Can you explain about how to convert the last 3 bytes of data from unsigned integer to a character array?
Example:
unsigned int unint = some value;
unsigned char array[3];
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6What do you mean by the last three bytes?user703016– user7030162012-01-24 15:51:07 +00:00Commented Jan 24, 2012 at 15:51
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I guess he means the least significant 3 bytes but I might be wrong.Nick– Nick2012-01-24 15:53:23 +00:00Commented Jan 24, 2012 at 15:53
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4 Answers
It's more difficult if you have to convert it to an array, but if you just want to access the individual bytes, then you can do
char* bytes = (char*)&unint;
If you really do want to make an array (and therefore make a copy of the last 3 bytes, not leave them in place) you do
unsigned char bytes[3]; // or char, but unsigned char is better
bytes[0] = unint >> 16 & 0xFF;
bytes[1] = unint >> 8 & 0xFF;
bytes[2] = unint & 0xFF;
Comments
You can do using it the bitwise right shift operator:
array[0] = unint;
array[1] = unint >> 8;
array[2] = unint >> 16;
The least signifcant byte of uint is stored in the first element of the array.
1 Comment
David Heffernan
@Nick In a char array of length 3?
Depending on your needs, you may prefer an union:
typedef union {
unsigned int unint;
unsigned char array[3];
} byteAndInt;
or bit-shift operations:
for(int i=0; i<3; i++)
array[i] = (unint>>8*i) & 0xFF;
The former is not endian-safe.
5 Comments
Seth Carnegie
Isn't it UB to write to one member of a union and read the other without writing to it first?
David Heffernan
@Seth UB by the standard but well-defined by all compilers I've ever encountered.
mouviciel
An union specifies that the same memory zone can be interpreted several ways. This is the very purpose of it to write one field and read another one. Of course, executing the same code on a big-endian machine and on a little-endian one gives different results. but this is a portability issue, not UB.
Daniel Fischer
@Seth Not anymore. In C11, well, in n1570, fn 95 in 6.5.2.3 says "the appropriate part of the object representation of the value is reinterpreted as an object representation in the new type as described in 6.2.6 (a process sometimes called ‘‘type punning’’). This might be a trap representation".
Seth Carnegie
@DanielFischer oh, I didn't know about C11, I thought the last one was C99. Thanks.
If by last three, you mean lsb+1, lsb+2 and msb (in other words every byte other than the lsb), then you can use this.
unsigned int unint = some value;
unsigned char * array = ( (unsigned char*)&some_value ) + 1;
