C, 22 bytes

f(){return(int**)0+8;}

C, stand-alone, 34 bytes

main(){printf("%d\n",(int**)0+8);} 

C, fun with Unicode, 30 code-points, 34 bytes(UTF-8)

main(){puts((int**)U" ㈳㐶"+1);}

note: non-competing

C, 22 bytes

f(){return(int**)0+8;}

This is a pointer-size based answer that assumes a native binary. The 0 is cast to int** (address 0x0). Then we add 8 to 0, which, in C advances, the pointer by sizeof(int*)*8. 4 bytes * 8 bits = 32, 8 bytes * 8 bits = 64. So we get (int**)0x20 and 0x40 which are then implicitly cast as integers by returning them from an implicitly int-returning function.

C, stand-alone, 34 bytes

main(){printf("%d\n",(int**)0+8);} 

C, fun with Unicode, 30 code-points, 34 bytes(UTF-8)

main(){puts((int**)U" ㈳㐶"+1);}

C, 22 bytes

f(){return(int**)0+8;}

C, stand-alone, 34 bytes

main(){printf("%d\n",(int**)0+8);} 

C, fun with Unicode, 30 code-points, 35 bytes(UTF-8)

main(){puts((int**)U" ㈳㐶"+1);}

note: non-competing

C, 22 bytes

f(){return(int**)0+8;}

C, stand-alone, 34 bytes

main(){printf("%d\n",(int**)0+8);} 

C, fun with Unicode, 30 code-points, 34 bytes(UTF-8)

main(){puts((int**)U" ㈳㐶"+1);}

note: non-competing

C, 22 bytes

f(){return(int**)0+8;}

C, stand-alone, 34 bytes

main(){printf("%d\n",(int**)0+8);} 

C, fun with Unicode, 35 bytes

main(){puts((int**)U" ㈳㐶"+1);}

note: non-competing

C, 22 bytes

f(){return(int**)0+8;}

C, stand-alone, 34 bytes

main(){printf("%d\n",(int**)0+8);} 

C, fun with Unicode, 30 code-points, 35 bytes(UTF-8)

main(){puts((int**)U" ㈳㐶"+1);}

note: non-competing