CODEWITHSUNDEEP
clanguage-lawyer
Johan Bezem
Johan Bezem

Reputation: 2595

Can the type difference between constants 32768 and 0x8000 make a difference?

The Standard specifies that hexadecimal constants like 0x8000 (larger than fits in a signed integer) are unsigned (just like octal constants), whereas decimal constants like 32768 are signed long. (The exact types assume a 16-bit integer and a 32-bit long.) However, in regular C environments both will have the same representation, in binary 1000 0000 0000 0000. Is a situation possible where this difference really produces a different outcome? In other words, is a situation possible where this difference matters at all?

Upvotes: 15

Views: 9340

Answers (5)

Johan Bezem
Johan Bezem

Reputation: 2595

Yes, it can matter. If your processor has a 16-bit int and a 32-bit long type, 32768 has the type long (since 32767 is the largest positive value fitting in a signed 16-bit int), whereas 0x8000 (since it is also considered for unsigned int) still fits in a 16-bit unsigned int.

Now consider the following program:

int main(int argc, char *argv[])
{
  volatile long long_dec = ((long)~32768);
  volatile long long_hex = ((long)~0x8000);

  return 0;
}

When 32768 is considered long, the negation will invert 32 bits, resulting in a representation 0xFFFF7FFF with type long; the cast is superfluous. When 0x8000 is considered unsigned int, the negation will invert 16 bits, resulting in a representation 0x7FFF with type unsigned int; the cast will then zero-extend to a long value of 0x00007FFF. Look at H&S5, section 2.7.1 page 24ff.

It is best to augment the constants with U, UL or L as appropriate.

Upvotes: 8

Keith Thompson
Keith Thompson

Reputation: 263517

Assuming int is 16 bits and long is 32 bits (which is actually fairly unusual these days; int is more commonly 32 bits):

printf("%ld\n", 32768);  // prints "32768"
printf("%ld\n", 0x8000); // has undefined behavior

In most contexts, a numeric expression will be implicitly converted to an appropriate type determined by the context. (That's not always the type you want, though.) This doesn't apply to non-fixed arguments to variadic functions, such as any argument to one of the *printf() functions following the format string.

Upvotes: 1

supercat
supercat

Reputation: 81247

Another examine not yet given: compare (with greater-than or less-than operators) -1 to both 32768 and to 0x8000. Or, for that matter, try comparing each of them for equality with an 'int' variable equal to -32768.

Upvotes: 1

undur_gongor
undur_gongor

Reputation: 15954

On a 32 bit platform with 64 bit long, a and b in the following code will have different values:

int x = 2;
long a = x * 0x80000000; /* multiplication done in unsigned -> 0           */
long b = x * 2147483648; /* multiplication done in long     -> 0x100000000 */

Upvotes: 1

yoyomommy
yoyomommy

Reputation: 127

The difference would be if you were to try and add a value to the 16 bit int it would not be able to do so because it would exceed the bounds of the variable whereas if you were using a 32bit long you could add any number that is less than 2^16 to it.

Upvotes: 0

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