在 C 中，我理解为什么不在指针返回函数中返回局部变量的地址，但我该如何解决？

Question

I have the following code:

int* foo(){
int x = 15;
return &x; }

Which I understand why not to do since the local variable address gets erased from the stack after the function finishes and it becomes a dangling pointer. The question is, how do I not make it a dangling variable without making xa static variable

Answer 1

Either allocate memory from the heap inside the function

int *f() {
  int *foo = malloc(sizeof(int));
  if(!foo) {
    // Do appropriate error handling here
  }
  return foo;
}

Don't forget to free it at some point though.

Or you pass in a pointer to a variable living outside the function:

void f(int *foo) {
  *foo = 42;
}

void g() {
  int goo;
  f(&goo);
}

Answer 2

The blessed ways are:

• return a value and not an address

   int foo(){ int x = 15; return x; }

• have the caller to provide an address

   int *foo(int *x) { *x = 15; return x; }

  or

   void foo(int *x) { *x = 15; }

• Return dynamic (allocated) memory:

   int *foo() { int *x = malloc(sizeof(*x)); // should test valid allocation but omitted for brievety *x = 15; return x; }

  Beware, the caller will take ownership or the allocated memory and is responsable to free it later.