CODEWITHSUNDEEP
cprintfoutputunsignedsigned
Rocko
Rocko

Reputation: 166

unexpected behavior of unsigned variables when assigned negative values

Here is a code snippet:

unsigned short a=-1;
unsigned char b=-1;
char c=-1;
unsigned int x=-1;
printf("%d %d %d %d",a,b,c,x);

Hhy the output is this:

65535 255 -1 -1

?

Can anybody please analyze this ?

Upvotes: 1

Views: 338

Answers (2)

user2217371
user2217371

Reputation: 11

Basically, you should not assign negative values to "unsigned" variables. You are trying to play tricks on the compiler, and who knows what it will do. Now, "char n = -1;" is OK because "n" can legitimently take on negative values and the compiler knows how to treat it.

Upvotes: 0

Dariusz
Dariusz

Reputation: 22311

You are printing the values using %d which is for signed numbers. The value is "converted" to a signed number (it actually stays the same bitwise, but the first bit is interpreted differently).

As for unsigned char and short - they are also converted to 32 bit int, so the values fit in it.

Had you used %lld (and cast the value as long long, otherwise it could be unspecified behavior) even the last two numbers may get printed as unsigned.

Anyway, use %u for unsigned numbers.

How does it work?

Bit value of 255 is 11111111. If treated like an unsigned number, it will be 255. If treated as a signed number - it'll be -1 (the first bit usually determines sign). When you pass the value to %d in printf, the value is converted to a 32 bit integer, which looks like this: 00000000000000000000000011111111. Since the first bit is 0, the value is printed simply as 255. It works similarily for your short.

The situation is different for 32 bit integer. It is immediately assigned a 11111111111111111111111111111111 value, which stands for -1 in singed notation. And since you have used %d in your printf, it is interpreted as -1.

Upvotes: 6

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