CODEWITHSUNDEEP
cbinaryprintfbit-shift
Syf Illis
Syf Illis

Reputation: 145

Display MSB to LSB

I'm having a hard time understanding my issue:

uint8_t nal_type=6;

for(i=7;i!=0;i--){
    printf("%d",(nal_type>>i)&0x01U);}
printf("\n");

I would expect the following code to display the binary value from MSB to LSB. But it displays the following:

0000011

Can someone enlighten me ?

Upvotes: 0

Views: 763

Answers (1)

Paul R
Paul R

Reputation: 213060

It's just a simple mistake in your loop, so you're not seeing the LS bit (bit 0) - make it:

for (i = 7; i >= 0; i--)
{           ^^^^^^
    printf("%d", (nal_type >> i) & 0x01U);
}

This will then give the output as:

00000110

which I think is what you are looking for (MSB to LSB).

Upvotes: 2

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