Unsigned Integer mystery

[patelss23]

Hello all,
how are you?

let me put a code snippet here. I used microsoft vc++ 6.0 on my windows xp machine.

/**********************************

unsigned int x1 = 10;
unsigned int y1 = 20;
 
printf("\n (x1-y1) = %d",(x1-y1));  //prints -10
 
if((x1-y1) < 0)
  printf("\n(x1 - y1) < 0");
else
 printf("\n(x1 - y1) > 0"); //prints (x1 - y1) > 0
 

*************************************/

Can any one please tell me what happens when we write
printf ("\n (x1-y1) = %d",(x1-y1)); ?

Does compiler create any temporary variable and assigns value of (x1-y1) to it?

if yes, this temporary variable would be signed int by default ?
if no, then why does it print (x1 - y1) > 0 ?

Thank you in advance,
Sanket

[Ganon11]

When you use the %d format specifier, you are telling printf to print that value as if it were a signed integer. Later, when your if statement checks x1-y1 (or y1-x1), you aren't telling the program to treat them as anything, so it treats the result as an unsigned integer - which cannot be less than 0.

[patelss23]

When you use the %d format specifier, you are telling printf to print that value as if it were a signed integer. Later, when your if statement checks x1-y1 (or y1-x1), you aren't telling the program to treat them as anything, so it treats the result as an unsigned integer - which cannot be less than 0.

Thank you Ganon11.

I got the point. That means When we compare them or use them in any expression, it will be treated as unsigned. as an example

unsigned int x1;
for(x1 = 10; x1>=0; x1--){
        printf("\nx1 = %d  : ",x1);
    }
 

This for loop will never end as x1 is unsigned and it wont be less than 0.
m I going right way ?

Thank you,
Sanket

[Ganon11]

I'm pretty sure you are correct.

[patelss23]

I'm pretty sure you are correct.

Thank you so much for clearing my concepts.

[weaknessforcats]

Of course s1 can be negative. If it contains 0 and you --x, you have -1.

Unfortunately, the sign bit is part of the data so you now have a really big positive number.

That's why the loop runs forever. It should be for(x1 = 10; x1>0; x1--) to avoid the run-on.