Limit integer to bounds

Question

I'm trying to make sure that int x is greater or equal than 0 but smaller than 1080 (screen size in this case).

I came up with this

int x = 123;
x = std::min(std::max(x, 0), 1080);

This seems ugly. Is there a better way to achieve this?

Looks fine to me. Write a wapper function... http://www.boost.org/doc/libs/1_54_0/libs/algorithm/doc/html/algorithm/Misc.html — Karoly Horvath, Jan 08 '17 at 17:30
Do it explicitly `if (x >= 0 && x < 1080) { // your code here. }` — 0x1337, Jan 08 '17 at 17:31
I would take the code to assure the range, and wrap it up in a template, something like [this](http://stackoverflow.com/a/3181826/179910). — Jerry Coffin, Jan 08 '17 at 17:34

score 8 · Accepted Answer · answered Jan 08 '17 at 17:32

8

If you live in the future, you can use std::clamp from C++17:

x = std::clamp(x, 0, 1080);

answered Jan 08 '17 at 17:32

melpomene

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Plus one from the future! There is no harm in including good snippets from future standards. – Bathsheba Jan 08 '17 at 17:34
Ok thanks, this is mostly what I was looking for. I'll keep the min / max combo until future comes – Eric Jan 08 '17 at 17:34
@Eric: huh?? Just write your own wrapper then replace it with clamp when the time comes... – Karoly Horvath Jan 08 '17 at 17:36

Nick · Answer 2 · 2017-01-08T17:37:02.923

2

Naive solution looks ok too:

int x = 123;
if (x < 0)
    x = 0;
if (x > 1080)
    x = 1080;

or wrap everything in a function:

template<typename T>
T between(const T x, const T min, const T max){
   if (x < min)
       return min;

   if (x > max)
       return max;

   return x;
}

int x = between(123, 0, 1080);

edited Jan 08 '17 at 17:37

answered Jan 08 '17 at 17:34

Nick

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code fixed, sorry – Nick Jan 08 '17 at 17:38
1

Reasonably simple to write and ultra simple to read... When questions are really simple no need for super complex C++17 things. My prefered way! – Serge Ballesta Jan 08 '17 at 17:40

Bathsheba · Answer 3 · 2017-01-08T17:33:00.087

1

Use an unsigned as the type for x. That automatically constrains it to be non-negative.

Then you're left with just the call to std::min which is palatable to me at least.

Building a class that takes an int on construction and has a conversion operator to int is also plausible but requires a fair bit of boilerplate.

edited Jan 08 '17 at 17:33

answered Jan 08 '17 at 17:30

Bathsheba

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This is a good idea, but unfortunately x is fetched through an API callback so I can't change the type easily. Of course I could cast int to unsigned int, but this is ugly as well imo – Eric Jan 08 '17 at 17:31

score -1 · Answer 4 · answered Jan 08 '17 at 17:32

-1

Why not just do something like this?

int x = 123; /* Or some arbitrary number */
bool bIsInRange = (x >= 0) && (x < 1080);
if(!bIsInRange){
   std::cout << "Variable 'x' is not between 0 and 1080." << std::endl;
   return 1;
} // else, x is fine. Continue operations as normal.

answered Jan 08 '17 at 17:32

Spencer D

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That looks super ugly. – melpomene Jan 08 '17 at 17:33
@melpomene, however, it is also standard, readable, and extremely understandable for anyone of any experience level. Plus, doing it in this normal way does not require additional calls to functions such as `min`, `max`, or `clamp`, which only serve to add overhead/expense to what would otherwise be a simple and essentially weightless task. That being said, I will agree that it is by no means *"beautiful."* – Spencer D Jan 08 '17 at 17:40
I don't find it readable. The main issue is the `bIsInRange` variable, which serves no purpose and has a horrible name. As for overhead, I'd expect the compiler to inline min/max/clamp. (And if we're talking efficiency, that `cout << "..." << endl` construct is probably 100 times slower than a plain function call.) – melpomene Jan 08 '17 at 17:43

Limit integer to bounds

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