C的memset,memcpy,strcpy 的区别 及memset memcpy memmove源码
2013-08-20 20:52 youxin 阅读(1353) 评论(0) 编辑 收藏 举报extern void *memcpy(void *dest,void *src,unsigned int count);
#include <string.h>
功能:由src所指内存区域复制count个字符串到dest所指内存区域.
说明:src和dest所指内存区域不能重叠,函数返回指向dest的指针.
memset
extern void *memset(void *buffer,int c,int count);
#include <string.h>
功能:把buffer所指内存区域的前count个字节设置成字符c
说明:返回指向buffer的指针.
二.区别
memset 用来对一段内存空间全部设置为某个字符,一般用于在对定义的字符串初始化为' '或者'\0';
例: char a[100];
memset(a,'\0',sizeof(a));
memcpy 是用来做内存拷贝,可以用来拷贝任何数据类型的对象,可以指定拷贝的数据长度;
例:
char a[100],b[50];
memcpy(b,a,sizeof(b)); //注意:如果使用sizeof(a),会造成内存溢出
mem是一段内存,他的长度,必须自己记住.memcpy是见着什么拷贝什么。
strcpy 就只能拷贝字符串,它遇到'\0'就结束拷贝;
例:char a[100],b[50];
strcpy(a,b);
如用strcpy(b,a)要注意a中的字符串长度(第一个'\0'之前) 是否超过50,如果超过,则会造成b的
内存溢出.它是不会拷贝'\0'的,所以一般还有加一个语句:
*a='\0';
三.使用技巧
memset 可以方便的清空一个数据结构的变量或数组.
如:
struct sample_struct
{
char csName[16];
int iSeq;
int iType;
};
对于变量
struct sample_struct stTest;
一般情况下,初始化stTest的方法:
stTest.csName[0]='\0';
stTest.iSeq=0;
stTest.iType=0;
而用memset:
memset(&stTest,0,sizeof(struct sample_struct));
如果是数组:
struct sample_struct TEST[100];
memset(TEST,0,sizeof(struct sample_struct)*100);
strcpy的原型为
char *strcpy(char *dest, const char *src)
而memcpy是给定来源和目标后,拷贝指定大小n的内存数据,而不管拷贝的内容是什么(不仅限于字符)
memcpy的原型为
void *memcpy(void *dest, const void *src, size_t n);
memcpy源码:
/*** *memcpy.c - contains memcpy routine *Purpose: * memcpy() copies a source memory buffer to a destination buffer. * Overlapping buffers are not treated specially, so propogation may occur. * **********/ #include <cruntime.h> #include <string.h> #pragma function(memcpy) /*** *memcpy - Copy source buffer to destination buffer * *Purpose: * memcpy() copies a source memory buffer to a destination memory buffer. * This routine does NOT recognize overlapping buffers, and thus can lead * to propogation. * * For cases where propogation must be avoided, memmove() must be used. * *Entry: * void *dst = pointer to destination buffer * const void *src = pointer to source buffer * size_t count = number of bytes to copy * *Exit: * Returns a pointer to the destination buffer * *Exceptions: *******************************************************************************/ void * __cdecl memcpy ( void * dst, const void * src, size_t count ) { void * ret = dst; #if defined (_M_IA64) { __declspec(dllimport) void RtlCopyMemory( void *, const void *, size_t count ); RtlCopyMemory( dst, src, count ); } #else /* defined (_M_IA64) */ /* * copy from lower addresses to higher addresses */ while (count--) { *(char *)dst = *(char *)src; dst = (char *)dst + 1; src = (char *)src + 1; } #endif /* defined (_M_IA64) */ return(ret); }
实现很简单,count一直减,把src值赋给dest,唯一值得注意的是函数原型。
void * __cdecl memcpy ( void * dst, const void * src, size_t count )
可以看到参数为void 类型的指针,也返回一个void *指针。
我们可以利用void *类型指针为任何类型实现copy,如结构体:
struct A{ int x; int y; }; A *a=new A(); A *b=new A(); a->x=5; a->y=3; memcpy(b,a,sizeof(A)); printf("%d%d\n",b->x,b->y);
输出53。
memset源码:
/*** *char *memset(dst, val, count) - sets "count" bytes at "dst" to "val" * *Purpose: * Sets the first "count" bytes of the memory starting * at "dst" to the character value "val". * *Entry: * void *dst - pointer to memory to fill with val * int val - value to put in dst bytes * size_t count - number of bytes of dst to fill * *Exit: * returns dst, with filled bytes * *Exceptions: * *******************************************************************************/ void * __cdecl memset ( void *dst, int val, size_t count ) { void *start = dst; #if defined (_M_IA64) || defined (_M_AMD64) { __declspec(dllimport) void RtlFillMemory( void *, size_t count, char ); RtlFillMemory( dst, count, (char)val ); } #else /* defined (_M_IA64) || defined (_M_AMD64) */ while (count--) { *(char *)dst = (char)val; dst = (char *)dst + 1; } #endif /* defined (_M_IA64) || defined (_M_AMD64) */ return(start); }
1.memmove
函数原型:void *memmove(void *dest, const void *source, size_t count)
返回值说明:返回指向dest的void *指针
参数说明:dest,source分别为目标串和源串的首地址。count为要移动的字符的个数
函数说明:memmove用于从source拷贝count个字符到dest,如果目标区域和源区域有重叠的话,memmove能够保证源串在被覆盖之前将重叠区域的字节拷贝到目标区域中。
2.memcpy
函数原型:void *memcpy(void *dest, const void *source, size_t count);
返回值说明:返回指向dest的void *指针
函数说明:memcpy功能和memmove相同,但是memcpy中dest和source中的区域不能重叠,否则会出现未知结果。
3.两者区别
函数memcpy() 从source 指向的区域向dest指向的区域复制count个字符,如果两数组重叠,不定义该函数的行为。
而memmove(),如果两函数重叠,赋值仍正确进行。
memcpy函数假设要复制的内存区域不存在重叠,如果你能确保你进行复制操作的的内存区域没有任何重叠,可以直接用memcpy;
如果你不能保证是否有重叠,为了确保复制的正确性,你必须用memmove。
memcpy的效率会比memmove高一些,如果还不明白的话可以看一些两者的实现:
![](https://www.cnblogs.com/Images/OutliningIndicators/ExpandedBlockStart.gif)
void *memmove(void *dest, const void *source, size_t count)
{
assert((NULL != dest) && (NULL != source));
char *tmp_source, *tmp_dest;
tmp_source = (char *)source;
tmp_dest = (char *)dest;
if((dest + count<source) || (source + count) <dest))
{// 如果没有重叠区域
while(count--)
*tmp_dest++ = *tmp_source++;
}
else
{ //如果有重叠
tmp_source += count - 1;
tmp_dest += count - 1;
while(count--)
*--tmp_dest = *--tmp;
}
return dest;
}
void *memcpy(void *dest, const void *source, size_t count)
{
assert((NULL != dest) && (NULL != source));
char *tmp_dest = (char *)dest;
char *tmp_source = (char *)source;
while(count --)//不对是否存在重叠区域进行判断
*tmp_dest ++ = *tmp_source ++;
return dest;
}
memmove普通实现
memmove - Copy source buffer to destination buffer
;
;Purpose:
; memmove() copies a source memory buffer to a destination memory buffer.
; This routine recognize overlapping buffers to avoid propogation.
; For cases where propogation is not a problem, memcpy() can be used.
;
; Algorithm:
void* memmove(void* dest, void* source, size_t count)
{
void* ret = dest;
if (dest <= source || dest >= (source + count))
{
//Non-Overlapping Buffers
//copy from lower addresses to higher addresses
while (count --)
*dest++ = *source++;
}
else
{
//Overlapping Buffers
//copy from higher addresses to lower addresses
dest += count - 1;
source += count - 1;
while (count--)
*dest-- = *source--;l
}
return ret;
}
另一种实现:
void* mymemcpy( void* dest, const void* src, size_t count )
{
char* d = (char*)dest;
const char* s = (const char*)src;
// int n = (count + 7) / 8; // count > 0 assumed
int n = count >> 3;
switch( count & 7 )
{
do { *d++ = *s++;
case 7: *d++ = *s++;
case 6: *d++ = *s++;
case 5: *d++ = *s++;
case 4: *d++ = *s++;
case 3: *d++ = *s++;
case 2: *d++ = *s++;
case 1: *d++ = *s++;
case 0 } //while (--n > 0);
while (n-- > 0)
}
return dest;
}