How to simplify if statements - c

Question

How can I simplify below if statements?, I'm trying to achieve the possibly most efficient code.
// doSomething based on x value and y value
void doSomething(x int, y int) {
//x not zero and y not zero
if (x != 0 && y != 0) {
//do a
//do b if x greater or equal y
//else do c
if (x >= y) {
//do b
} else (x < y) {
//do c
}
return;
}
//do b if x not zero and y zero
if (x != 0 && y == 0) {
//do b
return;
}
//do e if both x and y zero
if (x == 0 && y == 0) {
//do e
return;
}
}
What is the most concise and efficient way to simplify?

Answer 1

As it stands, your code is equivalent to:
if (x == 0)
{
if (y == 0)
do e
}
else if (y != 0)
{
if (x < y)
do a,c
else
do a,b
}
In particular:
if x=0 && y!=0 or y=0 && x!=0, then do nothing;
the two branches //do b if x not zero and y zero and //do d if x zero and y not zero are unreachable in your code.

Answer 2

History:
The code in the question was changed several times. This answer is based on the original version of the code. That code is shown below, minimally modified so that it's compilable C code:
void original(int x, int y)
{
if (x == y && x != 0 && y != 0) {
// do a
// do b
return;
}
if (x < y && x != 0 && y != 0) {
// do a
// do c
return;
}
if (x > y && x != 0 && y != 0) {
// do a
// do b
return;
}
if (x != 0 && y == 0) {
// do a
// do b
return;
}
if (x == 0 && y != 0) {
// do a
// do d
return;
}
if (x == 0 && y == 0) {
// do e
return;
}
}
Motivation:
The advantage of the original code is that it clearly covers all possible input combinations.
The first three if statements handle the cases where x and y are non-zero (with x==y, x<y, and x>y handled separately). Then the remaining three if statements handle the cases where one or both are zero. That covers everything.
The disadvantages of the original code are:
Many of the comparisons are repeated several times, e.g. y != 0 is checked four times. Given that branching statements tend to be time consuming in contemporary (year 2021) processors, redundant comparisons are something to be avoided 1.
The code violates the DRY principle. Specifically, do a appears in five locations, and do b appears in three locations. So any refactoring of the code should attempt to eliminate the repetition.
1) The optimizer may be able reduce the number of comparisons. However, that's hard to test without real code.
Refactoring with nested if statements:
If we focus on the DRY principle first, we notice that a appears in 5 of 6 code blocks. Only the case for e doesn't invoke a. So the logical choice is to handle e first. This results in the following code structure:
if (x == 0 && y == 0) {
// do e
} else {
// do a
// everything else (note that either x is not zero, or y is not zero)
}
Within the "everything else", we have do b three times, do c once, and do d once. So the next step is to handle c and d, leaving only b. The refactored code looks like this:
void nestedIf(int x, int y)
{
if (x == 0 && y == 0) {
// do e
} else {
// do a
if (x == 0) {
// do d
} else {
if (y != 0 && x < y) {
// do c
} else {
// do b
}
}
}
}
Refactoring with early return statements:
Some coding standards insist that a function shall only return from one location. The nested if code meets that requirement, but at the cost of being a little messy due to nesting. Using early returns eliminates the nesting, and may allow the code for a thru e to be implemented inline without additional functions (assuming they're only a few lines each). The refactored code looks like this:
void earlyReturn(int x, int y)
{
if (x == 0 && y == 0) {
// do e
return;
}
// do a
if (x == 0) { // x is zero and y is not zero
// do d
return;
}
if (y != 0 && x < y) { // x and y are both not zero, and x < y
// do c
return;
}
// do b
}

Answer 3

assuming a(), b(), and e() have the same type (fxtype)
fxtype *fx[4] = { e, NULL, b, a }; // NULL when x==0, y!=0
fx[2*!!x + !!y](x, y);
Change a to include c() and b()
int a(int x, int y) {
// do previous a()
if (x < y) c(x, y); else b(x, y);
return 0;
}