C专家编程学习中的笔记
}c;
&char = 134518468,&double = 134518472,&short = 134518480,&char = 134518482,&short = 134518484,&char = 134518486.sizeof(c) = 20.注意如果按以前我的认为，这里将会是16。
struct node_tag10{
char test0;
double test1;
short test2;
char test3;
short test4;
char test5;
char test6;
}c;
&char = 134518480,&double = 134518484,&short = 134518492,&char = 134518494,&short = 134518496,&char = 134518498,&char = 134518499.sizeof(c) = 20.注意如果按以前我的认为，这里将会是16。
struct node_tag10{
char test0;
short test1;
double test2;
short test3;
char test4;
short test5;
char test6;
char test7;
}c;
&char = 134518508,&short = 134518510,&double = 134518512,&short = 134518520,&char = 134518522,&short = 134518524,&char = 134518526,&char = 134518527.sizeof(c) = 20.
struct node_tag10{
short test0;
double test1;
short test2;
char test3;
short test4;
char test5;
char test6;
}c;
&short = 134518480,&double = 134518484,&short = 134518492,&char = 134518494,&short = 134518496,&char = 134518498,&char = 13451899.sizeof(c) = 20.
由上面这些可得，第一个分配地址总是能被4整除的，32位机，便于寻址。后面出现的类型（如int,double等）只要所占字节数大于或等于4个字节，也将会从能被4整除的地址开始，且与前面地址是连续的。下面这个能很好说明：
struct node_tag10{
char test0;
double test1;
short test2;
char test3;
short test4;
char test5;
}c;
&char = 134518468,&double = 134518472,&short = 134518480,&char = 134518482,&short = 134518484,&char = 134518486.sizeof(c) = 20.注意如果按以前我的认为，这里将会是16。
最后还是做一下Visual C++6.0上的测试:
sizechar = 1;sizeshort = 2;sizeint = 4;sizelong = 4;sizefloat = 4;sizedouble = 8;sizelongdouble = 8.
struct node_tag10{
short test0;
double test1;
short test2;
char test3;
short test4;
char test5;
char test6;
}c;
&short = 4339136,&double = 4339144,&short = 4339152,&char = 4339154,&short = 4339156,&char = 4339158,&char = 4339159.sizeof(c) = 24.与虚拟机上不同，无论第一个类型变成char,short,int,double都会占8个字节，除非一个也不要，就不会占8个字节。
将double改成int,前面第一个类型（char,short）会只占4个字节了。若改为short，前面的char又只占两上字节了，这，总之一句话，程序里面太多的神奇。
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下午又得
出一点结论：
编译器存在字对齐问题，将以结构体中占字节最多的类型来对齐。（不同的平台，会有区别，就如是我的虚拟机（有double存在也会像样int以4个字节做为分配单无）与VC++6.0（有double存在将会以8个字节做为分配单元）结果不一样，这里我们以VC++6.0为准，在不同的平