Is multiple variable comparision inline undefined behavior?

Question

I was telling a friend of mine (which is learning C) that he couldn't do multiple variables comparision at once:

int main(){
    int a[4];

    scanf("%d %d %d %d", &a[0], &a[1], &a[2], &a[3]);

    if(a[0] < a[1] < a[2] < a[3]){
        printf("OK!\n");

    }

    else{
        printf("I've told ya\n");

    }

}

So, to prove I was right I've coded the program above and then I've executed it with 1 2 3 4 . Surprisingly it printed OK! . And so I didn't know what to tell him, because I was sure it wasn't right.

Finally, is it or is it not undefined behavior?

Answer 1

No, it's well-defined. It simply has different semantics to what you're expecting.

The expression is evaluated as follows:

if (((a[0] < a[1]) < a[2]) < a[3]) {

Each comparison produces a boolean ( 0 or 1 ) outcome.

The (boolean) result of a[0] < a[1] is compared to a[2] , and the (boolean) result of that comparison is compared to a[3] .

I am sure there are some legitimate use cases, but they are rare at best.

The correct way to express what you're trying to express is

if (a[0] < a[1] && a[1] < a[2] && a[2] < a[3]) {

Answer 2

It's not undefined behavior, it just doesn't do what you think it does. It's equivalent to

(((a[0] < a[1]) < a[2]) < a[3])

1 for true, 0 for false. So if a[0] is less than a[1], then it's comparing a[2] to 1, otherwise, to zero. And so on.

Answer 3

It's not undefined behavior , it's unexpected-to-you behavior .

In C and C++ you cannot do math-like comparisons like that. You can compare two things at a time. So, if you want a < b < c < d , you must write:

if (a < b && b < c && c < d)
    ...

Answer 4

Since you brought C++ into the mix, you could very simply do

if ( std::is_sorted(a, a+4) )
    puts("OK!");

Answer 5

It is defined behaviour, but doesn't do what you expect, so don't do that.

Your expression becomes:

((a[0] < a[1]) < a[2]) < a[3]

where each x < y is turned into either 1 or 0 - which means that it's ALWAYS true if a[3] is greater than 1, and possibly true if a[3] is greater than 0.

Answer 6

It isn't undefined behavior, but it doesn't do what he thinks it does. < is left-to-right associative, so first a[0] < a[1] is evaluted, resulting in true . Then true < a[2] (equivalent to 1 < a[2] ) is evaluated, resulting in true . And so on. You only got the right result by coincidence. For example, "2 1 0 1" would also yield true. Use either

a[0] < a[1] && a[1] < a[2] && a[2] < a[3]

or more generally, if your array is long:

for(i=1,ok=1;i<n&&ok;i++)ok &= a[i-1] < a[i];

Answer 7

The order of evaluation of the < operator, when appearing like this, is well-defined. The code you have written is equivalent to:

bool x;

x = a[0] < a[1];

if(x == true)
{
  x = true < a[2]; 
}
else
{
  x = false < a[2]; 
}

if(x == true)
{
  x = true < a[3]; 
}
else
{
  x = false < a[3]; 
}

if(x == true)
{
  printf("OK!\n");
}
else
{
  printf("I've told ya\n");
}

true always evaluates to the integer 1, false to 0. As you can tell, this code makes no sense at all.