如果我给一个无符号变量赋一个负值会发生什么？

我很想知道如果给无符号变量赋一个负值会发生什么。

代码看起来有点像这样。

unsigned int nVal = 0;
nVal = -5;

它没有给我任何编译错误。当我运行程序时，nVal被赋予了一个奇怪的值！有没有可能某个2的补数值被赋给了 nVal？

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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

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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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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.

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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

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最佳答案

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.

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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

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