Difference between signed and unsigned data types?

Question

main()
{
    char i=255;
    printf("\n%x\n",i);
}

output: ffffffff

main()
{
    u_char i=255;
    printf("\n%x\n",i);
}

output: ff

What is happening here? Kindly explain the output to me with some good links. This is a very basic thing I guess and I am getting really confused...

Answer 1

What you are seeing here is caused by two things:

• 255 does not fit in the range of char (note that it is implementation-defined whether char is equivalent to signed char or unsigned char , but evidently on your platform it is signed char ). The resulting behaviour is implementation-defined, but typically it will wrap round and become -1; see two's complement .
• integer promotion , because printf() is a variable-argument function . Integral-type arguments (like char ) are automatically promoted to int .

So printf() sees an int with a value of -1 , and prints its hexadecimal representation accordingly.

For the unsigned case, there is no wrap-around. printf() sees an int with a value of 255 , and prints its hexadecimal representation accordingly (omitting the leading zeros).

Answer 2

The C compiler has to expand the value passed to printf (this is called "promotion"), because printf is a variadic function (it can be called with differing arguments). For values of type char , the promoted value is of type int . Since your compiler's char type seems to be signed, the promoted value is sign extended. In binary:

char i = 255           // or: 11111111 in binary
int promoted_i = -1    // or: 11....11111 (usually, 32 or 64 ones)

In the unsigned case, no sign-extension happens:

char u = 255           // or: 11111111 in binary, same as above but with different interpretation
unsigned int pu = i    // or: 00....0011111111 (8 ones, preceded by the appropriate number of zeroes) 

Answer 3

char i = 255; invokes implementation-defined behavior by converting a value to a signed type into which it does not fit (assuming char is only 8 bits and plain char is signed, both of which are also implementation-specific.

Answer 4

当你将一个8位有符号变量设置为它不能容纳的值255时，在这种情况下它似乎将负（高端）标志设置为1，因此如果它被签名则该值为-1 ，然后它被转换为整数-1，即ffffffff 。