C hardcoded array as memcpy parameter

If you use C99 or later, you can use compound literals. (N1256 6.5.2.5)

#include <stdio.h>
#include <string.h>
int main(void){
    char dest[5] = {0};
    memcpy(dest, (char[]){0xE3,0x83,0xA2,0xA4,0xCB} ,5);
    for (int i = 0; i < 5; i++) printf("%X ", (unsigned int)(unsigned char)dest[i]);
    putchar('\n');
    return 0;
}

UPDATE: This worked for C++03 and C++11 on GCC, but are rejected with -pedantic-errors option. This means this is not a valid solution for standard C++.

#include <cstdio>
#include <cstring>
int main(void){
    char dest[5] = {0};
    memcpy(dest, (const char[]){(char)0xE3,(char)0x83,(char)0xA2,(char)0xA4,(char)0xCB} ,5);
    for (int i = 0; i < 5; i++) printf("%X ", (unsigned int)(unsigned char)dest[i]);
    putchar('\n');
    return 0;
}

points are:

• Make the array const, or taking address of temporary array will be rejected.
• Cast numbers to char explicitly, or the narrowing conversion will be rejected.