C ++函數中的默認參數化是否將默認值推送到堆棧？

Question

如果一個函數有20個參數（指針）（示例中使用了4個），並且所有參數的默認值都指向NULL，那是否意味着每次調用此函數都會在運行時執行PUSH堆棧NULL值？

這樣的函數示例可能如下所示：

function test(val1=NULL, val2=NULL, .... val20=NULL)

我要求最大限度地提高速度，並在函數調用期間減少#cycle。

測試案例（這里有一些額外的東西）

// ArgsListTest.cpp
// @author Mathew Kurian

#include "stdafx.h"
#include <iostream>
#include <time.h>

using namespace std;

// Look at the difference in psuedo-assembly code
// My knowledge in compiler is little, but I can see that
// that there are unncessary cycles being wasted for this part.

// **** With array (this example) *****
// LOAD         Reg, memAddressOfArray
// WRITETOMEM   Reg, ptrToVar1
// INCREMENT    sp
// WRITETOMEM   Reg, ptrToVar2
// DECREMENT    sp
// PUSH         ptrToArray
// JUMP         test

//  ***** IDEALLY SUPPOSED TO LIKE THIS *****
// PUSH        ptrToVar2
// PUSH        ptrToVar1
// JUMP        test

class Base
{
public:
    virtual int test(void* arguments[])
    {
        cout << "Base function being called. VTable lookup ignored since there is no virtual." << endl;
        cout << *static_cast<int*>(arguments[0]) << endl;   // Parameter 1 (Thinks there is only 1 parameter!)

        int x = 5;      x += 1; // Random math to prevent optimizations. (I hope)
        return x;
    }
};

class Derived : public Base
{
public:
    virtual int test(void* arguments[])
    {
        // cout << "Derived function being called. VTable lookup during runtime. Slight overhead here!" << endl;

        // cout << *static_cast<string*>(arguments[0]) << endl; // Parameter 1
        // cout << *static_cast<int*>(arguments[1]) << endl;   // Parameter 2

        int x = 5;      x += 1;
        return x;
    }
};

class Base2
{
public:
    virtual int test(void* arg1 = NULL, void* arg2 = NULL, void* arg3 = NULL, void* arg4 = NULL)
    {
        // cout << "Base2 function being called. VTable lookup ignored since there is no virtual." << endl;

        // cout << *static_cast<string*>(arg1) << endl; // Parameter 1
        // cout << *static_cast<int*>(arg2) << endl;   // Parameter 2
        int x = 5;      x += 1;
        return x;
    }

    virtual int test2()
    {
        int x = 5;      x += 1;
        return x;
    }
};

int _tmain(int argc, _TCHAR* argv[])
{
    Base * base = new Derived;
    Base2 * base2 = new Base2;

    int r = 0;
    string * str = new string("sunny");
    int * vale = new int(20);
    int iterations = 1000000000;

    //================================================================================

    printf("Using No-Parameters [%d iterations]\n", iterations);

    clock_t tStart = clock();

    for (int x = 0; x < iterations; x++)
    {
        r = base2->test2();
    }

    printf("Time taken: %.9fs\n", (double)(clock() - tStart) / CLOCKS_PER_SEC);

    //================================================================================

    printf("Using Array [%d iterations]\n", iterations);

    tStart = clock();

    for (int x = 0; x < iterations; x++)
    {
        void * arguments[] = { str, vale };
        r = base->test(arguments);
    }

    printf("Time taken: %.9fs\n", (double)(clock() - tStart) / CLOCKS_PER_SEC);

    //================================================================================

    printf("Using Default-Parameters [%d iterations]\n", iterations);

    tStart = clock();

    for (int x = 0; x < iterations; x++)
    {
        r = base2->test(str, vale);
    }

    printf("Time taken: %.9fs\n", (double)(clock() - tStart) / CLOCKS_PER_SEC);

    //================================================================================

    // cout << "NOTE: Derived class has no extra methods although the parameter counts are different.\n      Parent class doesn't even realize parameter 1 exists!" << endl;

    std::getchar();

    return 0;
}

輸出（不會產生邏輯感）

為什么第一次測試慢於第二次？ 如何創建一個數組實際上比PUSH NULL更快？

Answer 1

也許虛擬方法會影響您的結果。 我將您的方法復制到全局范圍，演示了不同的性能順序。

Using No-Parameters [1000000000 iterations]
Time taken: 24.831000000s
Using Array [1000000000 iterations]
Time taken: 24.730000000s
Using Default-Parameters [1000000000 iterations]
Time taken: 25.241000000s
Using No-Parameters [1000000000 iterations] on int testA()
Time taken: 21.664000000s
Using Array [1000000000 iterations] on int testB(void* arguments[])
Time taken: 22.384000000s
Using Default-Parameters [1000000000 iterations] int testC(void* arg1 = NULL, ...)
Time taken: 22.329000000s

編輯：

testB中數組賦值的相同測試移出for循環的范圍：

Using No-Parameters [1000000000 iterations]
Time taken: 24.713000000s
Using Array [1000000000 iterations]
Time taken: 24.686000000s
Using Default-Parameters [1000000000 iterations]
Time taken: 25.225000000s
Using No-Parameters [1000000000 iterations] on int testA()
Time taken: 21.653000000s
Using Array [1000000000 iterations] on int testB(void* arguments[])
Time taken: 21.896000000s
Using Default-Parameters [1000000000 iterations] int testC(void* arg1 = NULL, ...)
Time taken: 22.353000000s

Answer 2

也許。 另一個簡單的解決方案是擁有多個入口點：

  # Pseudocode
  PUSH nullptr # Default argument 20
  PUSH nullptr # Default argument 19
  PUSH nullptr # Default argument 18
  ...
  ENTRYPOINT(test)

如果您有N個默認參數，則跳轉到常規入口點前面的地址N指令。 好處：推送這些默認參數的代碼現在可以在調用者之間共享。 但是，需要一點鏈接器智能。