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

后端 未结 5 1897
攒了一身酷
攒了一身酷 2021-01-04 04:30

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

相关标签:
5条回答
  • 2021-01-04 05:02

    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.

    0 讨论(0)
  • 2021-01-04 05:05

    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 */
    
    0 讨论(0)
  • 2021-01-04 05:07

    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.

    0 讨论(0)
  • 2021-01-04 05:10

    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.

    0 讨论(0)
  • 2021-01-04 05:21

    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.

    0 讨论(0)
提交回复
热议问题