相关练习和测试代码 C指针 1."带*类型" 的特征探测：宽度 宽度探测 带一个*的探测 源代码 #include "stdafx.h" #include string.h void fun() { char* a = (char*)1; short* b = (short*)2; int* c =(int*)3; printf("%d
相关练习和测试代码
C指针1."带*类型" 的特征探测：宽度
宽度探测
带一个*的探测
源代码
#include "stdafx.h"
#include <string.h>
void fun()
{
char* a = (char*)1;
short* b = (short*)2;
int* c =(int*)3;
printf("%d %d %d\n",a,b,c);
}
int main(int argc, char* argv[])
{
fun();
return 0;
}
反汇编代码
9: char* a = (char*)1;
0040B818 mov dword ptr [ebp-4],1
10: short* b = (short*)2;
0040B81F mov dword ptr [ebp-8],2
11: int* c =(int*)3;
0040B826 mov dword ptr [ebp-0Ch],3
上述显示都是站4个字节
带两个*的探测
源代码
#include "stdafx.h"
#include <string.h>
void fun()
{
char** a = (char**)1;
short** b = (short**)2;
int** c =(int**)3;
printf("%d %d %d\n",a,b,c);
}
int main(int argc, char* argv[])
{
fun();
return 0;
}
反汇编代码
9: char** a = (char**)1;
0040B818 mov dword ptr [ebp-4],1
10: short** b = (short**)2;
0040B81F mov dword ptr [ebp-8],2
11: int** c =(int**)3;
0040B826 mov dword ptr [ebp-0Ch],3
上述显示还是都站4个字节
带两个以上*的探测
源代码
#include "stdafx.h"
#include <string.h>
void fun()
{
char******* a = (char*******)1;
short******* b = (short*******)2;
int******* c =(int*******)3;
printf("%d %d %d\n",a,b,c);
}
int main(int argc, char* argv[])
{
fun();
return 0;
}
反汇编代码
9: char******* a = (char*******)1;
0040B818 mov dword ptr [ebp-4],1
10: short******* b = (short*******)2;
0040B81F mov dword ptr [ebp-8],2
11: int******* c =(int*******)3;
0040B826 mov dword ptr [ebp-0Ch],3
上述显示还是都站4个字节
结论：
不管什么类型，不管带多少个*，她们所站的字节大小都是4个字节；
2."带*类型"的特征探测：声明
上述1题目中已经探测了，任何类型都可以带*，加了*就是新类型，然后可以加很多个* ，而且写法是标准最好推荐的写法是类似:int* x; 这样方便阅读
结论：
(1)带有*的变量类型的标准写法：变量类型* 变量名
(2)任何类型都可以带* 加上*以后是新的类型
(3)*可以是任意多个
3."带*类型" 的特征探测：赋值
上述1题目已经探测了，测试赋值操作；
结论：
1.带*类型的变量赋值时只能使用"完整写法".
2.带*类型的变量宽度永远是4字节,无论类型是什么,无论有几个*.
4."带*类型" 的特征探测：++ --
带一个*的测试
源代码
#include "stdafx.h"
#include <string.h>
void fun()
{
char* a;
short* b;
int* c;
a = (char*)100;
b = (short*)100;
c = (int*)100;
a++;
b++;
c++;
printf("%d %d %d\n",a,b,c);
}
int main(int argc, char* argv[])
{
fun();
return 0;
}
运行结果：101 102 104
反汇编代码
9: char* a;
10: short* b;
11: int* c;
12:
13: a = (char*)100;
0040B818 mov dword ptr [ebp-4],64h
14: b = (short*)100;
0040B81F mov dword ptr [ebp-8],64h
15: c = (int*)100;
0040B826 mov dword ptr [ebp-0Ch],64h
16:
17:
18: a++;
0040B82D mov eax,dword ptr [ebp-4]
0040B830 add eax,1
0040B833 mov dword ptr [ebp-4],eax
19: b++;
0040B836 mov ecx,dword ptr [ebp-8]
0040B839 add ecx,2
0040B83C mov dword ptr [ebp-8],ecx
20: c++;
0040B83F mov edx,dword ptr [ebp-0Ch]
0040B842 add edx,4
0040B845 mov dword ptr [ebp-0Ch],edx
带两个*的测试
#include "stdafx.h"
#include <string.h>
void fun()
{
char** a;
short** b;
int** c;
a = (char**)100;
b = (short**)100;
c = (int**)100;
a++;
b++;
c++;
printf("%d %d %d\n",a,b,c);
}
int main(int argc, char* argv[])
{
fun();
return 0;
}
运行结果：104 104 104
反汇编代码
9: char** a;
10: short** b;
11: int** c;
12:
13: a = (char**)100;
0040B818 mov dword ptr [ebp-4],64h
14: b = (short**)100;
0040B81F mov dword ptr [ebp-8],64h
15: c = (int**)100;
0040B826 mov dword ptr [ebp-0Ch],64h
16:
17:
18: a++;
0040B82D mov eax,dword ptr [ebp-4]
0040B830 add eax,4
0040B833 mov dword ptr [ebp-4],eax
19: b++;
0040B836 mov ecx,dword ptr [ebp-8]
0040B839 add ecx,4
0040B83C mov dword ptr [ebp-8],ecx
20: c++;
0040B83F mov edx,dword ptr [ebp-0Ch]
0040B842 add edx,4
0040B845 mov dword ptr [ebp-0Ch],edx
带两个以上*的测试
源代码
#include "stdafx.h"
#include <string.h>
void fun()
{
char********* a;
short********* b;
int********* c;
a = (char*********)100;
b = (short*********)100;
c = (int*********)100;
a++;
b++;
c++;
printf("%d %d %d\n",a,b,c);
}
int main(int argc, char* argv[])
{
fun();
return 0;
}
反汇编代码
9: char********* a;
10: short********* b;
11: int********* c;
12:
13: a = (char*********)100;
0040B818 mov dword ptr [ebp-4],64h
14: b = (short*********)100;
0040B81F mov dword ptr [ebp-8],64h
15: c = (int*********)100;
0040B826 mov dword ptr [ebp-0Ch],64h
16:
17:
18: a++;
0040B82D mov eax,dword ptr [ebp-4]
0040B830 add eax,4
0040B833 mov dword ptr [ebp-4],eax
19: b++;
0040B836 mov ecx,dword ptr [ebp-8]
0040B839 add ecx,4
0040B83C mov dword ptr [ebp-8],ecx
20: c++;
0040B83F mov edx,dword ptr [ebp-0Ch]
0040B842 add edx,4
0040B845 mov dword ptr [ebp-0Ch],edx
运行结果：104 104 104
结论：
1.不带*类型的变量，++或者-- 都是加1 或者减1
2.带*类型的变量，可是进行++ 或者 --的操作
3.带*类型的变量，++ 或者 -- 新增(减少)的数量是去掉一个*后变量的宽度
5."带*类型" 的特征探测：加上/减去 一个整数
带一个*的测试
源代码
#include "stdafx.h"
#include <string.h>
void fun()
{
char* a;
short* b;
int* c;
a = (char*)100;
b = (short*)100;
c = (int*)100;
a = a + 6;
b = b + 6;
c = c + 6;
printf("%d %d %d\n",a,b,c);
}
int main(int argc, char* argv[])
{
fun();
return 0;
}
运行结果：106 112 124
反汇编代码
9: char* a;
10: short* b;
11: int* c;
12:
13: a = (char*)100;
0040B818 mov dword ptr [ebp-4],64h
14: b = (short*)100;
0040B81F mov dword ptr [ebp-8],64h
15: c = (int*)100;
0040B826 mov dword ptr [ebp-0Ch],64h
16:
17: a = a + 6;
0040B82D mov eax,dword ptr [ebp-4]
0040B830 add eax,6
0040B833 mov dword ptr [ebp-4],eax
18: b = b + 6;
0040B836 mov ecx,dword ptr [ebp-8]
0040B839 add ecx,0Ch
0040B83C mov dword ptr [ebp-8],ecx
19: c = c + 6;
0040B83F mov edx,dword ptr [ebp-0Ch]
0040B842 add edx,18h
0040B845 mov dword ptr [ebp-0Ch],edx
带两个*的测试
源代码
#include "stdafx.h"
#include <string.h>
void fun()
{
char** a;
short** b;
int** c;
a = (char**)100;
b = (short**)100;
c = (int**)100;
a = a + 6;
b = b + 6;
c = c + 6;
printf("%d %d %d\n",a,b,c);
}
int main(int argc, char* argv[])
{
fun();
return 0;
}
运行结果：124 124 124
反汇编代码
9: char** a;
10: short** b;
11: int** c;
12:
13: a = (char**)100;
0040B818 mov dword ptr [ebp-4],64h
14: b = (short**)100;
0040B81F mov dword ptr [ebp-8],64h
15: c = (int**)100;
0040B826 mov dword ptr [ebp-0Ch],64h
16:
17: a = a + 6;
0040B82D mov eax,dword ptr [ebp-4]
0040B830 add eax,18h
0040B833 mov dword ptr [ebp-4],eax
18: b = b + 6;
0040B836 mov ecx,dword ptr [ebp-8]
0040B839 add ecx,18h
0040B83C mov dword ptr [ebp-8],ecx
19: c = c + 6;
0040B83F mov edx,dword ptr [ebp-0Ch]
0040B842 add edx,18h
0040B845 mov dword ptr [ebp-0Ch],edx
带两个以上*的测试
源代码
#include "stdafx.h"
#include <string.h>
void fun()
{
char********* a;
short********* b;
int********* c;
a = (char*********)100;
b = (short*********)100;
c = (int*********)100;
a = a + 6;
b = b + 6;
c = c + 6;
printf("%d %d %d\n",a,b,c);
}
int main(int argc, char* argv[])
{
fun();
return 0;
}
运行结果：124 124 124
反汇编代码
9: char********* a;
10: short********* b;
11: int********* c;
12:
13: a = (char*********)100;
0040B818 mov dword ptr [ebp-4],64h
14: b = (short*********)100;
0040B81F mov dword ptr [ebp-8],64h
15: c = (int*********)100;
0040B826 mov dword ptr [ebp-0Ch],64h
16:
17: a = a + 6;
0040B82D mov eax,dword ptr [ebp-4]
0040B830 add eax,18h
0040B833 mov dword ptr [ebp-4],eax
18: b = b + 6;
0040B836 mov ecx,dword ptr [ebp-8]
0040B839 add ecx,18h
0040B83C mov dword ptr [ebp-8],ecx
19: c = c + 6;
0040B83F mov edx,dword ptr [ebp-0Ch]
0040B842 add edx,18h
0040B845 mov dword ptr [ebp-0Ch],edx
结论：
1.带*类型的变量可以加、减一个整数，但不能乘或者除.
2.带*类型变量与其他整数相加或者相减时：
带*类型变量 + N = 带*类型变量 + N*(去掉一个*后类型的宽度)
带*类型变量 - N = 带*类型变量 - N*(去掉一个*后类型的宽度)
所以计算结果就是：带*变量的值加上或者减去去掉一个*之后的类型宽度乘以一个整数；
例如：
上面测试加法的时候，两个以上*的计算操作：100+4x6=124
上面测试减法的时候，两个以上*的计算操作：100-4x6=76
6."带*类型" 的特征探测：求差值
带一个*的测试
源代码
#include "stdafx.h"
#include <string.h>
void fun()
{
char* a;
char* b;
a = (char*)200;
b = (char*)100;
int c = a - b;
printf("%d\n",c);
}
int main(int argc, char* argv[])
{
fun();
return 0;
}
运行结果：100
反汇编代码
9: char* a;
10: char* b;
11:
12: a = (char*)200;
0040B818 mov dword ptr [ebp-4],0C8h
13: b = (char*)100;
0040B81F mov dword ptr [ebp-8],64h
14:
15: int c = a - b;
0040B826 mov eax,dword ptr [ebp-4]
0040B829 sub eax,dword ptr [ebp-8]
0040B82C mov dword ptr [ebp-0Ch],eax
带两个*的测试
源代码
#include "stdafx.h"
#include <string.h>
void fun()
{
char** a;
char** b;
a = (char**)200;
b = (char**)100;
int c = a - b;
printf("%d\n",c);
}
int main(int argc, char* argv[])
{
fun();
return 0;
}
运行结果：25
反汇编代码
9: char** a;
10: char** b;
11:
12: a = (char**)200;
0040B818 mov dword ptr [ebp-4],0C8h
13: b = (char**)100;
0040B81F mov dword ptr [ebp-8],64h
14:
15: int c = a - b;
0040B826 mov eax,dword ptr [ebp-4]
0040B829 sub eax,dword ptr [ebp-8]
0040B82C sar eax,2
0040B82F mov dword ptr [ebp-0Ch],eax
带两个以上*的测试
源代码
#include "stdafx.h"
#include <string.h>
void fun()
{
char************ a;
char************ b;
a = (char************)200;
b = (char************)100;
int c = a - b;
printf("%d\n",c);
}
int main(int argc, char* argv[])
{
fun();
return 0;
}
运行结果：25
反汇编代码
9: char************ a;
10: char************ b;
11:
12: a = (char************)200;
0040B818 mov dword ptr [ebp-4],0C8h
13: b = (char************)100;
0040B81F mov dword ptr [ebp-8],64h
14:
15: int c = a - b;
0040B826 mov eax,dword ptr [ebp-4]
0040B829 sub eax,dword ptr [ebp-8]
0040B82C sar eax,2
0040B82F mov dword ptr [ebp-0Ch],eax
结论：
1.两个类型相同的带*类型的变量可以进行减法操作.
2.想减的结果要除以去掉一个*的数据的宽度.
例如：
上面的计算操作：200-100=100 100/4=25
7."带*类型" 的特征探测：比较
#include "stdafx.h"
#include <string.h>
void fun()
{
char************** a ;
char************** b ;
a = (char**************)200;
b = (char**************)100;
if(a>b)
{
printf("6\n");
}
else
{
printf("8\n");
}
}
int main(int argc, char* argv[])
{
fun();
return 0;
}
结论：
带*的变量，如果类型相同，可以做大小的比较。
课后练习
1.char类型占几字节？char*类型占几字节？int*****占几字节？
char 占用1字节
（32位下）char* 站用4字节
（32位下）int***** 站用4字节
2、char** arr[10] 占多少个字节？
4x10=40
（32位下）char*** arr[10]占用40个字节
3、自定义结构体如下：
struct Student
{
int x;
int y;
};
正常不带*测试代码
#include "stdafx.h"
#include <string.h>
struct student
{
int x;
int y;
};
void fun()
{
student s;
s.x = 100;
s.y = 200;
printf("%d %d",s.x,s.y);
}
int main(int argc, char* argv[])
{
fun();
return 0;
}
带一个*的测试
源代码
#include "stdafx.h"
#include <string.h>
struct student
{
int x;
int y;
};
void fun()
{
student* s;
s = (student*)100;
s++;
printf("%d",s);
}
int main(int argc, char* argv[])
{
fun();
return 0;
}
运算结果；108
反汇编代码：
4: #include "stdafx.h"
5: #include <string.h>
6:
7: struct student
8: {
9: int x;
10: int y;
11: };
12:
13:
14: void fun()
15: {
0040B800 push ebp
0040B801 mov ebp,esp
0040B803 sub esp,44h
0040B806 push ebx
0040B807 push esi
0040B808 push edi
0040B809 lea edi,[ebp-44h]
0040B80C mov ecx,11h
0040B811 mov eax,0CCCCCCCCh
0040B816 rep stos dword ptr [edi]
16: student* s;
17:
18: s = (student*)100;
0040B818 mov dword ptr [ebp-4],64h
19:
20: s++;
0040B81F mov eax,dword ptr [ebp-4]
0040B822 add eax,8
0040B825 mov dword ptr [ebp-4],eax
21:
22: printf("%d",s);
0040B828 mov ecx,dword ptr [ebp-4]
0040B82B push ecx
0040B82C push offset string "6\n" (0042001c)
0040B831 call printf (00401110)
0040B836 add esp,8
23: }
0040B839 pop edi
0040B83A pop esi
0040B83B pop ebx
0040B83C add esp,44h
0040B83F cmp ebp,esp
0040B841 call __chkesp (00401190)
0040B846 mov esp,ebp
0040B848 pop ebp
0040B849 ret
带两个*的测试
源代码
#include "stdafx.h"
#include <string.h>
struct student
{
int x;
int y;
};
void fun()
{
student** s;
s = (student**)100;
s++;
printf("%d",s);
}
int main(int argc, char* argv[])
{
fun();
return 0;
}
运算结果：104
带两个以上*的测试
源代码
#include "stdafx.h"
#include <string.h>
struct student
{
int x;
int y;
};
void fun()
{
student******* s;
s = (student*******)100;
s++;
printf("%d",s);
}
int main(int argc, char* argv[])
{
fun();
return 0;
}
运算结果：104
进行加上或者减去一个整数测试
带一个*测试
源代码
#include "stdafx.h"
#include <string.h>
struct student
{
int x;
int y;
};
void fun()
{
student* s;
s = (student*)100;
s = s + 2;
printf("%d",s);
}
int main(int argc, char* argv[])
{
fun();
return 0;
}
运算结果：116
带两个*测试
#include "stdafx.h"
#include <string.h>
struct student
{
int x;
int y;
};
void fun()
{
student** s;
s = (student**)100;
s = s + 2;
printf("%d",s);
}
int main(int argc, char* argv[])
{
fun();
return 0;
}
运算结果：108
带两个以上*测试
源代码
#include "stdafx.h"
#include <string.h>
struct student
{
int x;
int y;
};
void fun()
{
student********* s;
s = (student*********)100;
s = s + 2;
printf("%d",s);
}
int main(int argc, char* argv[])
{
fun();
return 0;
}
加法：100+2x4=108
减法：100-2x4=92
相减进行测试
源代码
两个以上*测试
#include "stdafx.h"
#include <string.h>
struct student
{
int x;
int y;
};
void fun()
{
student********* s1;
student********* s2;
int x;
s1 = (student*********)200;
s2 = (student*********)100;
x = s1 - s2;
printf("%d",x);
}
int main(int argc, char* argv[])
{
fun();
return 0;
}
运算结果：25
通过测试两个*的计算结果也是25
一个*测试
源代码
#include "stdafx.h"
#include <string.h>
struct student
{
int x;
int y;
};
void fun()
{
student* s1;
student* s2;
int x;
s1 = (student*)200;
s2 = (student*)100;
x = s1 - s2;
printf("%d",x);
}
int main(int argc, char* argv[])
{
fun();
return 0;
}
运算结果：12
200-100=100 100/8=12.5 取12
8的由来是结构体里面两个整形int，所站字节数是8
结论：
结构体定义带*的操作跟其他类型一样，当定义两个*以上的类型，都按照4个字节宽度进行计算，当定义一个*的时候按照结构体内部实际字节站用的计算
迷茫的人生，需要不断努力，才能看清远方模糊的志向！