义行为吗?"，答案是否；这是未指定的行为，而不是未定义的行为".

The title of the question is "Is this undefined behaviour in C?", and the answer is "No; it is unspecified behaviour, not undefined behaviour".

对于固定的代码2 ，该函数还具有未指定的行为:静态变量r的值通过递归调用进行更改，因此对评估顺序的更改可能会更改结果.

For Code 2 as fixed, the function also has unspecified behaviour: the value of the static variable r is changed by the recursive call, so changes to the evaluation order could change the result.

对于代码2 (最初与int f(static int n) { … }一起显示)，该代码不会(或至少不应)编译.函数的参数定义中唯一允许的存储类是register，因此static的出现应该给您带来编译错误.

For Code 2, as originally shown with int f(static int n) { … }, the code does not (or, at least, should not) compile. The only storage class permitted in the definition of an argument to a function is register, so the presence of static should be giving you compilation errors.

ISO/IEC 9899:2011§6.7.6.3函数声明符(包括原型) ¶2在参数声明中唯一出现的存储类说明符是register.

ISO/IEC 9899:2011 §6.7.6.3 Function declarators (including prototypes) ¶2 The only storage-class specifier that shall occur in a parameter declaration is register.

在macOS Sierra 10.12.2上使用GCC 6.3.0进行编译，如下所示(请注意，无需额外的警告):

Compiling with GCC 6.3.0 on macOS Sierra 10.12.2, like this (note, no extra warnings requested):

$ gcc -O ub17.c -o ub17
ub17.c:3:27: error: storage class specified for parameter ‘n’
 int foo(static int n)
                    ^

否；它根本不像所示那样编译-至少对于我而言，不是使用现代版本的GCC.

No; it doesn't compile at all as shown — at least, not for me using a modern version of GCC.

但是，假设该问题是固定的，该函数还具有 undefined 未指定的行为:静态变量r的值通过递归调用进行更改，因此对评估顺序的更改可能会更改结果.

However, assuming that is fixed, the function also has undefined unspecified behaviour: the value of the static variable r is changed by the recursive call, so changes to the evaluation order could change the result.

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