CODEWITHSUNDEEP
c++bit
magu_
magu_

Reputation: 4856

Bits apear twice in size_t

This might be a very stupid question, so please be gentle. If I run the following code:

#include <climits>
#include <cstdint>
#include <stdio.h>

typedef uint64_t obs;

int main () {
    printf("Size : %i\n",sizeof(obs)*CHAR_BIT);

    obs value = 1 << 3;
    printf("Number: %zu\n",value);

    printf("Bits : ");
    for (size_t ind = 0; ind < sizeof(obs)*CHAR_BIT; ind++) {
        if (value & (1 << ind)) {
            printf("%zu ",ind);
        }
    }
}

For various typedefs I get the following result for 64 bit datatypes (I run on a 64Bit system):

uint64_t / size_t / long unsigned

Size : 64
Number : 8
Bits : 3 35

and the following for other lengths:

uint32_t / uint16_t / uint8_t

Size : 32 / 16 / 8
Number : 8
Bits : 3

If if changing the shift the 64Bit type seems to have a "mirror counterpart" which is shifted by 32Bit. The same holds true when I change the value of value. Is there a reason for this or am I missing something?

using gcc on Win7

Upvotes: 2

Views: 110

Answers (1)

Rahul Tripathi
Rahul Tripathi

Reputation: 172378

The size_t is architecture dependant so on a 32-bit system size_t will likely be at least 32-bits wide. On a 64-bit system it will likely be at least 64-bit wide.

According C99 standard:

The value of the result is implementation-defined, and its type (an unsigned integer type) is size_t, defined in <stddef.h> (and other headers)

Also change

if (value & (1 << ind))

to 

if (value & (1LL << ind))

as 1 is a 32-bit integer

Upvotes: 2

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