overflow

[Christian Christmann]

Hi,

I would like to hear your opinion about the C standards.

When I have a

signed char a = 130;

I explicitly generate an overflow since the maximum value a singned char
can accommodate is "127". Is this behavior of the overflow defined in any
way or do I get an undefined state of "a" due to the C standards?

Thank you
Chris

 

[pete]

Hi,

I would like to hear your opinion about the C standards.

When I have a

signed char a = 130;

I explicitly generate an overflow
since the maximum value a singned char
can accommodate is "127".

That's the minimum maximum value.
SCHAR_MAX is higher than that, if CHAR_BIT is greater than 8.

Is this behavior of the overflow defined in any way

That's not overflow.
That's the assignment of an out of range value.

or do I get an undefined state of "a" due to the C standards?

Overflow must be the result of an arithmetic operation.

This is overflow:

signed char a = SCHAR_MAX;

++a;

Undefined behavior of part of the code, means that
the behavior of the entire program is undefined.

 

[usr.root]

oh,please test a=255;and a=256,you will found that it's not overflow
but show you a wrong glance.

 

[Tim Rentsch]

This is overflow:

signed char a = SCHAR_MAX;

++a;

That's not overflow either, unless SCHAR_MAX == INT_MAX.

 

[Keith Thompson]

I would like to hear your opinion about the C standards.

When I have a

signed char a = 130;

I explicitly generate an overflow since the maximum value a singned char
can accommodate is "127". Is this behavior of the overflow defined in any
way or do I get an undefined state of "a" due to the C standards?

Assuming SCHAR_MAX==127 (which is common but not universal), the
expression 130 is of type int, so it's implicitly converted to type
signed char. To find out what happens, we have to read C99 6.3.1.3,
which covers conversion of signed and unsigned integers:

When a value with integer type is converted to another integer
type other than _Bool, if the value can be represented by the
new type, it is unchanged.

Otherwise, if the new type is unsigned, the value is converted
by repeatedly adding or subtracting one more than the maximum
value that can be represented in the new type until the value is
in the range of the new type.

Otherwise, the new type is signed and the value cannot be
represented in it; either the result is implementation-defined
or an implementation-defined signal is raised.

The third paragraph applies here; either an implementation-defined
value is assigned to a, or an implementation-defined signal is raised.
(The wording about signals is new in C99.)

It's likely that the implementation-defined value will be -126, but
the standard doesn't guarantee that.

The rules are different for conversions and for arithmetic operations.
If an arithmetic operation overflows for a signed type, the behavior
is undefined.

 

[Christian Christmann]

The third paragraph applies here; either an implementation-defined value
is assigned to a, or an implementation-defined signal is raised. (The
wording about signals is new in C99.)

Thank you very much. This was an illuminating answer and help
me a lot.

Regards,
Chris

 

[pete]

That's not overflow either, unless SCHAR_MAX == INT_MAX.

Good point!

This is overflow:

int a = INT_MAX;

++a;

 

[Jack Klein]

oh,please test a=255;and a=256,you will found that it's not overflow
but show you a wrong glance.

Please don't post nonsense in this group, with improper grammar,
punctuation, and capitalization to boot.