Compare signed to unsigned

If you have this code:

main(){
    unsigned int x = 5;
    int y = -3;
    if (y>x)
        printf("y is bigger\n\r");
}

What is the output of this?

• int = signed int
• when comparing unsigned to signed, the signed is casted into unsigned
• in our case
• x = 00000005
• y = FFFFFFFD   (MSB = sign bit = 1, the other bits shall be the 2's complement of 3)
• casting u into unsigned, then comparing, y > x --> true
• the output: "y is bigger"

What about normal operation?

main(){

int x;

int y = -10;

unsigned int z = 4;

x = y + z;

printf("%d\n\r",x);

}

// -6?

// Also y will be casted to unsigned, then added to z,  the result will be correct, thanks to the 2's complement representation

// y + z = FFFFFFF6 + 00000004 = FFFFFFFA = - 6


Ex:

main(){

short x;
 long x2;

short y = -10;

unsigned short z = 4;

x = y + z;
 x2 = y + z;
 x3 = y + (short)z;

printf("%d, %d, %d\n\r",x, x2, x3);

}

in 16 bit machine:

• y + z = 0xFFF6 + 0x0004 = 0xFFFA
• x = 0xFFFA
• Since x is signed, then x = - 6
• x2 = 0x0000FFFA
• Since x2 is signed, then x = + 65530
• No sign extension happened. because (y + z) is unsigned
• x3 = 0xFFFFFFFA
• Since x3 is signed, then x = - 6
• sign extension happened because (y + z) is signed

in 32 bit machine:

• y + z = 0xFFFFFFF6 + 0x00000004 = 0xFFFFFFFA
• x = x2 = x3 = - 6