Question

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 didn'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?

Was it helpful?

Solution

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.

OTHER TIPS

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

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

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.

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.

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