具有相同逻辑的两段代码。 一个提供预期的输出，而另一个则没有

Question

char xtime(char m)
{
    //calculates the m value by checking m,if m is less than 0x80 hexadecimal
    // then it is left shifted else it is left shifted and xor'ed with 0x1b.
    if(m<0x80)
    {
        m<<=1;
    } else {
        m=(((m)<<1)^0x1b);
    }

    printf("%#01x ",m&0xff);
    return m;
}

This code doesnt show the expected output if m=0x80(which is 0x1b),it gives the output as 0 in hexadecimal.

#define xtime(a) (((a)<0x80)?(a)<<1:(((a)<<1)^0x1b) )

This code works and gives the expected result .

Could you please help as to what is wrong with the functional code and how is it solved in the second code.

Answer 1

Assuming that in your environment

• char is signed
• char is 8-bit long
• two's compliment is used to express negative integers
• if a signed integer to convert to cannot store original value, upper bits are simply thrown away

0x80 is too big to store into char variable, and it will be interpreted as -128 . -128 is smaller than 0x80 , so m <<= 1; is executed. The result of this shift is -256 , and its binary reprensentation is 0xffffff00 , and m will get the last 8 bits, which is 0 . That is what you get.

The macro will work if 0x80 is passed to a , because the calculation will be done using int , and int can hold integers upto at least 32767 .