What happens if I assign a negative value to an unsigned variable?

I was curious to know what would happen if I assign a negative value to an unsigned variable.

The code will look somewhat like this.

unsigned int nVal = 0;
nVal = -5;

It d开发者_如何学编程idn't give me any compiler error. When I ran the program the nVal was assigned a strange value! Could it be that some 2's complement value gets assigned to nVal?

For the official answer - Section 4.7 conv.integral

"If the destination type is unsigned, the resulting value is the least unsigned integer congruent to the source integer (modulo 2ⁿ where n is the number of bits used to represent the unsigned type). [ Note: In a two’s complement representation, this conversion is conceptual and there is no change in the bit pattern (if there is no truncation). —end note ]

This essentially means that if the underlying architecture stores in a method that is not Two's Complement (like Signed Magnitude, or One's Complement), that the conversion to unsigned must behave as if it was Two's Complement.

It will assign the bit pattern representing -5 (in 2's complement) to the unsigned int. Which will be a large unsigned value. For 32 bit ints this will be 2^32 - 5 or 4294967291

You're right, the signed integer is stored in 2's complement form, and the unsigned integer is stored in the unsigned binary representation. C (and C++) doesn't distinguish between the two, so the value you end up with is simply the unsigned binary value of the 2's complement binary representation.

It will show as a positive integer of value of max unsigned integer - 4 (value depends on computer architecture and compiler).

BTW
You can check this by writing a simple C++ "hello world" type program and see for yourself

Yes, you're correct. The actual value assigned is something like all bits set except the third. -1 is all bits set (hex: 0xFFFFFFFF), -2 is all bits except the first and so on. What you would see is probably the hex value 0xFFFFFFFB which in decimal corresponds to 4294967291.

When you assign a negative value to an unsigned variable then it uses the 2's complement method to process it and in this method it flips all 0s to 1s and all 1s to 0s and then adds 1 to it. In your case, you are dealing with int which is of 4 byte(32 bits) so it tries to use 2's complement method on 32 bit number which causes the higher bit to flip. For example:

┌─[student@pc]─[~]
└──╼ $pcalc 0y00000000000000000000000000000101      # 5 in binary
        5                       0x5                     0y101
┌─[student@pc]─[~]
└──╼ $pcalc 0y11111111111111111111111111111010      # flip all bits  
      4294967290      0xfffffffa      0y11111111111111111111111111111010
┌─[student@pc]─[~]
└──╼ $pcalc 0y11111111111111111111111111111010 + 1  # add 1 to that flipped binarry
      4294967291      0xfffffffb      0y11111111111111111111111111111011

In Windows and Ubuntu Linux that I have checked assigning any negative number (not just -1) to an unsigned integer in C and C++ results in the assignment of the value UINT_MAX to that unsigned integer.

Compiled example link.