您不能从 function 内部重新分配 C 中的 memory 块吗？

Question

So, while writeing a program I realised that when using realloc in a function outside of main if the original block of memory was declaired in main it doesnt seem to keep the changes outside of the function. EG

void main()
{

    int *ptr;

    //allocates memory
    ptr = calloc(4, sizeof(int));

    exampleFunction(&ptr);

} //end main


//this function reallocates the memory block of ptr
void exampleFunction(int *ptr)
{

    ptr = realloc(ptr, (sizeof(int) * 10));

} // end exampleFunction

Do I need to do something different or should this work fine? Also this is just example code and is not intended to be runnable

Extra info I am using MinGW on windows 10

Answer 1

You passes to the function the expression &ptr that has the type int ** .

exampleFunction(&ptr);

But the function parameter has the type int * .

void exampleFunction(int *ptr)

So the function declaration and its call do not make sense.

You have to declare and define the function at least like

//this function reallocates the memory block of ptr
void exampleFunction( int **ptr)
{

    *ptr = realloc( *ptr, (sizeof(int) * 10));

}

Though it will be better to use a temporary pointer with the call of realloc because the function can return NULL . In this case the original value of *ptr will be lost.

So you should declare the function like

//this function reallocates the memory block of ptr
int exampleFunction( int **ptr)
{
    int *tmp = realloc( *ptr, (sizeof(int) * 10));

    int success = tmp != NULL;

    if ( success ) *ptr = tmp;

    return success;

}

Answer 2

You can write like this.

void main()
{

    int *ptr;

    //allocates memory
    ptr = calloc(4, sizeof(int));

   ptr= exampleFunction(ptr);

}

int * exampleFunction(int *ptr)
{
    ptr = realloc(ptr, (sizeof(int) * 10));
  return(ptr);
}