I would like to initialize a 16-byte array of hexadecimal values, particularly the 0x20 (space character) value.
What is the correct way?
unsigned char a[16] = {0x20};
or
unsigned char a[16] = {"0x20"};
Thanks
6 Answers
Defining this, for example
unsigned char a[16] = {0x20, 0x41, 0x42, };
will initialise the first three elements as shown, and the remaining elements to 0.
Your second way
unsigned char a[16] = {"0x20"};
won't do what you want: it just defines a nul-terminated string with the four characters 0x20, the compiler won't treat it as a hexadecimal value.
2 Comments
Kingamere
Ahh ok I see. And what if I want the ASCII value of 0x20? Like I want the keyboard space that that hex value gives.
Weather Vane
@Kingamere the ASCII value is 0x20 (hex) or 32 (decimal). If you print that as an
int you'll get 32 and if you print as char you'll get ` ` (space).There is a GNU extension called designated initializers.
This is enabled by default with gcc
With this you can initialize your array in the form
unsigned char a[16] = {[0 ... 15] = 0x20};
answered Oct 31, 2015 at 18:31
Raghu Srikanth Reddy
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1 Comment
too honest for this site
Designated initialisers are not a gcc extension, but standard C. Allowing a range is an extension, however is..
unsigned char a[16] = {0x20,0x20,0x20,0x20,0x20,0x20,0x20,0x20,0x20,0x20,0x20,0x20,0x20,0x20,0x20,0x20};
or
unsigned char a[16] = "\x20\x20\x20\x20\x20\x20\x20\x20\x20\x20\x20\x20\x20\x20\x20\x20";
1 Comment
Missing User
This violates DRY and adds adds unneccasary bytes, no sane programmer should do this.
I don't know if this is what you were looking for, but if the size of the given array is going to be changed (frequently or not), for easier maintenance you may consider the following method based on memset()
#include <string.h>
#define NUM_OF_CHARS 16
int main()
{
unsigned char a[NUM_OF_CHARS];
// initialization of the array a with 0x20 for any value of NUM_OF_CHARS
memset(a, 0x20, sizeof(a));
....
}
Comments
The first way is correct, but you need to repeat the 0x20, sixteen times. You can also do this:
unsigned char a[16] = " ";
5 Comments
Alex Lop.
This solution will lead to undefined behavior!
"<16 spaces>" will actually occupy 17 bytes (the last one for '\0').
user3386109
@AlexLop. Negative, the C specification allows a character array to be initialized using a string literal that is the same size as the array. In that case the NUL terminator is dropped.
Alex Lop.
Could you please point me to the documentation which will support your statement?
user3386109
Section 6.7.9 paragraph 14, which says (emphasis added): "Successive bytes of the string literal (including the terminating null character if there is room or if the array is of unknown size) initialize the elements of the array."
Alex Lop.
You are right, my mistake. Could you please make a small edit in your answer so I could "undo" the downvote?
I would use memset.
No need to type out anything and the size of the array is only provided once at initialisation.
#include <string.h>
int main(void)
{
unsigned char b[16];
memset(b, 0x20, sizeof(b));
}
0x2is not the space character, but an integer.0x200; I meant0x20, of course). If you mean space, you should use a space character constant:' '.