Unsigned and Signed int and printf

Question

I understand that I am assigning signed int a value that is larger than what it can handle. Also I should be using %d for signed and %u for unsigned. Similarly I should not be assigning -ve value to unsigned. But if I make such assignments and use printf as below, I get the results show below.

My understanding is that in each case, the number of converted to its two's compliment binary representation which is same for -1 or 4294967295. That is why %u for signed prints 4294967295 by ignoring -ve leftmost bit. When used %d for signed int, it uses left most bit as -ve flag and prints -1. Similarly %u for unsigned prints unsigned value but %d causes it to treat the number as signed and thus prints -1. Is that correct?

signed int si = 4294967295;
unsigned int  ui = 4294967295;

printf("si = u=%u d=%d\n", si, si);
printf("ui = u=%u d=%d\n", ui, ui);

Output:

si = u=4294967295 d=-1
ui = u=4294967295 d=-1

signed int si = -1;
unsigned int  ui = -1;

printf("si = u=%u d=%d\n", si, si);
printf("ui = u=%u d=%d\n", ui, ui);

Output:

si = u=4294967295 d=-1
ui = u=4294967295 d=-1

Note as my answer to this question explains -1 converted to unsigned will always be the max unsigned value for that type ... so assigning -1 to an unsigned value is always well defined behavior. — Shafik Yaghmour
But just re-interpreting a negative integral value as an unsigned integral value is not well-defined. Just as re-interpreting a too-big unsigned integral value as a signed integral value is not. — Deduplicator
You might find "std::numeric_limits" useful. Changing from uin32_t to uint64_t is much more readable. std::numeric_limits<int32_t>::max(), becomes std::numeric_limits<int64_t>::max() etc. — 2785528

Clifford Clifford · Accepted Answer · 2014-11-30T11:08:26

That is why %u for signed prints 4294967295 by ignoring -ve leftmost bit. When used %d for signed int, it uses left most bit as -ve flag and prints -1.

In the case of unsigned, the "leftmost" or most significant bit is not ignored, and is not negative; rather it has a place value of 2³¹.

In the negative case, the sign bit is not a flag; instead it is a bit with a place value of -2³¹.

In both cases the value of the integer is equal to the sum of the place values of all the binary digits (bits) set to 1.

The encoding of signed values in this way is known as two's complement. It is not the only possible encoding; what you described is known as sign and magnitude for example, and one's complement is another possibility. However, these alternative encodings are seldom encountered in practice, not least because two's complement is how arithmetic works on modern hardware in all but perhaps the most arcane architectures.

Unsigned and Signed int and printf

2 Answers