Brad Bourn wrote:

I have a function say like this

func(unsigned int foo) { unsigned int i, retval = 0; for (i = 0; i < foo; i++) { retval++; }

return retval; }

and I call it with

val = func(4 - 3);

Seems like gcc WILL pass the -1 to func, and func just get stuck in a loop becase i starts at 0 (which is alread greater than a negative number)

This not what I would have expected.

Is this by design?

When you pass the -1 to the function, it will be implicit cast to an unsigned int (2^32 in this case). This is a consequence of the design of your function. When you say func(unsinged int foo) you are saying that this function only receives unsigned integers. When you pass a signed integer to it, it will break, since you are violating the signature. You must assure that foo is unsigned or change your function: func (int foo) { int i; unsigned int retval = 0; for (i = 0; i < foo; ++i) { retval++; } return retval; } Notice that you also must say that the function returns an unsigned int, or it will be implicit returning an int and similar errors might occur. []s Davi de Castro Reis

On Friday 11 February 2005 10:54 am, Davi de Castro Reis wrote:

Notice that you also must say that the function returns an unsigned int, or it will be implicit returning an int and similar errors might occur.

Is there a way to ASSERT that the value is not negative? B-)

I have a function say like this

func(unsigned int foo) { unsigned int i, retval = 0; for (i = 0; i < foo; i++) { retval++; }

return retval; }

and I call it with

val = func(4 - 3);

Seems like gcc WILL pass the -1 to func, and func just get stuck in a loop becase i starts at 0 (which is alread greater than a negative number)

This not what I would have expected.

Is this by design? I was thinking unsigned long in my previous response. :-) When you pass a -1 into this function, as a result of the conversion rules defined by the standard, it will pass a 0xFFFFFFFF the decimal value is 4294967295. The argument to a function may be an expression. In your case, this expression will evaluate to an unsigned in. if you pass, (4 - 3), you are

On Friday 11 February 2005 12:44 pm, Brad Bourn wrote: passing 1. You've got to be very careful with expressions when you are mixing signed and unsigned, especially in a 64 bit system where your int is smaller than the long. -- Jerry Feldman Boston Linux and Unix user group http://www.blu.org PGP key id:C5061EA9 PGP Key fingerprint:053C 73EC 3AC1 5C44 3E14 9245 FB00 3ED5 C506 1EA9

Thanks for the info usefull stuff! B-) On Friday 11 February 2005 11:40 am, Jerry Feldman wrote:

On Friday 11 February 2005 1:06 pm, Brad Bourn wrote:

...

On Friday 11 February 2005 10:54 am, Davi de Castro Reis wrote:

...

Notice that you also must say that the function returns an unsigned int, or it will be implicit returning an int and similar errors might occur.

Is there a way to ASSERT that the value is not negative?

Yes and no. In your function, you are using all unsigned ints, which include the range from 0 to 2^32. In your code you can test for it: assert(i <= INT_MAX); This will assert for the range of values > INT_MAX, or 2147483648 -> 44294967295

-- Jerry Feldman Boston Linux and Unix user group http://www.blu.org PGP key id:C5061EA9 PGP Key fingerprint:053C 73EC 3AC1 5C44 3E14 9245 FB00 3ED5 C506 1EA9

Brad, On Friday 11 February 2005 09:44, Brad Bourn wrote:

...

val = func(4 - 3);

Seems like gcc WILL pass the -1 to func, and func just get stuck in a loop becase i starts at 0 (which is alread greater than a negative number)

The last time I checked, 4 - 3 is 1, not -1.

...

RRS