Reading and changing bits in C uint32_t

Question

I am using this code

int get_bit(int n, int bitnr) {
int mask = 1 << bitnr;
int masked_n = n & mask;
int thebit = masked_n >> bitnr;
return thebit;
}

void printbits(uint32_t bits) {
    int i;
    for (i = 0; i < 32; i++)
        printf("%d", get_bit(bits, i));
    printf("\n");
}

to get and print the bits of a uint32_t, and in another function this code

uint32_t bits= 0;
bits|= 1<< 0;

to change the most significant bit (left-most) from 0 to 1.

the problem is when printing bits using the printbits function, it prints them right, but when using printf("%#x", bits); I'm getting the hex value of the bits as if they are read from right to left!

so the printbits gives me '10000000000000000000000000000000' but the hex value printed is the value of '00000000000000000000000000000001'.

Help appreciated

score 2 · Accepted Answer · edited Jun 20 '20 at 09:12

2

Changing the Most significant bit:

This line:

bits |= 1<< 0;

changes the least significant bit (LSB). 1 << 0 equals 1, which is not very significant :).

But, if you do:

bits |= 1 << 31;

or

bits |= 0x80000000;

You would actually change the most significant bit (MSB).

Printing the number in binary:

Your code is actually printing the number from right to left. You must change your loop to decrement.

for (i = 31; i >= 0; i--)

And if printing in the wrong way is fine for you (who knows...), try this:

uint32_t n = 41;
while (n) {
    printf("%d", n & 1);
    n >>= 1;
}

This can be easily adapted to print in the correct way using a recursive function:

void printbits(uint32_t n) {
    if (n) {
        printbits(n >> 1);
        printf("%d", n & 1);
    }
}

This algorithm works for any base with minor modifications.

edited Jun 20 '20 at 09:12

Community

1
1

answered Nov 15 '13 at 11:44

Maxime Chéramy

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Thank's, I actually used this answer: http://stackoverflow.com/questions/47981/how-do-you-set-clear-and-toggle-a-single-bit-in-c-c – codeoverflow Nov 15 '13 at 11:54
In your answer, bit 0 is the LSB. – Maxime Chéramy Nov 15 '13 at 11:55
One more thing, when I use my printbits it does a -1000... with yours not, and when i use my getbit(0) to get the MSB, its also a -1, any idea why? – codeoverflow Nov 15 '13 at 12:56
Try to change `thebit = masked_n >> bitnr;` to `thebit = masked_n & bitnr;` and change the type `int` into `uint32_t`. – Maxime Chéramy Nov 15 '13 at 12:59
Hmm, this gives me a 0: uint32_t get_bit(uint32_t n, int bitnr) { int mask = 1 << bitnr; int masked_n = n & mask; uint32_t thebit = masked_n & bitnr; return thebit; } – codeoverflow Nov 15 '13 at 13:03
Hum yes, sorry, forget what I said. Keep the code, only change the type of return. – Maxime Chéramy Nov 15 '13 at 13:25
it still does a -1, but can't i just do return -1*thebit for example? – codeoverflow Nov 15 '13 at 17:34
let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/41300/discussion-between-codeoverflow-and-maxime) – codeoverflow Nov 15 '13 at 23:50

score 1 · Answer 2 · answered Nov 15 '13 at 11:42

1

This code bits|= 1<< 0; sets the least not the most significant bit in an integer. Same goes true for your get_bit function - it numbers the bits from right to left.

answered Nov 15 '13 at 11:42

Ivaylo Strandjev

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user694733 · Answer 3 · 2013-11-15T11:51:18.777

1

(1 << 0) is the least significant bit. (1 << 31) would be the most significant bit.

Your print function is printing bits in ascending order (wrong way). You need to reverse your for loop:

for (i = 31; i >= 0; i--)

edited Nov 15 '13 at 11:51

answered Nov 15 '13 at 11:42

user694733

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score 0 · Answer 4 · answered Nov 15 '13 at 11:42

0

This is changing the first digit on the right (least important bit) :

 uint32_t bits= 0;
 bits|= 1<< 0;

Use something like :

uint32_t bits= 0;
bits |= 0x8000;

answered Nov 15 '13 at 11:42

Joze

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Reading and changing bits in C uint32_t

4 Answers4

Changing the Most significant bit:

Printing the number in binary: