CODEWITHSUNDEEP
ctypesbitwise-operatorsbit
user2756257
user2756257

Reputation:

unsigned integer in C printing incorrectly

With the following code I declare an unsigned int and assign it the value of 236. I then take the 1's complement of it and assign that to a separate variable. When printed with printf, I expect the 2nd variable to print as "19", but its printing as "4294967059". Why? Doesn't the ~ bitwise operator take the value of the first variable (base 2) and "flip" the bits (1's complement), resulting in "19" in base 10? Ints on my machine are 32-bit, and I assume this has something to do with 2^32-1 (4294967295), but I haven't figured it out

unsigned a = 236; // binary of this 11101100 = 236 base 10
unsigned b = ~a; // 1's complement to 00010011 = 19 base 10
printf("a: %u b: %u",a,b); // prints 236 and 4294967059.  WHY?

Upvotes: 2

Views: 335

Answers (2)

Eric Z
Eric Z

Reputation: 14505

The binary of a is 00000000 00000000 00000000 11101100 if unsigned int is 4 bytes long. ~a is 11111111 11111111 11111111 00010011, which is 4294967295.

You can use unsigned char to represent a single byte(most likely it's 8 bits).

unsigned char a = 236;
unsigned char b = ~a;  // b = 19

Upvotes: 4

P0W
P0W

Reputation: 47784

"Ints on my machine are 32-bit"

When you say int on your machine is 32 bit why you just consider 8 bit ?

236 => 00000000000000000000000011101100

1's complement of which is

11111111111111111111111100010011 => 4294967059

Upvotes: 2

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