how do you do a type casting in one line in c?

January 28, 2022

unsigned char num[4]={0,2}; //512 little endian
unsigned int * ptr = num;

printf("%u\n", *ptr); // 512

//trying to do the same underneath in one line but it dosen't work    
printf("%u\n", (unsigned int *)num); //

>Solution :

The same as

unsigned int * ptr = num; printf("%u\n", *ptr); // !! nonportable, nonsafe (UB-invoking)

in one line would be

printf("%u\n", *(unsigned int *)num); // !! nonportable, nonsafe (UB-invoking)

However both versions are nonportable and unsafe. They invoke undefined behavior by violating the strict aliasing rule and possibly by creating a pointer that’s not suitably aligned.

You can do it safely with memcpy:

#include <stdio.h>
#include <string.h>

int main(){
    unsigned char num[4]={0,2}; //512 little endian

    #if 0
    //the unsafe versions
    unsigned int * ptr = num; printf("%u\n", *ptr);
    printf("%u\n", *(unsigned int *)num);
    #endif

    //the safe version using a compound literal as an anonymous temporary;
    //utilizes how memcpy returns the destination address
    printf("%u\n",*(unsigned*){memcpy(&(unsigned){0},&num,sizeof(num))});

}