I am using the following to print out numbers from an array in hex:
char buff[1000];
// Populate array....
int i;
for(i = 0; i < 1000; ++i)
{
printf("[%d] %02x\n", i,buff[i]);
}
but I sometimes print weird values:
byte[280] 00
byte[281] 00
byte[282] 0a
byte[283] fffffff4 // Why is this a different length?
byte[284] 4e
byte[285] 66
byte[286] 0a
Why does this print 'fffffff4'?
Use %02hhx as the format string.
From CppReference, %02x accepts unsigned int. When you pass the arguments to printf(), which is a variadic function, buff[i] is automatically converted to int. Then the format specifier %02x makes printf() interprets the value as int, so potential negative values like (char)-1 get interpreted and printed as (int)-1, which is the cause of what you observed.
It can also be inferred that your platform has signed char type, and a 32-bit int type.
The length modifier hh will tell printf() to interpret whatever supplied as char type, so %hhx is the correct format specifier for unsigned char.
Alternatively, you can cast the data to unsigned char before printing. Like
printf("[%d] %02x\n", i, (unsigned char)buff[i]);
This can also prevent negative values from showing up as too long, as int can (almost) always contain unsigned char value.
See the following example:
#include <stdio.h>
int main(){
signed char a = +1, b = -1;
printf("%02x %02x %02hhx %02hhx\n", a, b, a, b);
return 0;
}
The output of the above program is:
01 ffffffff 01 ff
%02x accepts unsigned int, which causes your buff[i] to be promoted to int" This is somewhat misleading. Promotion happens because of default argument promotions on variable argument functions. Format string conversion specifiers have nothing to do with it.
int before printf is entered, and interpretation to unsigned int of happens inside printf because of %x.Your platform apparantly has signed char. On platforms where char is unsigned the output would be f4.
When calling a variadic function any integer argument smaller than int gets promoted to int.
A char value of f4 (-12 as a signed char) has the sign bit set, so when converted to int becomes fffffff4 (still -12 but now as a signed int) in your case.
%x02 causes printf to treat the argument as an unsigned int and will print it using at least 2 hexadecimal digits.
The output doesn't fit in 2 digits, so as many as are required are used.
Hence the output fffffff4.
To fix it, either declare your array unsigned char buff[1000]; or cast the argument:
printf("[%d] %02x\n", i, (unsigned char)buff[i]);
// Populate array....char buff[1000] = {0};to avoid using any un-init-ed value.charis a signed type (andintis 32 bits).