c - 编译器如何理解给定函数是按值传递还是按引用传递？

Question

C 编译器如何理解给定函数是按值传递还是按引用传递？如果变量的指针作为整数(按值传递)传递给函数会发生什么？或者这在 C 中可能吗？

即:将一个变量的地址复制到另一个变量(int)，并将该变量传递给函数。这里被调用的函数将获得一个地址作为普通的整数参数。这在C语言中可能吗？如果不是为什么？

Answer 1

在 C 中，只能按值传递(仅在 C++ 中支持按引用传递)。

标准变量和指针变量都是按值传递的。因此，编译器不得对引用或值进行任何检查。

标准变量和指针变量是一样的，都存储值，但是指针变量只能存储地址值，并且支持解引用运算符“*”，该运算符获取指针变量指向的值。

因此，当您按值传递(C 仅支持按值)指针变量时，指针变量的值将被复制到本地指针变量(在函数堆栈中实例化)中，该变量是功能。因此，您可以使用解引用运算符访问复制到本地指针变量中的地址值。

添加评论 2014 年 8 月 22 日

我可以用一个例子更好地解释:

void FuncSet10(int* PtrArgument)
{
    // The PtrArgument is a local pointer variable
    // it is allocated in the stack. The compiler
    // copies the MyIntVariable address in this variable

    // Using dereferencing operator I can access to
    // memory location pointed by PtrArgument
    *PtrArgument = 10;


}

void FuncSet20(int IntArgument)
{
    // The IntArgument is a local variable
    // allocated in the stack. The compiler
    // copies the MyIntVariable address in this variable

    // this code emulates the dereferencing operator
    // and so poiter mechanism.
    *((int*)IntArgument) = 20;
}

void FuncMovePointer(int* pointer)
{
    int a = *pointer++;  // now a contains 1
    int b = *pointer++;  // now b contains 2
    int c = *pointer++;  // now c contains 3

    printf("a contains %d\n", a);
    printf("b contains %d\n", b);
    printf("c contains %d\n", c);

    // The *Ptr now points to fourth (value 4) element of the Array
    printf("pointer points to %d\n", *pointer);
}


int _tmain(int argc, _TCHAR* argv[])
{

    int Array[] = {1, 2, 3, 4}; // Array definition
    int* pointer = Array;       // pointer points to the array

    int MyIntVariable;

    // First example for explanation of
    // the pointer mechanism and compiler actions

    // After calling this function MyIntVariable contains 10
    FuncSet10(&MyIntVariable);

    printf("MyIntVariable contains %d\n", MyIntVariable);

    // After calling this function  MyIntVariable contains 20
    // This code emulate pointer mechanism
    FuncSet20((int)&MyIntVariable);

    printf("MyIntVariable contains %d\n", MyIntVariable);

    // Second example to demonstrate that a pointer
    // is only a local copy in a called function

    // Inside function the pointer is incremented with ++ operator
    // so it will point the next element before returning by function
    // (it should be 4)
    FuncMovePointer(pointer);

    // But this is not true, it points always first element! Why?
    // Because the FuncMovePointer manages a its local copy!

    printf("but externally the FuncMovePointer it still points to first element %d\n", *pointer);

    return 0;
}

希望上面的代码可以帮助您更好地理解。

安杰洛