reinterpret_cast 用法
语法:
reinterpret_cast<type-name>(expression)
如果 type-name 和 expression 的位数一样,那么就能进行这种转换。reinterpret_cast 的安全性完全由程序员控制。
C语言的强制类型转换有时会忽略这一限制:转换源与转换目标的位数是否相同。例如,long 可以强制转换为 int,即“长数”强制转换为“短数”。char 可以强制转换为 short,即“短数”转换为“长数”。但是,在64位X86平台上,指针值不能强制转换为 int ,只能强制转换为 long 。
请看示例代码:
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#include<cstdio> int main(){
printf (" sizeof short , int , long , long long , float , double , void * : \
%lu, %lu, %lu, %lu, %lu, %lu, %lu\n", \
sizeof ( short ), sizeof ( int ), sizeof ( long ), sizeof ( long long ), sizeof ( float ), sizeof ( double ), sizeof ( void *));
long ldata = 0x1234567890abcdef;
long *pl = &ldata;
printf ( "pl=%p\n" , pl);
void *pvoid = pl;
printf ( "pvoid=%p\n" , pvoid);
struct stFormat{
int start;
int end;
};
stFormat *pst = reinterpret_cast <stFormat *>(pl);
printf ( "pst: %p, %#x, %#x\n" , pst, pst->start, pst->end);
void (*pf)( void ) = reinterpret_cast < void (*)( void )>(pl);
printf ( "pf=%p\n" , pf);
#if 0 //error: invalid cast from type ‘long int’ to type ‘int’
int idata1 = reinterpret_cast < int >(ldata);
printf ( "idata1 = %#x\n" , idata1);
#endif int idata2 = int (ldata);
printf ( "idata2 = %#x\n" , idata2);
char ch = 'a' ;
short sht1 = short (ch);
printf ( "sht1 = %#x\n" , sht1);
#if 0 //error: cast from ‘long int*’ to ‘short int’ loses precision [-fpermissive]
short sht2 = short (pl);
printf ( "sht2 = %#x\n" , sht2);
#endif long lval = long (pl);
printf ( "lval = %lx\n" , lval);
#if 0 //error: cast from ‘long int*’ to ‘int’ loses precision [-fpermissive]
int ival = int (pl);
printf ( "ival = %d\n" , ival);
#endif } |
测试结果:
sizeof short, int, long, long long, float, double, void * : 2, 4, 8, 8, 4, 8, 8
pl=0x7fff9ad82500
pvoid=0x7fff9ad82500
pst: 0x7fff9ad82500, 0x90abcdef, 0x12345678
pf=0x7fff9ad82500
idata2 = 0x90abcdef
sht1 = 0x61
lval = 7fff9ad82500
本文转自FrankNie0101 51CTO博客,原文链接:http://blog.51cto.com/frankniefaquan/1939619,如需转载请自行联系原作者