c - is there any consequence if I do assignment but not memcpy after malloc -

in following program:

int main() {   struct node node;   struct node* p = (struct node*) malloc(sizeof(struct node));   *p =node;   printf("%d\n", *p->seq); } 

usually did memcpy(p, node, sizeof(node))

now tried code above, , works fine, i'm afraid there consequence or faulty stuff if assignment not memcpy after malloc. there or assignment correct? thanks!

jesus ramos correct:

1) *p =node; copies in "node" "*p"

2) not need extraneous "memcpy()"

3) must allocate "*p" (with "malloc()") before copy.

here standalone test:

// c source #include <stdio.h> #include <malloc.h>  struct node {     int a;     int b;     struct node *next; };  int main() {   struct node node;   struct node *p = malloc(sizeof(struct node));   *p = node;   return 0; }   # resulting assembler main:         leal    4(%esp), %ecx         andl    $-16, %esp         pushl   -4(%ecx)         pushl   %ebp         movl    %esp, %ebp         pushl   %ecx         subl    $20, %esp         movl    $12, (%esp)         call    malloc         movl    %eax, -8(%ebp)         movl    -8(%ebp), %edx         movl    -20(%ebp), %eax         movl    %eax, (%edx)         movl    -16(%ebp), %eax         movl    %eax, 4(%edx)         movl    -12(%ebp), %eax         movl    %eax, 8(%edx)         movl    $0, %eax         addl    $20, %esp         popl    %ecx         popl    %ebp         leal    -4(%ecx), %esp         ret