A simple numerical comparison Problem

# Include <stdio. h>

Int main (void)
{

Int I =   0 xaabbccdd ;

Char * P = (Char * ) & I;

If (P [ 0 ] =   0xdd )
Printf ( " Equal " );
Else
Printf ( " Not equal " );
Return   0 ;
}

 
ProgramThe intention is to output equal when running on a little-endian machine (Intel), but output "not equal" and the expected idea after running.

Where is the problem? Type escalation rules in incorrect C.

In the IF (P [0] = 0xdd) Statement, the right operand of the operator "=" is an integer constant, and the default type is int. The left operand is a char type, therefore, the type from Char to int needs to be upgraded. Since the value of P [0] Is 0xdd (the symbol bit is 1), according to the extension rules of the symbol bit when the data width is extended, P [0] is extended to (INT) 0 xffffffdd. It is not equal to (INT) 0xdd, so "not equal" is output ".