bit shift for unsigned int, why negative?

Question

Code:

unsigned int i = 1<<31;
printf("%d\n", i);

Why the out put is -2147483648, a negative value?

Updated question:

#include <stdio.h>

int main(int argc, char * argv[]) {
    int i = 1<<31;
    unsigned int j = 1<<31;
    printf("%u, %u\n", i, j);
    printf("%d, %d\n", i, j);

    return 0;
}

The above print:

2147483648, 2147483648
-2147483648, -2147483648

So, does this means, signed int & unsigned int have the same bit values, the difference is how you treat the 31st bit when convert it to a number value?

Because %d is the format character for "signed int". If you tell the computer it's a signed int and give it an unsigned int that is too big to fit into an int, then it's entirely your fault. — gnasher729, Dec 23 '14 at 08:28
@gnasher729 Or it's the language's fault for not understanding its own types and letting you print a type as if it were another type without any complaints. — endolith, Apr 05 '15 at 15:36
@endolith `printf()` is part of the standard library, not of the language itself. — 12431234123412341234123, Sep 14 '20 at 17:40
`1<<31` is a `int` and UB when `INT_MAX<1UL<<31`, which is the case when `int is 32 bit or shorter. Better use `1UL<<31` which uses `long` which is at least 32 bit and `unsigned`. — 12431234123412341234123, Sep 14 '20 at 17:42

thepace · Accepted Answer · 2014-12-23T09:25:04.630

5

%d prints the int version of the unsigned int i. Try %u for unsigned int.

printf("%u\n", i);

int main(){
    printf("%d, %u",-1,-1);
    return 0;
}

Output: -1, 4294967295

i.e The way a signed integer is stored and how it gets converted to signed from unsigned or vice-versa will help you. Follow this.

To answer your updated question, its how the system represents them i.e in a 2's complement (as in the above case where -1 =2's complement of 1 = 4294967295.

edited Dec 23 '14 at 09:25

answered Dec 23 '14 at 08:29

thepace

2,221
1
13
21

1

Quoted in your example: `printf("%d, %u",-1,-1);` is undefined behavior. – Gyapti Jain Dec 23 '14 at 12:49
I checked it in codepad. But I hope you get the essence. – thepace Dec 23 '14 at 14:37
@thepace I checked the link to wiki, it's great ! Now I understand the history of signed number representation by computer, and why 2'complement is widely used in now days. Thanks. – Eric Dec 25 '14 at 08:30

score 4 · Answer 2 · edited May 23 '17 at 10:33

printf("%d\n", i); invokes UB. i is unsigned int and you try to print it as signed int. Writing 1 << 31 instead of 1U << 31 is undefined too.

Print it as:

printf("%u\n", i);

or

printf("%X\n", i);

About your updated question, it also invokes UB for the very same reasons (If you use '1U' instead of 1, then for the reason that an int is initialized with 1U << 31 which is out of range value. If an unsigned is initialized with out of range value, modular arithmetic come into picture and remainder is assigned. For signed the behavior is undefined.)

Understanding the behavior on your platform
On your platform, int appears to be 4 byte. When you write something like 1 << 31, it converts to bit patters 0x80000000 on your machine.

Now when you try to print this pattern as signed, it prints signed interpretation which is -2³¹ (AKA INT_MIN) in 2s completement system. When you print this as unsigned, you get expected 2³¹ as output.

Learnings
1. Use 1U << 31 instead of 1 << 31
2. Always use correct print specifiers in printf.
3. Pass correct argument types to variadic functions.
4. Be careful when implicit typecast (unsigned -> signed, wider type -> narrow type) takes place. If possible, avoid such castings completely.

Hi downvoter, please leave your kind comment about what is wrong in the answer. — Gyapti Jain, Dec 23 '14 at 11:55
" When you write something like `1 << 31`, it converts to bit patters 0x80000000 on your machine." No it is UB, the compiler may generate something completely different, because UB is UB. `1U<<32==0x80000000` when `UINT_MAX>=0x80000000`, without UB. — 12431234123412341234123, Sep 14 '20 at 17:46

Mitch Wheat · Answer 3 · 2014-12-23T08:51:03.717

4

Use '%u' for unsigned int

 printf("%u\n", i);

.......................................................

Response to updated question: any sequence of bits can be interpreted as a signed or unsigned value.

edited Dec 23 '14 at 08:51

answered Dec 23 '14 at 08:27

Mitch Wheat

295,962
43
465
541

score 3 · Answer 4 · answered Dec 23 '14 at 08:27

3

Try

printf("%u\n", i);

By using %d specifier, printf expects argument as int and it typecast to int. So, use %u for unsigned int.

answered Dec 23 '14 at 08:27

doptimusprime

9,115
6
52
90

Jens Gustedt · Answer 5 · 2018-12-06T22:57:48.603

3

You are not doing the shift operation on an unsigned but on a signed int.

1 is signed.
the shift operation then shifts into the sign bit of that signed (assuming that int is 32 bit wide), that is already undefined behavior
then you assign whatever the compiler thinks he wants for that value to an unsigned.
then you print an unsigned as a signed, again no defined behavior.

edited Dec 06 '18 at 22:57

answered Dec 23 '14 at 11:57

Jens Gustedt

76,821
6
102
177

score 2 · Answer 6 · edited Dec 23 '14 at 13:14

2

%u is for unsigned int.
%d is for signed int.

in your programs output :

2147483648, 2147483648     (output for unsigned int)
-2147483648, -2147483648   (output for signed int )

edited Dec 23 '14 at 13:14

Gyapti Jain

4,056
20
40

answered Dec 23 '14 at 09:16

rabi shaw

441
1
3
14

bit shift for unsigned int, why negative?

6 Answers6