Char data type arithmetic expression

Question

int main()

{

        char a = 'P';  

        char b = 0x80;  

        printf("a>b  %s\n",a>b ? "true":"false");  

        return 0;

}

Why does it evaluates to true?

The ASCII value of `P` happens to be (decimal) 80. Did you confuse decimal `80` vs. hexadecimal `0x80`? — Keith Thompson, Sep 06 '13 at 18:17
read [What is -(-128) for signed single byte char in C?](http://stackoverflow.com/questions/17469804/what-is-128-for-signed-single-byte-char-in-c/17469941#17469941) — Grijesh Chauhan, Sep 07 '13 at 07:58

Eric Postpischil · Accepted Answer · 2013-09-06T17:50:01.037

10

On your system, char is signed. It is also eight bits, so 0x80 overflows what a signed 8-bit integer can represent. The resulting value is -128. Since P is some positive value, it is greater than -128.

C permits the char type to be signed or unsigned. This is a special (annoying) property, unlike other integer types such as int. It is often advisable to explicitly declare character types with unsigned char so that the behavior is more determined rather than implementation-dependent.

edited Sep 06 '13 at 17:50

answered Sep 06 '13 at 17:43

Eric Postpischil

195,579
13
168
312

Assuming ASCII, `P` is 0x50. – Dennis Meng Sep 06 '13 at 17:43
@user2712068 Assuming two's complement, it'll be -127. – Dennis Meng Sep 06 '13 at 17:45
@DennisMeng: I have no knowledge of whether the OP’s system is using ASCII, and this assumption is not necessary for the problem. Per C 2011 (N1570) 6.2.5 3, characters of the execution character set are nonnegative. – Eric Postpischil Sep 06 '13 at 17:47

Char data type arithmetic expression

1 Answers1