字符串拷贝,一般用memcpy就行了
但如果是一个字符串从自身的某一个位置 copy到另外一个位置时，就涉及到重叠(overlap)的情景，这时不能使用memcpy,
要使用一个临时数组，存到这个临时数组中，然后memcpy(dst, 临时数组）
上述这个过程，有现成的函数memmove，它能处理这种情景。
#include
#include <string.h>
#include
char string1[60] = "The quick brown dog jumps over the lazy fox";
char string2[60] = "The quick brown fox jumps over the lazy dog";
/* 1 2 3 4 5
* 12345678901234567890123456789012345678901234567890
*/
void main( void )
{
printf( "Function:\tmemcpy without overlap\n" );
printf( "Source:\t\t%s\n", string1 + 40 );
printf( "Destination:\t%s\n", string1 + 16 );
memcpy( string1 + 16, string1 + 40, 3 );
printf( "Result:\t\t%s\n", string1 );
printf( "Length:\t\t%d characters\n\n", strlen( string1 ) );
/* Restore string1 to original contents */
memcpy( string1 + 16, string2 + 40, 3 );
printf( "Function:\tmemmove with overlap\n" );
printf( "Source:\t\t%s\n", string2 + 4 );
printf( "Destination:\t%s\n", string2 + 10 );
memmove( string2 + 10, string2 + 4, 40 );
printf( "Result:\t\t%s\n", string2 );
printf( "Length:\t\t%d characters\n\n", strlen( string2 ) );
printf( "Function:\tmemcpy with overlap\n" );
printf( "Source:\t\t%s\n", string1 + 4 );
printf( "Destination:\t%s\n", string1 + 10 );
memcpy( string1 + 10, string1 + 4, 40 );
printf( "Result:\t\t%s\n", string1 );
printf( "Length:\t\t%d characters\n\n", strlen( string1 ) );
}
在linux下的结果是：
Vsles11:/mnt/Pub/1 # ./linux_memmove
Function: memcpy without overlap
Source: fox
Destination: dog jumps over the lazy fox
Result: The quick brown fox jumps over the lazy fox
Length: 43 characters
Function: memmove with overlap
Source: quick brown fox jumps over the lazy dog
Destination: brown fox jumps over the lazy dog
Result: The quick quick brown fox jumps over the lazy dog
Length: 49 characters
Function: memcpy with overlap
Source: quick brown dog jumps over the lazy fox
Destination: brown dog jumps over the lazy fox
Result: The quick quick quicn dog jumps over the lazy laz
Length: 50 characters
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