CODEWITHSUNDEEP
ccastingtype-conversionlanguage-lawyertype-safety
William Breathitt Gray
William Breathitt Gray

Reputation: 11986

Regarding type safety when storing an unsigned char value in char variable

I have a char array holding several characters. I want to compare one of these characters with an unsigned char variable. For example:

char myarr = { 20, 14, 5, 6, 42 };
const unsigned char foobar = 133;

myarr[2] = foobar;

if(myarr[2] == foobar){
    printf("You win a shmoo!\n");
}

Is this comparison type safe?

I know from the C99 standard that char, signed char, and unsigned char are three different types (section 6.2.5 paragraph 14).

In section 6.2.5 paragraph 15:

The implementation shall define char to have the same range, representation, and behavior as either signed char or unsigned char.

In section 6.3.1.3 paragraph 3:

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.

I'm afraid that if char is defined as a signed char, then myarr[2] = foobar could result in an implementation-defined value that will not be converted correctly back to the original unsigned char value; for example, an implementation may always result in the value 42 regardless of the unsigned value involved.

Also what is an implementation-defined signal; does this mean an implementation could simply end the program in this case?

In section 6.3.1.1 paragraph 1:

-- The rank of long long int shall be greater than the rank of long int, which shall be greater than the rank of int, which shall be greater than the rank of short int, which shall be greater than the rank of signed char.

-- The rank of any unsigned integer type shall equal the rank of the corresponding signed integer type, if any.

In section 6.2.5 paragraph 8:

For any two integer types with the same signedness and different integer conversion rank (see 6.3.1.1), the range of values of the type with smaller integer conversion rank is a subrange of the values of the other type.

In section 6.3.1 paragraph 2:

If an int can represent all values of the original type, the value is converted to an int; otherwise, it is converted to an unsigned int.

In section 6.3.1.8 paragraph 1:

Otherwise, both operands are converted to the unsigned integer type corresponding to the type of the operand with signed integer type.

The range of char is guaranteed to be the same range as that of signed char or unsigned char, which are both subranges of int and unsigned int respectively as a result of their smaller integer conversion rank.

Since, integer promotions rules dictate that char, signed char, and unsigned char be promoted to at least int before being evaluated, does this mean that char could maintain its "signedness" throughout the comparision?

For example:

signed char foo = -1;
unsigned char bar = 255;

if(foo == bar){
    printf("same\n");
}

UPDATE:

In section J.3.5 paragraph 1 regarding which cases result in implementation-defined values and behavior:

-- The result of, or the signal raised by, converting an integer to a signed integer type when the value cannot be represented in an object of that type (6.3.1.3).

For example, could the following code result in implementation-defined behavior since char could be defined as a signed integer type:

char blah = (char)255;

Upvotes: 3

Views: 1899

Answers (4)

William Breathitt Gray
William Breathitt Gray

Reputation: 11986

My original post is rather broad and consists of many specific questions of which I should have given each its own page. However, I address and answer each question here so future visitors can grok the answers more easily.

Answer 1

Question:

  • Is this comparison type safe?

The comparison between myarr[2] and foobar in this particular case is safe since both variables hold unsigned values. In general, however, this is not true.

For example, suppose an implementation defines char to have the same behavior as signed char, and int is able to represent all values representable by unsigned char and signed char.

char foo = -25;
unsigned char bar = foo;

if(foo == bar){
    printf("This line of text will not be printed.\n");
}

Although bar is set equal to foo, and the C99 standard guarantees that there is no loss of precision when converting from signed char to unsigned char (see Answer 2), the foo == bar conditional expression will evaluate false.

This is due to the nature of integer promotion as required by section 6.3.1 paragraph 2 of the C99 standard:

If an int can represent all values of the original type, the value is converted to an int; otherwise, it is converted to an unsigned int.

Since in this implementation int can represent all values of both signed char and unsigned char, the values of both foo and bar are converted to type int before being evaluated. Thus the resulting conditional expression is -25 == 231 which evaluates to false.

Answer 2

Question:

  • Nevertheless, can I safely convert between unsigned char and char, and back, without losing precision and without risking undefined (or implementation-defined) behavior?

You can safely convert from char to unsigned char without losing precision (nor width nor information), but converting in the other direction -- unsigned char to char -- can lead to implementation-defined behavior.

The C99 standard makes certain guarantees which enable us to convert safely from char to unsigned char.

In section 6.2.5 paragraph 15:

The implementation shall define char to have the same range, representation, and behavior as either signed char or unsigned char.

Here, we are guaranteed that char will have the same range, representation, and behavior as signed char or unsigned char. If the implementation chooses the unsigned char option, then the conversion from char to unsigned char is essentially that of unsigned char to unsigned char -- thus no width nor information is lost and there are no issues.

The conversion for the signed char option is not as intuitive, but is implicitly guaranteed to preserve precision.

In section 6.2.5 paragraph 6:

For each of the signed integer types, there is a corresponding (but different) unsigned integer type (designated with the keyword unsigned) that uses the same amount of storage (including sign information) and has the same alignment requirements.

In 6.2.6.1 paragraph 3:

Values stored in unsigned bit-fields and objects of type unsigned char shall be represented using a pure binary notation.

In section 6.2.6.2 paragraph 2:

For signed integer types, the bits of the object representation shall be divided into three groups: value bits, padding bits, and the sign bit. There need not be any padding bits; there shall be exactly one sign bit. Each bit that is a value bit shall have the same value as the same bit in the object representation of the corresponding unsigned type (if there are M value bits in the signed type and N in the unsigned type, then M <= N).

  1. First, signed char is guaranteed to occupy the same amount of storage as an unsigned char, as are all signed integers in respect to their unsigned counterparts.
  2. Second, unsigned char is guaranteed to have a pure binary representation (i.e. no padding bits and no sign bit).
  3. signed char is required to have exactly one sign bit, and no more than the same number of value bits as unsigned char.

Given these three facts, we can prove via pigeonhole principle that the signed char type has at most one less than the number of value bits as the unsigned char type. Similarly, signed char can safely be converted to unsigned char with not only no loss of precision, but no loss of width or information as well:

  • unsigned char has storage size of N bits.
    • signed char must have the same storage size of N bits.
  • unsigned char has no padding or sign bits and therefore has N value bits
  • signed char can have at most N non-padding bits, and must allocate exactly one bit as the sign bit.
    • signed char can have at most N-1 value bits and exactly one sign bit

All signed char bits therefore match up one-to-one to the respective unsigned char value bits; in other words, for any given signed char value, there is a unique unsigned char representation.

/* binary representation prefix: 0b */
(signed char)(-25)   = 0b11100111
(unsigned char)(231) = 0b11100111

Unfortunately, converting from unsigned char to char can lead to implementation-defined behavior. For example, if char is defined by the implementation to behave as signed char, then an unsigned char variable may hold a value that is outside the range of values representable by a signed char. In such cases, either the result is implementation-defined or an implementation-defined signal is raised.

In section 6.3.1.3 paragraph 3:

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.

Answer 3

Question:

  • Does this mean that it is not safe to store an unsigned value in a signed variable of the same type?

Trying to convert an unsigned type value to a signed type value can result in implementation-defined behavior if the unsigned type value cannot be represented in the new signed type.

unsigned foo = UINT_MAX;
signed bar = foo;    /* possible implementation-defined behavior */

In section 6.3.1.3 paragraph 3:

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.

An implementation-defined result would be any value returned within the range of values representable by the new signed type. An implementation could theoretically return the same value consistently (e.g. 42) for these cases and thus loss information occurs -- i.e. there is no guarantee that converting from unsigned to signed to back to unsigned will result in the same original unsigned value.

An implementation-defined signal is that which conforms to the rules laid out in section 7.14 of the C99 standard; an implementation is permitted to define additional conforming signals which are not explicitly enumerated by the C99 standard.

In this particular case, an implementation could theoretically raise the SIGTERM signal which requests the termination of the program. Thus, attempting to convert an unsigned type value to signed type could result in a program termination.

Answer 4

Question:

  • Does foo == bar evaluate to a false value, even if -1 is equivalent to 255 when an explicit (unsigned char) cast is used?

Consider the following code:

signed char foo = -1;
unsigned char bar = 255;

if((unsigned char)foo == bar){
    printf("same\n");
}

Although signed char and unsigned char values are promoted to at least int before the evaluation of a conditional expression, the explicit unsigned char cast will convert the signed char value to unsigned char before the integer promotions occur. Furthermore, converting to an unsigned value is well-defined in the C99 standard and does not lead to implementation-defined behavior.

In section 6.3.1.3 paragraph 2:

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

This the conditional expression essentially becomes 255 = 255 which evaluates to true. until the value is in the range of the new type.

Answer 5

Questions:

  • Does this mean that not even an explicit conversion is safe?

In general, an explicit cast to char for a value outside the range of values representable by signed char can lead to implementation-defined behavior (see Answer 3). A conversion need not be implicit for section 6.3.1.3 paragraph 3 of the C99 standard to apply.

Upvotes: 2

nio
nio

Reputation: 5289

I've tested your code and it doesn't compare (signed char)-1 and (unsigned char)255 the same. You should convert signed char into unsigned char first, because it doesn't use the MSB sign bit in operations.

I have bad experience with using signed char type for buffer operations. Things like your problem then happen. Then be sure you have turned on all warnings during compilation and try to fix them.

Upvotes: 0

Lochemage
Lochemage

Reputation: 3974

It has to do with how the memory for the char's are stored, in an unsigned char, all 8 bits are used to represent the value of the char while a signed char uses only 7 bits for the number and the 8'th bit to represent the sign.

For an example, lets take a simpler 3 bit value (I will call this new value type tinychar):

bits    unsigned  signed
000     0         0
001     1         1
010     2         2
011     3         3
100     4         -4
101     5         -3
110     6         -2
111     7         -1

By looking at this chart, you can see the difference in value between a signed and an unsigned tinychar based on how the bits are arranged. Up until you start getting into the negative range, the values are identical for both types. However, once you reach the point where the left-most bit changes to 1, the value suddenly becomes a negative for the signed. The way this works is if you reach the maximum positive value (3) and then add one more you end up with the maximum negative value (-4) and if you subtract one from 0 you will underflow and cause the signed tinychar to become -1 while an unsigned tinychar would become 7. You can also see the equivalence (==) between an unsigned 7 and the signed -1 tinychar because the bits are the same (111) for both.

Now if you expand this to have a total of 8 bits, you should see similar results.

Upvotes: 1

Drew McGowen
Drew McGowen

Reputation: 11706

"does this mean that char could maintain its 'signedness' throughout the comparison?" yes; -1 as a signed char will be promoted to a signed int, which will retain its -1 value. As for the unsigned char, it will also keep its 255 value when being promoted, so yes, the comparison will be false. If you want it to evaluate to true, you will need an explicit cast.

Upvotes: 1

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