java strcpy_如何实现函数strcpy，用C/C++/JAVA其中一种语言？

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概述

strlen (const char *str){const char *char_ptr;const unsignedlong int *longword_ptr;unsignedlong int longword, himagic, lomagic; /* Handle the first few characters by reading one character at a time.Do this until CHAR_PTR is aligned on a longword boundary. */for (char_ptr = str; ((unsignedlong int) char_ptr& (sizeof (longword) - 1)) != 0;++char_ptr)if (*char_ptr =='')return char_ptr - str; /* All these elucidatory comments refer to 4-byte longwords,but the theory applies equally well to 8-byte longwords. */ longword_ptr = (unsignedlong int *) char_ptr; /* Bits 31, 24, 16, and 8 of this number are zero. Call these bitsthe "holes." Note that there is a hole just to the left ofeach byte, with an extra at the end: bits: 01111110 11111110 11111110 11111111bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD The 1-bits make sure that carries propagate to the next 0-bit.The 0-bits provide holes for carries to fall into. */himagic = 0x80808080L;lomagic = 0x01010101L;if (sizeof (longword) > 4){/* 64-bit version of the magic. *//* Do the shift in two steps to avoid a warning if long has 32 bits. */himagic = ((himagic < 8)abort (); /* Instead of the traditional loop which tests each character,we will test a longword at a time. The tricky part is testingif *any of the four* bytes in the longword in question are zero. */for (;;){longword = *longword_ptr++; if (((longword - lomagic) & ~longword & himagic) != 0){/* Which of the bytes was the zero? If none of them were, it wasa misfire; continue the search. */ const char *cp = (const char *) (longword_ptr - 1); if (cp[0] == 0)return cp - str;if (cp[1] == 0)return cp - str + 1;if (cp[2] == 0)return cp - str + 2;if (cp[3] == 0)return cp - str + 3;if (sizeof (longword) > 4){if (cp[4] == 0)return cp - str + 4;if (cp[5] == 0)return cp - str + 5;if (cp[6] == 0)return cp - str + 6;if (cp[7] == 0)return cp - str + 7;}}}}