Why C compilers optimize switch and if differently

If you explicitely enumerate all the cases, gcc is very efficient :

int c(int num) {
    num &= 0xF;
    switch (num) {
        case 0: case 1: case 8: case 9: 
            return -1;
        case 4: case 5: case 12: case 13:
            return 1;
            case 2: case 3: case 6: case 7: case 10: case 11: case 14: case 15: 
        //default:
            return 0;
    }
}

is just compiled in a simple indexed branch :

c:
        and     edi, 15
        jmp     [QWORD PTR .L10[0+rdi*8]]
.L10:
        .quad   .L12
        .quad   .L12
        .quad   .L9
        .quad   .L9
        .quad   .L11
        .quad   .L11
        .quad   .L9
        .quad   .L9
        .quad   .L12
etc...

Note that if default: is uncommented, gcc turns back to its nested branch version.

C compilers have special cases for switch, because they expect programmers to understand the idiom of switch and exploit it.

Code like:

if (num == 0 || num == 1 || num == 8 || num == 9) 
    return -1;

if (num == 4 || num == 5 || num == 12 || num == 13)
    return 1;

would not pass review by competent C coders; three or four reviewers would simultaneously exclaim "this should be a switch!"

It's not worth it for C compilers to analyze the structure of if statements for conversion to a jump table. The conditions for that have to be just right, and the amount of variation that is possible in a bunch of if statements is astronomical. The analysis is both complicated and likely to come up negative (as in: "no, we can't convert these ifs to a switch").