Type conversion.

[arunkumar_gona]

--------------------------------------------------------------------------
unsigned int x = 100;
unsigned int y = 200;
float f;

f = (float)x-(float)y; // ok result -100.000

f = (float)(x-y); // Not ok , result 4.29497e+009
--------------------------------------------------------------------------
can anybody tell me why this behaviour by compiler

[Philip Nicoletti]

The first method converts x to float , converts y to float,
and does the subtraction using float arithmetic.

The second does the subtraction using "unsigned int" arithmetic,
and converts the result to float.

You are getting a "negative" number for the unsigned int subtraction.

[jwbarton]

You are getting an underflow when you subtract 200 from 100 as an unsigned int. Assuming a 32 bit integer, this results in a very large positive number (4294967196). You then cast this to a float, and get 4.29497e+009.

This is the same as doing:
unsigned int x = 100;
unsigned int y= 200;

unsigned int z = x - y; // this underflows
float f = (float) z;


When you put the float cast before each number, you are subtracting 2 floats. Since floats are a signed data type, they can handle going negative. This gives you the result that you are expecting.

Best regards,
John

[arunkumar_gona]

Thanks guys

[Chambers]

While the answers above are absolutely correct, just for those who still don't understand, using the type:

int x = 100;
int y = 200;

instead of:

unsigned int x = 100;
unsigned int y = 200;

and then casting a conversion to float:

float z = (float) (x-y);

should produce the desired result (-100.0) since x and y can now handle the negative SIGNature.

Alan.