The result of abs(-2147483648) is -2147483648, isn't it? it seems unacceptable.
printf("abs(-2147483648): %d\n", abs(-2147483648));
output:
abs(-2147483648): -2147483648
Asked
Active
Viewed 2.4k times
27
Victor S
- 4,021
- 15
- 43
- 54
-
1I believe this is undefined behavior. I don't have the C standard handy, so I cannot back it up. – Alexandre C. Jun 28 '12 at 10:53
-
5What do you expect it to be, given that `abs(int)` returns an `int`? – Philip Kendall Jun 28 '12 at 10:55
-
9Latest draft of C11 says (7.21.6.1, about abs and friends) "If the result cannot be represented, the behavior is undefined" – Alexandre C. Jun 28 '12 at 10:58
-
@PhilipKendall The absolute value can be a negative value? – Victor S Jun 28 '12 at 10:59
-
4linux man page says (man 3 abs): Trying to take the absolute value of the most negative integer is not defined. – Pat Jun 28 '12 at 11:07
-
You get the right value if int has more than 32 bits. But for 32 bit ints, what would you have it be, considering that 2147483648 cannot be represented in 32 bits? – Jim Balter Jun 28 '12 at 11:14
-
@JimBalter technically it can not be represented in 31 bits. The sign bit is what's causing this behaviour. – rubenvb Jun 28 '12 at 13:50
5 Answers
33
The standard says about abs():
The
abs,labs, andllabsfunctions compute the absolute value of an integerj. If the result cannot be represented, the behavior is undefined.
And the result indeed cannot be represented because the 2's complement representation of signed integers isn't symmetric. Think about it... If you have 32 bits in an int, that gives you 232 distinct values from INT_MIN to INT_MAX. That's an even number of values. So, if there's only one 0, the number of values greater than 0 cannot be the same as the number of values less than 0. And so there's no positive counterpart to INT_MIN with a value of -INT_MIN.
So, what's unacceptable is calling abs(INT_MIN) on your platform.
Spikatrix
- 20,225
- 7
- 37
- 83
Alexey Frunze
- 61,140
- 12
- 83
- 180
18
Negative numbers are usually represented whit binary complement.
To convert positive to negative it is used logic
x -> not(x)+1
For 8 bits arithmetic
01111111b is 127 and -127 becomes
10000000b + 1 = 10000001b
and to opposite direction -127
10000001b becomes
01111110b + 1 = 01111111b
What about -128?
-128 is 10000000b and there is no positive counterpart of it, because there is no 128 in 8 bits signed arithmetic.
10000000 -> 01111111 + 1 = 10000000 and -128 again
Same applies to original question
Luka Rahne
- 10,336
- 3
- 34
- 56
-
that's why 0 and the minimum value are always the same after negating in two's complement – phuclv Jan 13 '14 at 02:25
10
Since 2147483648 is greater than INT_MAX on your implementation, then abs(-2147483648) is undefined.
Alexandre C.
- 55,948
- 11
- 128
- 197
5
This is code in abs.c in GNU glibc source code.
/* Return the absolute value of I. */
int
DEFUN(abs, (i), int i)
{
return(i < 0 ? -i : i);
}
So,abs(-2147483648) return -(-2147483648) . In x86,it is implement by this two instruction
movl $-2147483648, %eax
negl %eax
negl instruction is implemented by this way: num=0-num; sbb is implemented by this way: Subtracts the source from the destination, and subtracts 1 extra if the Carry Flag is set. So abs(-2147483648) (hex is 0x80000000 ) --> -(-2147483648) --> 0-(-2147483648) becomes (0x80000000) finally.
details of negl instruction,please visit http://zsmith.co/intel_n.html#neg
details of sbb instruction ,please visit http://web.itu.edu.tr/kesgin/mul06/intel/instr/sbb.html
Mazheng
- 166
- 4
-5
Try this
printf("abs(-2147483648): %u\n", abs(-2147483648));
-
6This, my friend, has undefined behavior. You're printing a signed integer with the unsigned formatter. I -1 because it moreover does not answer the question. – Alexandre C. Jun 28 '12 at 11:16