memcpy函数的实现代码

memcpy函数的实现代码memcpy函数的实现代码(面试急用) void memcpy( void * dest, void * src , size_t n) char far * pd = (char far * )dest; char far * ps = (char far * )src; long len = (long)sizeof(size_t) * n; for( long i=0; i n; +i ) *pd+ = *ps+; TOP 精华推荐： 一段小代码，是否出现了野指针？ 请大家发表一下对野指针理解！ 改进一下，这样我觉得更安全一点 void memcpy( void * dest, void * src , size_t n) char far * pd = (char far * )dest; char far * ps = (char far * )src; long len = (long)sizeof(size_t) * n; while(len) while(*pd+ = *ps+)!= 0 ) len-; 有2个问题: 1. n不就是要拷贝的字节数?为什么还要乘以sizeof(size_t)? 是sizeof(char)吧? 2. 为什么要far指针呢?是不是要考虑跨段拷贝的情形? 你的方法我好像更不能理解, 为什么要判断 0 呢? 这又不是拷贝字符串. 程序是这样的： void _fmemcpy(void far *dest, void far* src, long n) while (n-=sizeof(char) *(char far*)dest)=*(char far*)src); 不过，在位语言中，far是不必要的 C里面定义的memcpy讲求的是快，没有保证任何的安全性。 抱歉，有个bug，改之 void _fmemcpy(void far *dest, void far* src, long n) do *(char far*)dest)=*(char far*)src); while (n-=sizeof(char); extern void *MEMCPY ( void *pDest, const void *pSrc, Length size ) Counter cNumBytes; char *pD, *pS; if (NULL = pSrc) | (NULL = pDest) return NULL; pD = ( char * ) pDest; pS = ( char * ) pSrc; cNumBytes = 0; while ( cNumBytes size ) *pD+ = *pS+; cNumBytes+; return pDest; void * memmove(void * dst, void * src, size_t count) void * ret = dst; if (dst = (src + count) /* * Non-Overlapping Buffers * copy from lower addresses to higher addresses */ while (count-) *dst+ = *src+; else /* * Overlapping Buffers * copy from higher addresses to lower addresses */ dst += count - 1; src += count - 1; while (count-) *dst- = *src-; return(ret); #include void * memcpy(void *s1, const void *s2, register size_t n) register char *p1 = s1; register const char *p2 = s2; if (n) n+; while (-n 0) *p1+ = *p2+; return s1; 看看真正的C/C+库的是现代码吧 Func memmove, _EXPFUNC, _RTLENTRYF, , , Link esi,edi mov edi,dest mov esi,src mov ecx,n mov edx,ecx cmp edi,esi ; dest source ? jb mov_fwd ; yes, do a simple forward move je mov_end ; dest = source ? do nothing std ; force esi and edi to decrement lea esi,esi+ecx-1 ; point to last byte of source buffer lea edi,edi+ecx-1 ; point to last byte of dest buffer and ecx,3 ; copy trailing bytes rep movsb mov ecx,edx ; get original count back shr ecx,1 ; calculate no. of longwords shr ecx,1 sub esi,3 ; point to previous longword in source sub edi,3 ; point to previous longword in dest rep movsd ; copy longwords cld ; clear direction flag jmp short mov_end mov_fwd: cld shr ecx,1 shr ecx,1 rep movsd ; copy longwords mov ecx,edx ; get original count back and ecx,3 ; calculate remainder byte count (0-3) rep movsb ; copy remaining bytes mov_end: mov eax,dest Unlink esi,edi Return EndFunc memmove 做几件事情： 1。内存自身拷贝问题 2。有相交重叠的两块内存的拷贝问题 3。拷贝速度的问题 对于拷贝速度问题，方法是32位机器上双字对齐边界开始拷贝，16位机器上字边界开始拷贝，把内存分为3段： 对齐边界之前的字节，对齐拷贝的字节，最后对齐边界之后的字节 第一段使用一个movsb和movsw完成任务，第二段采用movsd完成，第三段也采用一个movsw和movsb完成。 面试过程只要说出这三个要点就可以了，程序其实你可以不用编写了 刚才拷贝部分的等价操作如下 memmove(char * src,char *dest, int n) / = 2 ) *(short*)dest+ = *(short*)src+; if( k = 1 ) *dest+ = src+; 其他再加上一些if判断就可以了 的确memcpy讲求的是效率，下面的代码是c库的实现代码 看一看大师是怎样考虑的。 typedef int word; /* word used for optimal copy speed */ #define wsize sizeof(word) #define wmask (wsize - 1) void * memcpy(void *dst0, const void *src0, size_t length) char *dst = dst0; const char *src = src0; size_t t; if (length = 0 | dst = src) /* nothing to do */ goto done; /* * Macros: loop-t-times; and loop-t-times, t 0 */ #define TLOOP(s) if (t) TLOOP1(s) #define TLOOP1(s) do s; while (-t) if (unsigned long)dst (unsigned long)src) /* * Copy forward. */ t = (int)src; /* only need low bits */ if (t | (int)dst) & wmask) /* * Try to align operands. This cannot be done * unless the low bits match. */ if (t (int)dst) & wmask | length wsize) t = length; else t = wsize - (t & wmask); length -= t; TLOOP1(*dst+ = *src+); /* * Copy whole words, then mop up any trailing bytes. */ t = length / wsize; TLOOP(*(word *)dst = *(word *)src; src += wsize; dst += wsize); t = length & wmask; TLOOP(*dst+ = *src+); else /* * Copy backwards. Otherwise essentially the same. * Alignment works as before, except that it takes * (t&wmask) bytes to align, not wsize-(t&wmask). */ src += length; dst += length; t = (int)src; if (t | (int)dst) & wmask) if (t (int)dst) & wmask | length = wsize) t = length; else t &= wmask; length -= t; TLOOP1(*-dst = *-src); t = length / wsize; TLOOP(src -= wsize; dst -= wsize; *(word *)dst = *(word *)src); t = length & wmask; TLOOP(*-dst = *-src); done: return (dst0); VC: void* Dest=(void*)pSurf; void* Src = (void*)pBuffer; size_t nBytes = (size_t)buffLen; _asm push edi push esi mov esi, dword ptrSrc mov edi, dword ptrDest mov ecx, nBytes shr ecx, 6 ; nBytes / 64 align 4 Memcpy_Loop: movq mm0, qword ptresi movq mm1, qword ptresi+8*1 movq mm2, qword ptresi+8*2 movq mm3, qword ptresi+8*3 movq mm4, qword ptresi+8*4 movq mm5, qword ptresi+8*5 movq mm6, qword ptresi+8*6 movq mm7, qword ptresi+8*7 movq qword ptredi, mm0 movq qword ptredi+8*1, mm1 movq qword ptredi+8*2, mm2 movq qword ptredi+8*3, mm3 movq qword ptredi+8*4, mm4 movq qword ptredi+8*5, mm5 movq qword ptredi+8*6, mm6 movq qword ptredi+8*7, mm7 add esi, 64 add edi, 64 loop Memcpy_Loop mov ecx, nBytes and ecx, 63 cmp ecx, 0 je Memcpy_End rep movsb Memcpy_End: emms pop esi pop edi