标签：面试题
不使用sizeof,求某机器平台的int型整数位数(16,32,64)
思路,将整数变成字符串处理,并且在内存的末尾填零,即字符串结束符。
如 0x0012,0x00123456,0x00123456789ABCDE分别对应16,32,64位数; 小端机
代码
#include <stdio.h>
#include <string.h>
int main(int argc , char *argv[])
{
intn16 = 0x0012;
/*
此处用于填充内存防止出现内存连续
如, 32位机器上。
*/
intnIntNop1 = 0;
intnIntNop2 = 0;
intn32 = 0x00123456;
intnIntNop3 = 0;
intnIntNop4 = 0;
intn64 = 0x00123456789ABCDE;
intnInt16 = strlen((char *)&n16);
intnInt32 = strlen((char *)&n32);
intnInt64 = strlen((char *)&n64);
if(1== nInt16 && 2 == nInt32 && 2 == nInt64)
{
printf("16Bits\n");
}
if(1== nInt16 && 3 == nInt32 && 4 == nInt64)
{
printf("32Bits\n");
}
if(1== nInt16 && 3 == nInt32 && 7 == nInt64)
{
printf("64Bits\n");
}
return0;
}
反汇编分析:
5: int n16 = 0x0012;
00401028 mov dword ptr [ebp-4],12h
6: int nIntNop1 = 0;
0040102F mov dword ptr [ebp-8],0
7: int nIntNop2 = 0;
00401036 mov dword ptr [ebp-0Ch],0
8: int n32 = 0x00123456;
0040103D mov dword ptr [ebp-10h],123456h
9: int nIntNop3 = 0;
00401044 mov dword ptr [ebp-14h],0
10: int nIntNop4 = 0;
0040104B mov dword ptr [ebp-18h],0
11: int n64 = 0x00123456789ABCDE;
00401052 mov dword ptr [ebp-1Ch],789ABCDEh
12: int nInt16 = strlen((char *)&n16);
00401059 lea eax,[ebp-4]
0040105C push eax
0040105D call strlen (004011b0)
00401062 add esp,4
00401065 mov dword ptr[ebp-20h],eax
13:
14: int nInt32 = strlen((char *)&n32);
00401068 lea ecx,[ebp-10h]
0040106B push ecx
0040106C call strlen (004011b0)
00401071 add esp,4
00401074 mov dword ptr[ebp-24h],eax
15:
16: int nInt64 = strlen((char *)&n64);
00401077 lea edx,[ebp-1Ch]
0040107A push edx
0040107B call strlen (004011b0)
00401080 add esp,4
00401083 mov dword ptr[ebp-28h],eax
17: if(1 == nInt16 && 2 == nInt32 && 2 == nInt64)
00401086 cmp dword ptr [ebp-20h],1
0040108A jne main+95h (004010a5)
0040108C cmp dword ptr [ebp-24h],2
00401090 jne main+95h (004010a5)
00401092 cmp dword ptr [ebp-28h],2
00401096 jne main+95h (004010a5)
18: {
19: printf("16 Bits\n");
00401098 push offset string "16Bits\n" (00422034)
0040109D call printf (00401130)
004010A2 add esp,4
20: }
21: if(1 == nInt16 && 3 == nInt32 && 4 == nInt64)
004010A5 cmp dword ptr [ebp-20h],1
004010A9 jne main+0B4h (004010c4)
004010AB cmp dword ptr [ebp-24h],3
004010AF jne main+0B4h (004010c4)
004010B1 cmp dword ptr [ebp-28h],4
004010B5 jne main+0B4h (004010c4)
22: {
23: printf("32 Bits\n");
004010B7 push offset string "32Bits\n" (00422028)
004010BC call printf (00401130)
004010C1 add esp,4
24: }
25: if(1 == nInt16 && 3 == nInt32 && 7 == nInt64)
004010C4 cmp dword ptr [ebp-20h],1
004010C8 jne main+0D3h (004010e3)
004010CA cmp dword ptr [ebp-24h],3
004010CE jne main+0D3h (004010e3)
004010D0 cmp dword ptr [ebp-28h],7
004010D4 jne main+0D3h (004010e3)
26: {
27: printf("64 Bits\n");
004010D6 push offset string "64Bits\n" (0042201c)
004010DB call printf (00401130)
004010E0 add esp,4
28: }
29: return 0;
004010E3 xor eax,eax
30: }
004010E5 pop edi
004010E6 pop esi
004010E7 pop ebx
004010E8 add esp,68h
004010EB cmp ebp,esp
004010ED call __chkesp (00401230)
004010F2 mov esp,ebp
004010F4 pop ebp
004010F5 ret
Vc6.0下 32位平台
初始寄存器:执行5:xxx 语句之前
EAX = CCCCCCCC EBX = 7EFDE000
ECX= 00000000 EDX = 002A1810
ESI= 00000000 EDI = 0018FF48
EIP= 00401028 ESP = 0018FED4
EBP= 0018FF48 EFL = 00000202
初始内存
0018FEFB CC CC CC CC CC CC CC 烫烫烫.
0018FF02 CC CC CC CC CC CC CC 烫烫烫.
0018FF09 CC CC CC CC CC CC CC 烫烫烫.
0018FF10 CC CC CC CC CC CC CC 烫烫烫.
0018FF17 CC CC CC CC CC CC CC 烫烫烫.
0018FF1E CC CC CC CC CC CC CC 烫烫烫.
0018FF25 CC CC CC CC CC CC CC 烫烫烫.
0018FF2C CC CC CC CC CC CC CC 烫烫烫.
0018FF33 CC CC CC CC CC CC CC 烫烫烫.
0018FF3A CC CC CC CC CC CC CC 烫烫烫.
0018FF41 CC CC CC CC CC CC CC 烫烫烫.
0018FF48 88 FF 18 00 59 13 40 ....Y.@
之后 内存
0018FEFB CC CC CC CC CC CC CC 烫烫烫.
0018FF02 CC CC CC CC CC CC CC 烫烫烫.
0018FF09 CC CC CC CC CC CC CC 烫烫烫.
0018FF10 CC CC CC CC CC CC CC 烫烫烫.
0018FF17 CC CC CC CC CC CC CC 烫烫烫.
0018FF1E CC CC 04 00 00 00 03 烫..... 4是因为那个DEBC9A78的缘故最后不是0
0018FF25 00 00 00 0100 00 00 ....... 此处分别是 1 和3 即是32位下的
0018FF2C DE BC 9A 7800 00 00 藜歺... 那个没有放下的整数0x00123456789ABCDE
0018FF33 00 00 00 00 00 56 34 .....V4 截取了末位的八个字节
0018FF3A 12 00 00 00 00 00 00 ....... 32位整数0x00123456
0018FF41 00 00 00 12 00 00 00 .......
0018FF48 88 FF 18 00 59 13 40 ....Y.@
收尾之后
0018FEED CC CC CC CC CC CC CC 烫烫烫.
0018FEF4 CC CC CC CC CC CC CC 烫烫烫.
0018FEFB CC CC CC CC CC CC CC 烫烫烫.
0018FF02 CC CC CC CC CC CC CC 烫烫烫.
0018FF09 CC CC CC CC CC CC CC 烫烫烫.
0018FF10 CC CC CC CC CC CC CC 烫烫烫.
0018FF17 CC CC CC CC CC CC CC 烫烫烫.
0018FF1E CC CC 04 00 00 00 03 烫.....
0018FF25 00 00 00 01 00 00 00 .......
0018FF2C DE BC 9A78 00 00 00 藜歺...
0018FF33 00 00 00 00 00 56 34 .....V4
0018FF3A 12 00 00 00 00 00 00 .......
0018FF41 00 00 00 F2 10 40 00 .....@. 004010ED call __chkesp (00401230) 之后改变了值,估计是已经回收了内存,有待日后研究。
0018FF48 88 FF 18 00 59 13 40 ....Y.@
原来的错去程序
#include <stdio.h>
#include <string.h>
int main(int argc , char *argv[])
{
intn16 = 0x0012;
intn32 = 0x00123456;
intn64 = 0x00123456789ABCDE;
intnInt16 = strlen((char *)&n16);
intnInt32 = strlen((char *)&n32);
intnInt64 = strlen((char *)&n64);
if(1== nInt16)
{
printf("16Bits\n");
}
if(3== nInt32)
{
printf("32Bits\n");
}
if(7== nInt64)
{
printf("64Bits\n");
}
return0;
}
初始执行前
EAX = CCCCCCCC EBX = 7EFDE000
ECX= 00000000 EDX = 004F1810
ESI= 00000000 EDI = 0018FF48
EIP= 00401028 ESP = 0018FEE4
EBP= 0018FF48 EFL = 00000206
0018FEFB CC CC CC CC CC CC CC 烫烫烫.
0018FF02 CC CC CC CC CC CC CC 烫烫烫.
0018FF09 CC CC CC CC CC CC CC 烫烫烫.
0018FF10 CC CC CC CC CC CC CC 烫烫烫.
0018FF17 CC CC CC CC CC CC CC 烫烫烫.
0018FF1E CC CC CC CC CC CC CC 烫烫烫.
0018FF25 CC CC CC CC CC CC CC 烫烫烫.
0018FF2C CC CC CC CC CC CC CC 烫烫烫.
0018FF33 CC CC CC CC CC CC CC 烫烫烫.
0018FF3A CC CC CC CC CC CC CC 烫烫烫.
0018FF41 CC CC CC CC CC CC CC 烫烫烫.
0018FF48 88 FF 18 00 39 13 40 ....9.@
执行后
0018FF25 CC CC CC CC CC CC CC 烫烫烫.
0018FF2C CC CC CC CC 07 00 00 烫烫...
0018FF33 00 03 00 00 00 01 00 .......
0018FF3A 00 00 DE BC 9A 7856 ..藜歺V 执行n64 和 n32赋值之后 恍然大悟。。
0018FF41 34 12 00 12 00 00 00 4...... 对内存把握不够。。果断牺牲。
0018FF48 88 FF 18 00 39 13 40 ....9.@
若是64位机器则 包含两个0填充
目测是这样的
DE BC 9A78 56 34 12 00
00 00 00 00 00 00 0000填充的0
00 00 00 00 00 00 00 00
56 34 12 00 00 00 00 00
00 00 00 00 0000 00 00
00 00 00 00 00 0000 00
12 00 00 00 0000 00 00
若是16位机器 目测是这样的
DE BC
00 00
00 00
56 34
00 00
00 00
12 00
2015春季腾讯实习在线测试题之求int型字节长度,不准用sizeof
标签：面试题
原文地址：http://blog.csdn.net/u011185633/article/details/44975291
