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Comparing unsigned int and int without using translation operator

Everyone I looked around has a thread about my question, but I couldn't find.

unsigned int x = 5; 
int y = -3;
if(y<x)
   func1();
else
   func2();

func2

... But I want to func1

call.

I know that when comparing these values, I have to use the translation operator. But it is not allowed to use a translation operator or change the type of a variable.

How can I solve this problem?

+3


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4 answers


Check first if the y

value is negative, then knowing that, you know that x

there will always be more as it is not specified.

If y

not negative, compare directly with x

. I don't think this will cause a problem as there is no negative sign.



See example below:

if(y<0)
{
    //x>y
    func1();
}
else if (y<x)
{
    //lets say y=3, and x=5
    func1();
}
else
{
    func2();
}
+3


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You can write the condition in the if statement like this



if( y < 0 || y<x)
   func1();
else
   func2();
+2


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Use broader math for comparison. Listing is not used, the type of variables is not changed.
The optimized compiler will not do the multiplication, only the whole expansion.
You can use * 1LL

or instead+ 0LL 

int main(void) {
  long long ll = 1;
  unsigned int x = 5;
  int y = -3;
  // if (y < x)
  if (ll * y < ll * x)
    puts("func1();");
  else
    puts("func2();");
  return 0;
}

Output: func1();

long long

usually wider int

: See How to define integer types that are twice as wide as `int` and` unsigned`?

0


source


Try adding y

( ~y

), it becomes 2

unsigned int x = 5; 
int y = -3;
if(~y<x)
   func1();
else
   func2();
-1


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