什么样的strcopy才能得满分？
char* strcopy(char* ch1,char* ch2)
{
assert(ch1 && ch2);
if(!ch1 || !ch2)return ch1;
int size1=strlen(ch1);
int size2=strlen(ch2);
if(size1<size2)return ch1;
char* q=ch1,*p=ch2;
while(*(q++)=*(p++));
*q='\0';
return ch1;
}
这个可以吗？
如果不能大家可以发发自己的版本
[解决办法]
差远了···
1.变量的命名方式
2.第二个参数用const
3.if(size1<size2)return ch1;//返回ch1？？？
4.*q='\0';//不需要再补一个\0了
5.楼下大牛补
[解决办法]
page ,132 title strcat - concatenate (append) one string to another;***;strcat.asm - contains strcat() and strcpy() routines;; Copyright (c) Microsoft Corporation. All rights reserved.;;Purpose:; STRCAT concatenates (appends) a copy of the source string to the; end of the destination string, returning the destination string.;;******************************************************************************* .xlist include cruntime.inc .listpage;***;char *strcat(dst, src) - concatenate (append) one string to another;;Purpose:; Concatenates src onto the end of dest. Assumes enough; space in dest.;; Algorithm:; char * strcat (char * dst, char * src); {; char * cp = dst;;; while( *cp ); ++cp; /* Find end of dst */; while( *cp++ = *src++ ); ; /* Copy src to end of dst */; return( dst );; };;Entry:; char *dst - string to which "src" is to be appended; const char *src - string to be appended to the end of "dst";;Exit:; The address of "dst" in EAX;;Uses:; EAX, ECX;;Exceptions:;;*******************************************************************************page;***;char *strcpy(dst, src) - copy one string over another;;Purpose:; Copies the string src into the spot specified by; dest; assumes enough room.;; Algorithm:; char * strcpy (char * dst, char * src); {; char * cp = dst;;; while( *cp++ = *src++ ); ; /* Copy src over dst */; return( dst );; };;Entry:; char * dst - string over which "src" is to be copied; const char * src - string to be copied over "dst";;Exit:; The address of "dst" in EAX;;Uses:; EAX, ECX;;Exceptions:;******************************************************************************* CODESEG% public strcat, strcpy ; make both functions availablestrcpy proc \ dst:ptr byte, \ src:ptr byte OPTION PROLOGUE:NONE, EPILOGUE:NONE push edi ; preserve edi mov edi,[esp+8] ; edi points to dest string jmp short copy_startstrcpy endp align 16strcat proc \ dst:ptr byte, \ src:ptr byte OPTION PROLOGUE:NONE, EPILOGUE:NONE .FPO ( 0, 2, 0, 0, 0, 0 ) mov ecx,[esp+4] ; ecx -> dest string push edi ; preserve edi test ecx,3 ; test if string is aligned on 32 bits je short find_end_of_dest_string_loopdest_misaligned: ; simple byte loop until string is aligned mov al,byte ptr [ecx] add ecx,1 test al,al je short start_byte_3 test ecx,3 jne short dest_misaligned align 4find_end_of_dest_string_loop: mov eax,dword ptr [ecx] ; read 4 bytes mov edx,7efefeffh add edx,eax xor eax,-1 xor eax,edx add ecx,4 test eax,81010100h je short find_end_of_dest_string_loop ; found zero byte in the loop mov eax,[ecx - 4] test al,al ; is it byte 0 je short start_byte_0 test ah,ah ; is it byte 1 je short start_byte_1 test eax,00ff0000h ; is it byte 2 je short start_byte_2 test eax,0ff000000h ; is it byte 3 je short start_byte_3 jmp short find_end_of_dest_string_loop ; taken if bits 24-30 are clear and bit ; 31 is setstart_byte_3: lea edi,[ecx - 1] jmp short copy_startstart_byte_2: lea edi,[ecx - 2] jmp short copy_startstart_byte_1: lea edi,[ecx - 3] jmp short copy_startstart_byte_0: lea edi,[ecx - 4]; jmp short copy_start; edi points to the end of dest string.copy_start:: mov ecx,[esp+0ch] ; ecx -> sorc string test ecx,3 ; test if string is aligned on 32 bits je short main_loop_entrancesrc_misaligned: ; simple byte loop until string is aligned mov dl,byte ptr [ecx] add ecx,1 test dl,dl je short byte_0 mov [edi],dl add edi,1 test ecx,3 jne short src_misaligned jmp short main_loop_entrancemain_loop: ; edx contains first dword of sorc string mov [edi],edx ; store one more dword add edi,4 ; kick dest pointermain_loop_entrance: mov edx,7efefeffh mov eax,dword ptr [ecx] ; read 4 bytes add edx,eax xor eax,-1 xor eax,edx mov edx,[ecx] ; it's in cache now add ecx,4 ; kick dest pointer test eax,81010100h je short main_loop ; found zero byte in the loop; main_loop_end: test dl,dl ; is it byte 0 je short byte_0 test dh,dh ; is it byte 1 je short byte_1 test edx,00ff0000h ; is it byte 2 je short byte_2 test edx,0ff000000h ; is it byte 3 je short byte_3 jmp short main_loop ; taken if bits 24-30 are clear and bit ; 31 is setbyte_3: mov [edi],edx mov eax,[esp+8] ; return in eax pointer to dest string pop edi retbyte_2: mov [edi],dx mov eax,[esp+8] ; return in eax pointer to dest string mov byte ptr [edi+2],0 pop edi retbyte_1: mov [edi],dx mov eax,[esp+8] ; return in eax pointer to dest string pop edi retbyte_0: mov [edi],dl mov eax,[esp+8] ; return in eax pointer to dest string pop edi retstrcat endp end
[解决办法]
#include <assert.h>char *strcpy(char *strDest,const char * strSrc){ assert(strDest && strSrc); char *dest = strDest; while((*strDest++ = *strSrc++) != '\0'); return dest; }
[解决办法]
按楼主的做法 如果我定义了一个char c[1000];
然后c[0]=0;这样楼主的函数不就费了么
[解决办法]
大家认为有没有必要做自复制的判断？
if (strDest == strSrc) return strDest;
[解决办法]
我也来一遍吧strcpy
char* strcpy(char* strDest,char* strSrc)
{
assert((strDest != NULL) && (strSrc != NULL));
char *address=strDest;
while((*strDest++=*strSrc++) != '\0 ')
{
}
return address;
}
[解决办法]
char* strcpy(char* strdest,const char* strsrc)
{
assert(strdest&&strsrc);
if(strdest==strsrc)
return strdest;
char* pstr=strdest;
while((*pstr++=*strsrc++)!='\0')
{
NULL;
}
return strdest;
}