用C返回局部变量的地址 [英] Return address of local variable in C

问题描述

说我有以下两个功能：

1

int * foo()
{
  int b=8;
  int * temp=&b;
  return temp;
}

2

int * foo()
{
   int b=8;
   return &b;
}

我没有得到的第一个任何警告（比如函数返回一个局部变量的地址的），但我知道这是因为非法b 从堆栈中消失，我们只剩下一个指向不确定的内存。

I don't get any warning for the first one (e.g. function returns address of a local variable) but I know this is illegal since b disappears from the stack and we are left with a pointer to undefined memory.

所以，当我需要小心返回临时值的地址？

So when do I need to be careful about returning the address of a temporary value?

推荐答案

你没有得到的第一个片段一个警告的原因是因为你是不是（从编译器的角度来看）返回一个地址给一个局部变量。

The reason you are not getting a warning in the first snippet is because you aren't (from the compiler's perspective) returning an address to a local variable.

您在返回整型值*温度。尽管这个变量可能包含一个值，它是一个局部变量的地址是（在这个例子），编译器将不上浮code执行栈，看看是否属于这种情况。

You are returning the value of int * temp. Even though this variable might be (and in this example is) containing a value which is an address of a local variable, the compiler will not go up the code execution stack to see whether this is the case.

^{注意：这两个片段都同样糟糕，即使你的编译器不会警告你关于前者。不要使用这种方法。的}

您应该返回到本地变量地址时，一定要加倍小心;作为一项规则，你可以说你的从不应

---

You should always be careful when returning addresses to local variables; as a rule, you could say that you never should.

静态变量是一个完全不同的情况下，虽然，这是在<一个正在讨论href=\"http://stackoverflow.com/questions/453696/is-returning-a-pointer-to-a-static-local-variable-safe\">this螺纹。

static variables are a whole different case though, which is being discussed in this thread.

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