Comparison operation on unsigned and signed integers

Binary operations between different integral types are performed within a "common" type defined by so called usual arithmetic conversions (see the language specification, 6.3.1.8). In your case the "common" type is unsigned int. This means that int operand (your b) will get converted to unsigned int before the comparison, as well as for the purpose of performing subtraction.

When -1 is converted to unsigned int the result is the maximal possible unsigned int value (same as UINT_MAX). Needless to say, it is going to be greater than your unsigned 1000 value, meaning that a > b is indeed false and a is indeed small compared to (unsigned) b. The if in your code should resolve to else branch, which is what you observed in your experiment.

The same conversion rules apply to subtraction. Your a-b is really interpreted as a - (unsigned) b and the result has type unsigned int. Such value cannot be printed with %d format specifier, since %d only works with signed values. Your attempt to print it with %d results in undefined behavior, so the value that you see printed (even though it has a logical deterministic explanation in practice) is completely meaningless from the point of view of C language.

Edit: Actually, I could be wrong about the undefined behavior part. According to C language specification, the common part of the range of the corresponding signed and unsigned integer type shall have identical representation (implying, according to the footnote 31, "interchangeability as arguments to functions"). So, the result of a - b expression is unsigned 1001 as described above, and unless I'm missing something, it is legal to print this specific unsigned value with %d specifier, since it falls within the positive range of int. Printing (unsigned) INT_MAX + 1 with %d would be undefined, but 1001u is fine.

On a typical implementation where int is 32-bit, -1 when converted to an unsigned int is 4,294,967,295 which is indeed ≥ 1000.

Even if you treat the subtraction in an unsigned world, 1000 - (4,294,967,295) = -4,294,966,295 = 1,001 which is what you get.

That's why gcc will spit a warning when you compare unsigned with signed. (If you don't see a warning, pass the -Wsign-compare flag.)

 #include<stdio.h>
 int main()
 {
   int a = 1000;
   signed int b = -1, c = -2;
   printf("%d",(unsigned int)b);
   printf("%d\n",(unsigned int)c);
   printf("%d\n",(unsigned int)a);

   if(1000>-1){
      printf("\ntrue");
   }
   else 
     printf("\nfalse");
     return 0;
 }

For this you need to understand the precedence of operators

1. Relational Operators works left to right ... so when it comes

   if(1000>-1)

then first of all it will change -1 to unsigned integer because int is by default treated as unsigned number and it range it greater than the signed number

-1 will change into the unsigned number ,it changes into a very big number

You are doing unsigned comparison, i.e. comparing 1000 to 2^32 - 1.

The output is signed because of %d in printf.

N.B. sometimes the behavior when you mix signed and unsigned operands is compiler-specific. I think it's best to avoid them and do casts when in doubt.

Find a easy way to compare, maybe useful when you can not get rid of unsigned declaration, (for example, [NSArray count]), just force the "unsigned int" to an "int".

Please correct me if I am wrong.

if (((int)a)>b) {
    ....
}

while comparing a>b where a is unsigned int type and b is int type, b is type casted to unsigned int so, signed int value -1 is converted into MAX value of unsigned**(range: 0 to (2^32)-1 )** Thus, a>b i.e., (1000>4294967296) becomes false. Hence else loop printf("a is SMALL! %d\n", a-b); executed.