返回一个指向自动变量 [英] Returning a pointer to an automatic variable

问题描述

假设你有以下功能：

char *getp()
{
    char s[] = "hello";
    return s;
}

由于函数是在函数返回一个指针到一个局部变量来使用外，它会导致内存泄漏？

Since the function is returning a pointer to a local variable in the function to be used outside, will it cause a memory leak?

的 P.S。我仍然在学习C所以我的问题可能有点天真... 的

[更新]

所以，如果说你要返回一个新的的char [] 阵列（即也许对于一个子功能），你怎么只返回？它应该是指向一个外部变量？即的char [] 不在本地的功能？

[Update]
So, if say you want to return a new char[] array (ie maybe for a substring function), what do you return exactly? Should it be pointer to an external variable ? ie a char[] that is not local to the function?

推荐答案

这不会导致内存泄漏。它会导致悬挂引用。局部变量是在栈中分配，并会尽快，因为它超出范围释放。其结果是，当函数结束时，指针，你正在返回不再指向你自己的记忆。这不是一个内存泄漏（内存泄漏是当你分配一些内存，并且不释放它）。

It won't cause a memory leak. It'll cause a dangling reference. The local variable is allocated on the stack and will be freed as soon as it goes out of scope. As a result, when the function ends, the pointer you are returning no longer points to a memory you own. This is not a memory leak (memory leak is when you allocate some memory and don't free it).

[更新] ：
为了能够返回一个函数分配的数组，你应该分配它之外的堆栈（例如，在堆），如：

[Update]: To be able to return an array allocated in a function, you should allocate it outside stack (e.g. in the heap) like:

char *test() {
    char* arr = malloc(100);
    arr[0] = 'M';
    return arr;
}

现在，如果你不免费在调用函数的内存，你使用它后完成后，你有内存泄漏。

Now, if you don't free the memory in the calling function after you finished using it, you'll have a memory leak.

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