通过 C 中的函数传递指针变量

Question

我在这里遇到了一些奇怪的行为。任何帮助都会很棒。

我是这样开始的：

   int *event_positions = (int *) malloc(1 * sizeof(int)); // let us start with 1 and then add more within the method. This should continue until we have all the flags we want.  
   int number_of_flags = event_extractor(vocal_data, size, event_positions);


 // HERE I WOULD LIKE TO USE THE VALUES OF event_positions BUT THE ARE WEIRD I.E. THEY DON'T MATCH THE VALUES BEING PRINTED IN THE LAST METHOD. 

事件提取器然后将变量传递给另一个方法。简化后如下所示：

int event_extractor (int *audio_samples, unsigned int size_of_audio ,int *event_flags)
{

    int number_of_flags = apply_threshold (lopass_samples, length, event_flags);
    // PRINT ARRAY event_flags HERE 
    // VALUES ARE INCORRECT AND WEIRD

}

最后一种方法：

int apply_threshold (int *audio_samples, unsigned int size_of_audio, int *event_flags)
{
// DO SOME STUFF HERE.
// PRINT THE ARRAY WHICH SHOW THE  CORRECT VALUES.



}

我希望这很清楚。基本上我有一个数组，我将其作为参数传递，并且在方法完成后无法访问这些值。

---

编辑 1

第一个文件：

   int *event_positions = (int *) malloc(1 * sizeof(int)); // let us start with 1 and then add more within the method. This should continue until we have all the flags we want.  
   int number_of_flags = event_extractor(vocal_data, size, event_positions);

第二个文件：

  int apply_threshold (int *audio_samples, unsigned int size_of_audio, int *event_flags)
{


int flag = 0; // this will be the number of flags that I have 
bool run = true; // this will make sure that a minimum amount of time passes before I grab another flag. It's a guard.
int counter = 0; // this is the counter for the above guard. 





printf("\n\nCURRENT MINIMUM TIME:  20100 SAMPLES \n\n");

// event_flags[0] = 1; // this first one is a dud. within the loop we will automatically start adding flags


int threshold = calculate_threshold_value(audio_samples, size_of_audio);

printf("\n\n this is the threshold %d \n\n", threshold);

int length = (int)size_of_audio;

for (int i = 0; i < length; i++) 
{

    if (audio_samples[i] > threshold  && run) 
    {

        // ** is this realloc working ?
        event_flags = (int*)realloc(event_flags, sizeof(int) * (flag+1)); // reallocate the size of the array
        event_flags[flag] = i;
        // printf("FLAG CREATED! %i\n ", i);
        printf("EVENT FLAG %i  %i\n",flag, event_flags[flag] );
        flag++;
        run = false;

    }   

    if (!run) {
        counter++;
        if (counter > 20100) { // hardcode minimum size for now. 
            counter = 0;
            run=true;
        }
    }

}

printf("\n\n\n NUMBER OF EVENTS --- %d\n", flag);

for (int i = 0; i < flag; i++) {
    printf("FLAG %i  -- %d\n", i, event_flags[i]);
}



printf("\nFIVE samples before and after my second flag: \n 0 should indicate a reach in the threshold\n");

for (int i = 0; i <10 ; i++) {
    printf("VOCAL SAMPLE %i  %i \n", i-5,audio_samples[event_flags[1]+i-5] );
}


return flag;



}

---

编辑 2

我已经根据 Erik 的模型更新了我的代码。我的循环现在看起来像这样

    if (audio_samples[i] > threshold  && run) 
    {

        // ** is this realloc working ?
        // event_flags = (int*)realloc(event_flags, sizeof(int) * (flag+1));
        *event_flags = (int*)realloc(*event_flags, sizeof(int) * (flag+1)); // reallocate the size of the array
        *event_flags[flag] = i;
        // printf("FLAG CREATED! %i\n ", i);
        printf("EVENT FLAG %i  %i\n",flag, *event_flags[flag] );
        flag++;
        run = false;

    }   

现在我收到一个看起来像这样的错误。有什么想法吗？

Answer 1

您是否在 apply_threshold 中重新分配 event_flags？如果是这样，你需要让调用者取回更新后的指针。

类似：

int apply_threshold (int *audio_samples, unsigned int size_of_audio, int **event_flags) {
    *event_flags = realloc ...
}

...

int number_of_flags = apply_threshold (lopass_samples, length, &event_flags);

编辑：针对更新的问题：

event_flags = (int*)realloc(event_flags, sizeof(int) * (flag+1)); // reallocate the size of the array

这会更改 event_flags 指针的本地副本。来电者不会看到变化。使用我上面描述的方法。

EDIT2：更详细的示例。

void foo(int * v) {
    v = 0; // The local copy of main's myvar is now 0. main's actual myvar is unchanged
}

void bar(int ** v) {
    *v = 0; // Main's myvar is now 0, we have a pointer to it and can modify it.
}

int main() {
    int * myvar = (int *) malloc(1); // Allocate 1 byte and make myvar point at this byte.
    foo(myvar); // Call foo, passing a *copy of* myvar, which also points at the allocated byte
    bar(&myvar); // Call bar, passing a *pointer to* myvar, which again points to the allocated byte
}

EDIT3：针对新问题。

你的“长度”是整数个数还是字节数？您将其视为整数数，如果它确实是字节数，则可能会导致您的错误。

Answer 2

您需要传递指向指针的指针或指向指针的引用。 realloc 可能会将您的内存移动到另一个位置，而调用者不会注意到。

  realloc() changes the size of the memory block pointed to by ptr to size bytes.  The contents will be unchanged to the minimum of the old and new sizes; newly allocated memory will be uninitialized.  If ptr is NULL, then  the
   call  is  equivalent to malloc(size), for all values of size; if size is equal to zero, and ptr is not NULL, then the call is equivalent to free(ptr).  Unless ptr is NULL, it must have been returned by an earlier call to mal‐

loc()、calloc() 或 realloc()。 如果指向的区域被移动，则执行 free(ptr)。