打印指针时行为怪异 [英] Strange behaviour when printing pointers

问题描述

我有以下的code：

#include <stdio.h>

typedef struct {
  int* arg1;
  int  arg2;
} data;

int main(int argc, char** argv) {
  data d;
  printf("arg1: %p | arg2: %d\n", d.arg1, d.arg2);
}

输出最终被该 d.arg1 不是 NULL 和 D。 ARG2 0。例如：

The output ends up being that d.arg1 is not NULL and d.arg2 is 0. For example:

arg1: 0x7fff84b3d660 | arg2: 0

当我添加一个指向主，没有什么变化。然而，当我打印指针：

When I add a pointer to main, nothing changes. However, when I print that pointer:

#include <stdio.h>

typedef struct {
  int* arg1;
  int  arg2;
} data;

int main(int argc, char** argv) {
  data d;
  int* test;
  printf("arg1: %p | arg2: %d | test: %p\n", d.arg1, d.arg2, test);
}

输出总是导致：

arg1: (nil) | arg2: 4195264 | test: (nil)

为什么我遇到这种行为？我不知道如何打印另一个指针值改变不同的指针为NULL值。请注意，我现在用的编译器GCC 4.8.2。

Why am I experiencing this behaviour? I don't understand how printing another pointer value changes the value of a different pointer to NULL. Note that the compiler I am using is GCC 4.8.2.

推荐答案

您遇到此行为是因为你的程序调用的未定义行为即可。 D 在你的程序uinitialized因此其值为不确定这调用不确定的行为。

You are experiencing this behavior because your program invokes undefined behavior. d is uinitialized in your program and hence its value is indeterminate and this invokes undefined behavior.

C11 6.7.9初始化：

如果具有自动存储时间的对象没有明确初始化，它的价值是不确定的。

If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate.

这个答案：

这是不确定的值的甚至比未指定的更不确定。这是不是一个未确定的值或重新presentation一个陷阱。陷阱再presentation是对于一些魔法值，如果你尝试将其分配到任何东西，结果在不确定的行为当中，标准的说话。这不必是实际值;大概想它最好的办法就是如果C有异常，陷阱重新presentation将是一个例外。例如，如果你声明 INT I; 在块，没有初始化，变量的初始值i 是不确定的，也就是说如果你尝试将此分配给初始化之前别的东西，这种行为是未定义，编译器有权尝试说恶魔出鼻子的帽子戏法。当然，在大多数情况下，编译器会做那么戏剧性的东西/好玩，喜欢它初始化为0或其他一些随机的有效值，但不管它做什么，你无权反对。

An indeterminate value even more unspecified than unspecified. It's either an unspecified value or a trap representation. A trap representation is standards-speak for some magic value which, if you try to assign it to anything, results in undefined behavior. This wouldn't have to be an actual value; probably the best way to think about it is "if C had exceptions, a trap representation would be an exception". For example, if you declare int i; in a block, without an initialization, the initial value of the variable i is indeterminate, meaning that if you try to assign this to something else before initializing it, the behavior is undefined, and the compiler is entitled to try the said demons-out-of-nose trick. Of course, in most cases, the compiler will do something less dramatic/fun, like initialize it to 0 or some other random valid value, but no matter what it does, you're not entitled to object.

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