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As we know that the range of char is -128 to 127.

The 2's compliment of -128 and the binary of 128 is same, which is

10000000

So why the range of char is -128 to 127 but not -127 to 128 .
Where as in case of int , 128 and -128 both are different.

zee
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    The range of char is not always -128 to 127. Even if you discount the part of the C standard that says that it has 8 _or more_ bits, it can be signed or unsigned. Perhaps you meant `int8_t`? – Paul Hankin Feb 22 '14 at 08:38

3 Answers3

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In twos-complement notation, whenever the high-order bit is 1, the number is negative. So the biggest positive number is 

01111111 = 127

and the smallest negative number is

10000000 = -128

The same thing happens for int, but its range is much larger. The biggest positive int is

01111111 11111111 11111111 11111111 = 2147483647

and the smallest negative int is

10000000 00000000 00000000 00000000 = -2147483648

As you can see, in both cases, the smallest negative is 1 more than the biggest positive.

You might be wondering, If there are the same number of bits other than the sign bit, why are there more negative numbers than positives? That's because you're not counting 0. It has 0 in the high-order bit, so it's considered positive in this notation. When you include that, there are 128 negative chars, and 128 non-negative chars, and they balance equally.

Barmar
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  • take for example `4294967234`, its binary is `11111111111111111111111111000010`.Its high order bit is `1` , so how it is positive? – zee Feb 22 '14 at 08:47
  • That's an unsigned int, we're talking about signed int. – Barmar Feb 22 '14 at 08:48
  • or it's a `long long`, which is 64 bits. – Barmar Feb 22 '14 at 08:49
  • it is a unsigned it, so is the theory of high order bit is not applicable on unsigned int – zee Feb 22 '14 at 08:50
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    @zee yes, unsigned int as the name implies you can't have negative numbers in this variable. so you have all the 32bits(if its a 32bit variable) to represent a positive number. if you need to represent negative numbers, the computer needs to know if its negative, this is done by setting the sign bit to 1 and using 2's compliment. – tesseract Feb 22 '14 at 08:56
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A character has 8 bits. So the bit pattern for -128 and 128 are the same when you have 8 bits to deal with. (ignoring the sign bit)

Now the int. It has more bits so the MSB is not -128 but a larger -ve number depending on the platform.

It does exhibit the same problem at a "higher" number

Ed Heal
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Look 128 is

10000000

But by default char is signed compiler treats first bit as sign bit and treats the no. as negative and calculates its 2s complement ( because MSB is 1)

2s complement is
 01111111
      +        1
 10000000

So - 128 is stored To store +128 compiler have to consider 9 bits that is to interpret

010000000

Here MSB is 0 for positive no. But size is of 9 bits this shows why char only accomodate upto +127 and if we want to store +128 then -128 will be stored as explained above

OldSchool
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