c - strncpy() and memcpy() are different? -

strncpy() , memcpy() same?

because of fact strncpy() accepts char * parameter, icasts integer arrays char *.

why gets different output?

here code,

#define _crt_secure_no_warnings  #include <stdio.h> #include <string.h> #define size 5   void print_array(int *array, char *name ,int len) {     int i;     printf("%s ={ ",name);     for(i=0; i<len;i++)         printf("%d," ,array[i]);     printf("}\n"); }  int main(void){     char string1[size] = {'1','2','3','4','\0'};     char string2[size], string3[size];     int array1[size] = {1,2,3,4,5};     int array2[size],array3[size];      strncpy(string2,string1,sizeof(string1));     memcpy(string3, string1,sizeof(string1));     printf("string2 =%s \n",string2);     printf("string3 =%s \n",string3);      strncpy((char *)array2, (char*)array1 , sizeof(array1));     memcpy(array3,array1,sizeof(array1));     print_array(array2,"array2",size);     print_array(array3,"array3",size);      return 0; } 

it turns out

string2 =1234 string3 =1234 array2 ={ 1,0,0,0,0,} array3 ={ 1,2,3,4,5,} 

but why? gives different answer?

strncpy((char *)array2, (char*)array1 , sizeof(array1)); 

casting array1 char * doesn't want. there no characters @ location, there integers (little-endian seems).

what happening strncpy copies bytes integer array until reaches 0 byte, pretty soon. on other hand memcpy doesn't care 0 bytes , copies whole thing.

to put way, strncpy finds 0 byte "inside" first integer , stops copying; fill destination buffer zeros specified size.