Intel和微软同时出现的C语言面试题 Voy1
#pragma pack(8) Q\Wh]=}
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struct s1{
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short a; X})5XYvA*
long b; ^Gi9&fS,
}; 3PkVMX
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struct s2{ wnUuoX(
char c; ,5V w^@F
s1 d; |"}oGL6-
long long e; Ey|{yUmU+
}; &3gC&b^i
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#pragma pack() ]2E#P.-!b
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问 GmhfBW?
1.sizeof(s2) = ? P* X^)R
2.s2的s1中的a后面空了几个字节接着是b? oZ,J{I!L
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如果您知道答案请在讨论中写出,以下是部份网友的答案,供参考: 'Em3;`/C*+
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网友rwxybh(行云)的答案: TOT#l6yqdd
内存布局是 M(
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1*** 11** O06 2c)vIY
1111 **** *^y,Gg/
1111 1111 68*a'0
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所以答案就是24和3 A@uU*]TqJ8
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下面是一个测试的程序,试一试就知道了,我用的是VC2005 &u`EYxT
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#pragma pack(8) ,V^2Oa
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struct s1{ =kzuU1s
short a; // 2 BYtes rEHlo[7^
long b; // 4 Bytes QM('bbN
}; Qm_IU!b
struct s2{ `T\_Wje(
char c; // 1 Byte bv^wE,+?o
s1 d; // 8 Bytes f9K+o-P.h
long long e; // 8 Bytes 7D(Eo{ue
}; KvjsibI/Y
// 1*** 11** S>Z07d6 &
// 1111 **** g^l~AR
// 1111 1111 E3hXs6P
// ~P7zg!p/q
[][ze2+b
// 00 01 02 03 04 05 06 07 E"%dO
// 00 01 02 03 04 05 06 07 |LV}kG(2
// 00 01 02 03 04 05 06 07 *I:a\o~$[
// )\KU:_l
#pragma pack() FuC#w 9_
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int main(int argc, char* argv[]) mE\)j*Nnv
{ mzRH:HgN?
s2 a; 63E)RR_Lh
char *p = (char *)&a; 2c*w{\X
for(int i=0;i<24;++i) /
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p = (char)(i%8); B?%e-xV-
printf("%d\n",sizeof(a)); 15z(hzU?#
printf("c=0x%lx\n",a.c); R]&lVXyH
printf("d.a=0x%x\n",a.d.a); '4Drs}j5
printf("d.b=0x%x\n",a.d.b); E4$y|Ni"
printf("e=0x%llx\n",a.e); M3U?\g
return 0; ' hDs.Wnu
} *Sg6VGP
结果: c(b2f-0!4
24 l(Ya,/4
c=0x0 (:P#l&f
d.a=0x504 A("\m>g$b
d.b=0x3020100 ?[]jJ
e=0x706050403020100 wP7
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网友 redleaves (ID最吊的网友)的答案和分析: \3&