代码中指针的不同内存分配

Question

为什么这两个代码在其糊状的内存位置之间存在不同的内存空间差异？

查看函数_1并在代码1和2中输出，以更好地理解问题。

代码1

#include <iostream>
using namespace std;

static int *function_1()
{
static int *pointer_1;// main point to be looked at in code 1
static int variable_1 = 77;
pointer_1 = &variable_1;
return pointer_1;
}
static int *function_2()
{
static int variable_2 = 777;
static int *pointer_2 = &variable_2;
return pointer_2;
}
static int *function_3()
{
static int variable_3 = 77;
static int *pointer_3 = &variable_3;
return pointer_3;
}
int main()
{
cout << "function - 1 referenced = " << function_1() << "  derefered = " << * 
(function_1()) << " fuc1 size =" << sizeof(function_1()) << endl;
cout << "function - 2 referenced = " << function_2() << "  derefered = " << * 
(function_2()) << " fuc2 size =" << sizeof(function_2()) << endl;
cout << "function - 3 referenced = " << function_3() << "  derefered = " << * 
(function_3()) << " fuc2 size =" << sizeof(function_2()) << endl;
return 0;
}

输出-1

code 2
#include <iostream>
using namespace std;

static int *function_1()
{
static int variable_1 = 77;
static int* pointer_1 = &variable_1; //main point to be looked at code 2 
return pointer_1;
}
static int *function_2()
{
static int variable_2 = 777;
static int *pointer_2 = &variable_2;
return pointer_2;
}
static int *function_3()
{
static int variable_3 = 77;
static int *pointer_3 = &variable_3;
return pointer_3;
}
int main()
{
cout << "function - 1 referenced = " << function_1() << "  derefred = " << * 
(function_1()) << " fuc1 size =" << sizeof(function_1()) << endl;
cout << "function - 2 referenced = " << function_2() << "  derefred = " << * 
(function_2()) << " fuc2 size =" << sizeof(function_2()) << endl;
cout << "function - 3 referenced = " << function_3() << "  derefred = " << * 
(function_3()) << " fuc2 size =" << sizeof(function_2()) << endl;
return 0;
}

输出-2

“ alt = ”

输出

i.sstatic.net/g1fn7.png

2 “

> 4206612 //主要点

0x403020-4206624

在代码2中，我们有输出

hexaddress-转换后的十进制地址

0x403010-4206608

0x403020-4206624 //主要点

0x4030303030-40x4030-4206640

为什么在继续存在如此差异的情况下，只是在继续存在的位置，并且因为宣传而宣传，并且在宣传，并且最初是宣传，并且最初是宣传的，又是宣传的，一个步骤（代码2）和2步（代码1）

Answer 1

根据输出，指针的大小等于8，类型int的对象的大小等于4。

输出取决于如何在记忆中放置在功能中声明的静态存储持续时间的对象。

在第一个演示程序中，我们将

static int *pointer_1;// main point to be looked at in code 1
static int variable_1 = 77;
static int variable_2 = 777;
static int *pointer_2 = &variable_2;
static int variable_3 = 77;
static int *pointer_3 = &variable_3

Pointer Pointer_1指向变量变量_1，指针Pointer_2指向变量变量_2。 Between these two variables, variable_1 and variable_2, there is no gap due to possible alingment.因此，输出是

0x403010
0x403014

变量变量_2和变量_3之间的输出。结果，两个变量之间有12个字节（可变_2和8个字节占据的4个字节由PointerPointer_2。8 + 4 = 12占用）。因此，您的

0x403010
0x403014
0x403020

方式可以分析其他程序的输出，即由于变量后的指针，变量和指针之间可能存在差距。

例如，可以将Pointer_1之后放置的变量_2附加到4个字节上，以确保指针Pointer_2的对齐（使Pointer Pointer pointer_2可将内存的地址划分为8）

static int pointer_1 = &variable_1;
static int variable_2 = 777;
static int *pointer_2 = &variable_2;