函数调用中的Malloc似乎在返回时被释放? [英] Malloc inside a function call appears to be getting freed on return?

问题描述

我认为我已经将其归结为最基本的情况:

I think I've got it down to the most basic case:

int main(int argc, char ** argv) {
  int * arr;

  foo(arr);
  printf("car[3]=%d\n",arr[3]);
  free (arr);
  return 1;
}

void foo(int * arr) {
  arr = (int*) malloc( sizeof(int)*25 );
  arr[3] = 69;
}

输出是这样的:

> ./a.out 
 car[3]=-1869558540
 a.out(4100) malloc: *** error for object 0x8fe01037: Non-aligned pointer
                         being freed
 *** set a breakpoint in malloc_error_break to debug
>

如果有人能阐明我的理解失败的地方，将不胜感激.

If anyone can shed light on where my understanding is failing, it'd be greatly appreciated.

推荐答案

您按值传递指针，而不是按引用传递指针，因此，在foo内部对arr所做的任何操作都不会在foo函数之外产生任何影响. 正如m_pGladiator所写的那样，一种方法是这样声明对指针的引用(仅在C ++中可用，btw.C不知道引用):

You pass the pointer by value, not by reference, so whatever you do with arr inside foo will not make a difference outside the foo-function. As m_pGladiator wrote one way is to declare a reference to pointer like this (only possible in C++ btw. C does not know about references):

int main(int argc, char ** argv) {
  int * arr;

  foo(arr);
  printf("car[3]=%d\n",arr[3]);
  free (arr);
  return 1;
}

void foo(int * &arr ) {
  arr = (int*) malloc( sizeof(int)*25 );
  arr[3] = 69;
}

另一种(更好的恕我直言)的方法是不将指针作为参数传递而返回指针:

Another (better imho) way is to not pass the pointer as an argument but to return a pointer:

int main(int argc, char ** argv) {
  int * arr;

  arr = foo();
  printf("car[3]=%d\n",arr[3]);
  free (arr);
  return 1;
}

int * foo(void ) {
  int * arr;
  arr = (int*) malloc( sizeof(int)*25 );
  arr[3] = 69;
  return arr;
}

您可以将指针传递给指针.这就是引用传递的C方法.语法有点复杂，但是很好-这就是C的方式...

And you can pass a pointer to a pointer. That's the C way to pass by reference. Complicates the syntax a bit but well - that's how C is...

int main(int argc, char ** argv) {
  int * arr;

  foo(&arr);
  printf("car[3]=%d\n",arr[3]);
  free (arr);
  return 1;
}

void foo(int ** arr ) {
  (*arr) = (int*) malloc( sizeof(int)*25 );
  (*arr)[3] = 69;
}

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