dEEpEst Posted December 8, 2016 Share Posted December 8, 2016 function [h=1]memcpy[/h]void * memcpy ( void * destination, const void * source, size_t num ); Copy block of memory Copies the values of num bytes from the location pointed to by source directly to the memory block pointed to by destination. The underlying type of the objects pointed to by both the source and destination pointers are irrelevant for this function; The result is a binary copy of the data. The function does not check for any terminating null character in source - it always copies exactly num bytes. To avoid overflows, the size of the arrays pointed to by both the destination and source parameters, shall be at least numbytes, and should not overlap (for overlapping memory blocks, memmove is a safer approach). [h=3]Parameters[/h]destinationPointer to the destination array where the content is to be copied, type-casted to a pointer of type void*.sourcePointer to the source of data to be copied, type-casted to a pointer of type const void*.numNumber of bytes to copy. size_t is an unsigned integral type. [h=3]Return Value[/h]destination is returned. [h=3]Example[/h][TABLE=class: snippet] [TR] [TD=class: rownum, align: right]1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 [/TD] [TD=class: source]/* memcpy example */ #include #include struct { char name[40]; int age; } person, person_copy; int main () { char myname[] = "Pierre de Fermat"; /* using memcpy to copy string: */ memcpy ( person.name, myname, strlen(myname)+1 ); person.age = 46; /* using memcpy to copy structure: */ memcpy ( &person_copy, &person, sizeof(person) ); printf ("person_copy: %s, %d \n", person_copy.name, person_copy.age ); return 0; }[/TD] [TD=class: C_btnholder] [/TD] [/TR] [/TABLE] Output: [TABLE=class: snippet] [TR] [TD=class: output] person_copy: Pierre de Fermat, 46 [/TD] [/TR] [/TABLE] Link to comment Share on other sites More sharing options...
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