CODEWITHSUNDEEP
cpointersinttype-conversion
sin
sin

Reputation: 89

Type conversion

Can someone please tell me what do these lines of code do 

*(a++)  = (int)((value >> 16) & 0xFF) ;
*(a++)  = (int)((value >> 8) & 0xFF) ;  
*(a++)  = (int)((value & 0xFF)) ;

I understand that it checks the value, if it is much greater than 16 it converts it to type int and if it is much smaller than 8 does the same. But what does the
& 0xFF and *(a++) do?

Upvotes: 0

Views: 150

Answers (4)

Klamore74
Klamore74

Reputation: 578

*(a++)  = (int)((value >> 16) & 0xFF) ;

is like:

aIntValue = value/65536;
aIntBalue = a%256;

*(a++)  = (int)((value >> 8) & 0xFF) ;

is like:

aIntValue = value/256;
aIntValue = a%256;

*(a++)  = (int)((value & 0xFF)) ;

is like:

aIntValue = a%256;

At the end of the code, either code assign the aIntValut to the value pointed to the pointer 'a' and next the pointer is moved to the next element.

Upvotes: 1

reece
reece

Reputation: 8145

Given:

char data[10];

uint32_t value = 0x61626364; // 'abcd'

char *a = data;
*(a++) = (int)((value >> 24) & 0xFF);
*(a++) = (int)((value >> 16) & 0xFF);
*(a++) = (int)((value >> 8) & 0xFF);
*(a++) = (int)(value & 0xFF);
*(a++) = ':';
*((uint32_t *)a) = value;
a+=4;
*(a++) = 0;

printf("%s\n", data);

I get (on my intel box, which is a little endian system):

abcd:dcba

So this is ensuring that the bytes of an integer are in an platform-independent form (choosing big endian as the byte format).

Now, for:

*(a++) = (int)((value >> 16) & 0xFF);

we have:

0x61626364 -- value
0x00006162 -- value >> 16 : shifted 2 bytes
0x00000062 -- (value >> 16) & 0xFF : last byte only

Upvotes: 1

cnicutar
cnicutar

Reputation: 182619

I understand that it checks the value

It doesn't check anything, it's not like the << symbol in math which means "much smaller". To break down this line:

*(a++)  = (int)((value >> 16) & 0xFF);
  • (>>) shifts value 16 times to the right
  • (&) ands the result with 0xFF, thereby discarding everything to the left
  • Stores the result at the address pointed by a
  • Increments the pointer, making a point to some "next" element

Upvotes: 2

huseyin tugrul buyukisik
huseyin tugrul buyukisik

Reputation: 11910

(value>>16)

No it is not much greater. 

It is shift right by 16 bits.

But dividing it by 2 exatly 16 times makes it much smaller than before.

val&0xff makes a solution if it is divisible by 256. For example: if val&0xff is different than zero, than it is not divisible by 256

Upvotes: 1

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