void * mempcpy (void *dest, const void *src, size_t dimension)
The perform takes Three arguments:
1. dest :
It is a beginning pointer of a reminiscence block the place the reminiscence block pointed by src (2nd argument) can be copied. The pointer is asserted as void, so any kind of reminiscence block might be copied.
2. src :
It is a beginning pointer of the supply reminiscence block from the place the reminiscence block can be copied. The pointer is asserted as void, so any kind of reminiscence block might be copied.
3. dimension :
That is the dimensions of the reminiscence block in bytes.
The worth of the 2 pointer dest and src must be in such a manner that two reminiscence blocks don’t overlap. The scale of reminiscence blocks of supply and vacation spot should be not less than of dimension (third argument) bytes to keep away from overlapping conditions. If the 2 reminiscence blocks are overlapped then the habits of the memcpy() perform is undefined. When there’s a risk of overlapping, you need to use the memmove() library perform the place overlapping is effectively outlined. memmove() perform is slower in comparison with memcpy() perform.
As a result of worth of dimension, if the supply or vacation spot is accessed past their buffer size then the habits of the memcpy() perform is undefined.
The memcpy() perform doesn’t examine to terminate ‘’ character.
This perform returns the worth of vacation spot tackle dest. As the worth of dest is already out there so, it want to not retailer in any variable.
In Example1.c we’ve declared two-character array src and dest. The scale of the src is 6 and the dest is 13. First, we copied 6 characters ‘H’, ‘e’, ‘l’, ‘l’, ‘o’, ‘’ from src to dest ( Line 11 ). Within the second memcpy() perform copied Eight characters ‘ ’, ‘w’, ‘o’, ‘r’, ‘l’, ‘d’, ‘!’, ‘’ to the dest after 5 characters ( Line 15 ). Pictorially we will symbolize this as follows:
typedef struct scholar
std student1; // Declare student1 of kind std
std student2; // Declare student2 of kind std
// Assigning the worth of sudent1
student1.title = “Bamdev Ghosh”;
student1.id = 1105;
student1.age = 30;
printf(“Student1:ntName : %sntid : %dntage : %dn”,student1.title,
// Copy student1 to student2
memcpy(&student2, &student1, sizeof(student1));
printf(“nnStudent2:ntName : %sntid : %dntage : %dn”,
In Example2.c we’ve declared two construction student1 and student2 (Line 15 and 16). First, we initialize student1 (Line 19, 20, 21). After that, we use memcpy to repeat knowledge from student1 to student2.
On this article, we’ve realized the way to use the memcpy perform. We’ve seen that this perform can be utilized for any kind of reminiscence block however this perform has some limitations. So, it’s important to use this perform rigorously.
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