DEV Community

Cover image for How can you swap two variables without using a third?
Nick Karnik
Nick Karnik

Posted on • Updated on

How can you swap two variables without using a third?

Moore's law is just an observation and not a real law like the Law of Gravity.

Over the years, developers have taken the increasing CPU speeds and Memory sizes for granted. Here's a warm-up interview question that I ask every candidate.

Let's assume the working memory for your function is 8-Bytes. You are given two 32-Bit Integers and you need to swap them. In other words, how can you swap two variables without using a third?

Please take your time to solve this and refrain from looking up solutions online or for answers below. This is your first step in becoming a Computer Scientist!


If you accepted this challenge, ❤️ it and [follow me on Twitter](https://twitter.com/intent/follow?screen_name=theoutlander).

Top comments (122)

Collapse
 
sadarshannaiynar profile image
Adarsh
if (x == y) return;
x = x ^ y
y = x ^ y
x = x ^ y

Bitwise operators for the win.

Collapse
 
nestedsoftware profile image
Nested Software • Edited

This is a rather beautiful "bit" of problem solving :)

It will work for both integers and floating points, positive and negative, and there is no issue with overflow. If someone can figure this out on their own, I think that's quite impressive. Most of us just know it from having seen it before though.

Collapse
 
theoutlander profile image
Nick Karnik

That's true. My initial solution to it was using addition/subtraction and once I got the concept of diffs, I was able to deduce a solution using bit manipulation.

Collapse
 
ptdecker profile image
P. Todd Decker

JavaScript, by nature, stores all numbers as double-precision IEEE 754 floating point numbers. The 52 bits of the fractional portion is used to store integers. This results in 'maximum safe integer' of 2^53 - 1. Further complicating things is that JavaScript's bitwise operators only deal with the lowest 32 bits of JavaScript's 54-bit integers. So the bitwise approach will not work on big integers in JavaScript.

Collapse
 
d4vecarter profile image
Dave Carter™

I guess BigInt new primitive proposal will do the trick :)

Collapse
 
theoutlander profile image
Nick Karnik

The swap with bitwise operators will be the fastest at the hardware level. :)

Collapse
 
almostconverge profile image
Peter Ellis

Rather disappointingly, in most modern hardware it's the one with the temp variable that is fastest.

Thread Thread
 
theoutlander profile image
Nick Karnik

Depending on the architecture and compiler, they end up using three registers to do a swap under the hood I think.

Thread Thread
 
almostconverge profile image
Peter Ellis

Yes, but the idea is that you do this in assembly. It was a common trick because it saved a register and (at worst) it ran in the same number of cycles.

Collapse
 
codemouse92 profile image
Jason C. McDonald • Edited

When I started going through the comments, I was actually thinking "what about bitwise xor?" And sure enough, here it is. ^,^

Collapse
 
sadarshannaiynar profile image
Adarsh

Yes. Whenever you can achieve something using bitwise operations I always prefer that.

Collapse
 
cathodion profile image
Dustin King

x, y = y, x #

Collapse
 
cathodion profile image
Dustin King • Edited

To be a little bit sneaky, use part of the function itself as swap space. This would only work if that space could be after the return instruction.

In pseudo-C:

#define NOP {32 bits of assembly NOP instructions};

void swap(int* a, int* b){

    *&yep = *a;
    *a = *b;
    *b = *&yep;

    return;

    yep:
    NOP;
}

This isn't thread-safe though, and a decent operating system might throw exceptions if you try to have executable code modify itself.

Collapse
 
zhu48 profile image
Zuodian Hu

The NOPs don't actually give you any memory. Those NOPs are wherever the executable image was loaded in memory, and the variable yep is on the stack. Assuming your standard C memory model.

Thread Thread
 
theoutlander profile image
Nick Karnik

I misread the post on the phone, but I see what the code is doing. Although, I'm not sure what assigning to NOP does?

Thread Thread
 
cathodion profile image
Dustin King

C memory layout:

C memory layout (source)

The function code itself is in the text segment. The variables for a particular invocation of the function are on the stack. Padding out the function with NOP creates space within the function, which I'm using as a variable. Basically I'm interpreting the "working memory" part of the post to mean "variables allocated on the stack". Now that I think of it, what I'm doing is basically a static variable, which (according to the link) C puts in the initialized data segment if you do it the right way, but then it would legitimately be a variable so I'd lose.

It's probably not a legit way to do things (and might be impossible if the text segment is read-only as the link says it might be), but people had already posted the legit ways.

Thread Thread
 
theoutlander profile image
Nick Karnik

Thanks for that explanation. It's been a while since I've written C/C++, but this approach makes sense. I love the thought process of cleverly using that available memory, even if it doesn't work out. 👏👏👏

Collapse
 
zhu48 profile image
Zuodian Hu

The variable yep is declared within a function body without a static qualifier, which means it's what the C standard calls an automatic storage duration variable. This means its duration is between the curly braces, and most compilers will use a register or put it on the stack.

Being completely nitpicky here, but the variable yep is declared after its use and also without a type. I'm sure modern compilers wouldn't even let you do that.

Modern operating systems load the text segment into a virtual memory page marked as read-only. So yes, even if this compiles, it will generate a segfault on modern operating systems.

Thread Thread
 
cathodion profile image
Dustin King

yep isn't a variable, though, it's a label, representing an offset from the start of the function.

I agree compilers and OS's will tend to prevent this. I said as much before, and trying to create a proof of concept in assembly has born that out.

It wasn't a completely fruitless exercise though, as it was a chance to learn some assembly, which provided some insights about the challenge that I might make a post about. Basically, swapping using math might not actually be more memory-efficient (however you define that, and without the kind of cheating I've been talking about) than swapping using "a variable".

Collapse
 
theoutlander profile image
Nick Karnik

It creates more memory though.

Collapse
 
theodesp profile image
Theofanis Despoudis • Edited

Tricky bit: It can cause integer overflow if x + y > int.MAX

Collapse
 
drbearhands profile image
DrBearhands • Edited

assuming cyclic overflows, i.e. INTMAX + 1 = INTMIN:
if INTMAX - y < x (equivalent to x + y > INTMAX without cycles)
then x + y = INTMIN + y - (INTMAX - x - 1)
ergo, rewriting the equations as
x' = x + y
y' = x' - y
x'' = x' - y'
we get:
x' = INTMIN + y - (INTMAX - x - 1)
y' = x' - y
y' = INTMIN - (INTMAX - x - 1)
y' = INTMIN + 1 - INTMAX + x
y' = x
x'' = INTMIN + y - (INTMAX - x - 1) - y'
x'' = INTMIN + y - (INTMAX - x - 1) - x
x'' = INTMIN + y - INTMAX + x + 1 - x
x'' = INTMIN + 1 + y - INTMAX
x'' = y

Overflows schmoverflows.

EDIT: Might be worth noting that this approach and the bitwise operator are essentially the same thing. Combine two things in a way that is reversible if you have either of the two originals.

Thread Thread
 
theodesp profile image
Theofanis Despoudis

Unfortunately that works in theory but in practice and most compilers do not guarantee that behavior. For example:

Java: Will back cycle to -2147483648
Javascript: will eventually print infinity if the value is too high
Go: Will not compile

 
theodesp profile image
Theofanis Despoudis

That's the second step in becoming a computer scientist.

Thread Thread
 
lewiscowles1986 profile image
Lewis Cowles

limit x and y to 16 bit or store 32-bit numbers in 64-bit. max int for 64-bit signed or unsigned is twice that of it's 32-bit counterpart.

You cannot overcome size problem if it's expressed in bits. Only higher-order structures like arrays would enable that and then you are absolutely storing more than two variables (possibly more than one per entry if it's a pointer).

Thread Thread
 
theodesp profile image
Theofanis Despoudis • Edited

That defeats the purpose of doing that plus you are making a difficult work of complicating things. I suspect is better if you use a temp variable after all!

Thread Thread
 
lewiscowles1986 profile image
Lewis Cowles

I suspect it's better if you don't try to copy with only two memory areas, but if you're going to worry about overflowing, then doubling storage size is the least complex thing you could do, and compatible with various CPU operating modes.

Collapse
 
yaser profile image
Yaser Al-Najjar • Edited

Only two options, either using memory or CPU

  1. Memory = third variable
temp = x
x = y
y = temp
  1. CPU = Swap algorithms

like XOR swap (bullet-proof solution)

x ^= y
y ^= x
x ^= y

Or addition & subtraction (which is not bullet-proof when dealing with a 32-bit register that would need extra care with overflow)

x += y
y = x - y
x -= y
Collapse
 
dean profile image
dean

XOR swap is NOT bullet proof, you need to check if x == y first. If x == y, then the function sets both values to zero.

Collapse
 
johnfound profile image
johnfound • Edited

No, it will not. Let set x=5, y=5.

x = x xor y = 5 xor 5 = 0
y = y xor x = 5 xor 0 = 5
x = x xor y = 0 xor 5 = 5

It is a nop actually, but still correct.

Collapse
 
lexlohr profile image
Alex Lohr

Since xor and subtraction were already mentioned, may I chime in with a circular shift over the 64 bit value of the whole function memory?

Collapse
 
amexboy profile image
Amanu

This one is different, but how exactly would you implement it in a programming language?

Collapse
 
lexlohr profile image
Alex Lohr

Basically, it works like this (x64 assembler):

mov rax, [m64]
ror rax, 32
mov [m64], rax

Thread Thread
 
amexboy profile image
Amanu

No, I mean on a high level programming language which you have no idea where is stored.

Even for an assembly, the two variables could potentially be stored in two different registers. But we're going to assume not!

Cool idea anyways

Collapse
 
drbearhands profile image
DrBearhands

I love you

 
theoutlander profile image
Nick Karnik

That's awesome. I did not know about it. I wonder if it swaps values instead of pointters.

Found this on SO @ stackoverflow.com/questions/445179....

StackOverflow Answer

Thread Thread
 
almostconverge profile image
Peter Ellis

So now we've reduced the problem to "How do you swap the two topmost stack items?" :)

Collapse
 
zhu48 profile image
Zuodian Hu • Edited

Very contrived, only works on ARM, not even acceptable by all C compilers, but you never constrained the problem that way!

void swap(int* a, int* b) {
    __asm volatile (
        "MOV R2, R1 \n"
        "MOV R1, R0 \n"
        "MOV R0, R2 \n"
    );
}
Collapse
 
zhu48 profile image
Zuodian Hu

I went to bed, and realized that my answer is wrong. Exchanging the values in register is completely invisible to the function caller. Here's the correct answer.

void swap(int* a, int* b) {
    __asm volatile {
        "LDR R2, [R0] \n"
        "LDR R3, [R1] \n"
        "STR R2, [R1] \n"
        "STR R3, [R0] \n"
    }
}

Further, as R0-R3 are all caller-save registers in the ARM ABI, the function body, excluding the function entry stack shenanigans, uses zero memory.

Collapse
 
zhu48 profile image
Zuodian Hu

Oh and as prplz mentioned, any self-respecting compiler nowadays will use only registers any way you decide to implement this little stub of a function. Yes yes, the question is very contrived too, but hey.

Collapse
 
theoutlander profile image
Nick Karnik • Edited

Aren't you using a third register though?

Are you're thinking that you can do that because it's not in RAM?

Thread Thread
 
zhu48 profile image
Zuodian Hu

The way the question is formulated, you're limited by having a certain amount of memory. Any register use doesn't count against that at all.

Collapse
 
jjjjcccjjf profile image
endan
function swap (&$a, $b, $A) {
    $a = $b;
    return $A;

} 

$a = 7;
$b = 9;

$b = swap($a, $b, $a);

echo "a is " . $a; # 9
echo "b is " . $b; # 7

/* scratches head again */
Collapse
 
gmartigny profile image
Guillaume Martigny

As funny as it is, you declare a new pointer to the swap function allocating new memory.

Collapse
 
svenluijten profile image
Sven Luijten
// Since PHP 7.1:
[$b, $a] = [$a, $b];

// Or before 7.1:
list($b, $a) = [$a, $b];
Collapse
 
vycoder profile image
yev • Edited
x ^= y;
y ^= x;
x ^= y;
Collapse
 
dean profile image
dean • Edited
x := 5
y := 5
swap(&x, &y)
fmt.Println(x, y) // uh oh, they are both zero!
Collapse
 
vycoder profile image
yev

I've written my answer with C in mind, and it does explain how it works if you try the calculation by hand: onlinegdb.com/B1IY6hFK7

#include <stdio.h>

int main()
{
    int a = 77;
    int b = 42;
    printf("a: %d, b: %d \n", a, b);

    swap(&a, &b);
    printf("after swap \n");
    printf("a: %d, b: %d", a, b);

    return 0;
}

void swap(int *a, int *b) {
    *a ^= *b;
    *b ^= *a;
    *a ^= *b;
}

When both a and b has the same value: onlinegdb.com/r1WeypFF7

Result may vary depending on the language compiler, you might have to adjust the snippet according to your language's nuances.

Some comments may only be visible to logged-in visitors. Sign in to view all comments.