unsigned char ch = 212 ; And your printf will work. Even with ch changed to unsigned char , the behavior of the code is not defined by the C standard. This is because the unsigned char is promoted to an int (in normal C implementations), so an int is passed to printf for the specifier %u .
What you need to do is print out the char values individually as hex values. printf("hashedChars: "); for (int i = 0; i < 32; i++) { printf("%x", hashedChars[i]); } printf("\n"); Since you are using C++ though you should consider using cout instead of printf (it's more idiomatic for C++.
unsigned char is a character datatype where the variable consumes all the 8 bits of the memory and there is no sign bit (which is there in signed char). So it means that the range of unsigned char data type ranges from 0 to 255.
To print an unsigned int number, use the %u notation. To print a long value, use the %ld format specifier. You can use the l prefix for x and o, too. So you would use %lx to print a long integer in hexadecimal format and %lo to print in octal format.
Declare your ch
as
unsigned char ch = 212 ;
And your printf will work.
This is because in this case the char
type is signed on your system*. When this happens, the data gets sign-extended during the default conversions while passing the data to the function with variable number of arguments. Since 212 is greater than 0x80, it's treated as negative, %u
interprets the number as a large positive number:
212 = 0xD4
When it is sign-extended, FF
s are pre-pended to your number, so it becomes
0xFFFFFFD4 = 4294967252
which is the number that gets printed.
Note that this behavior is specific to your implementation. According to C99 specification, all char
types are promoted to (signed) int
, because an int
can represent all values of a char
, signed or unsigned:
6.1.1.2: If an
int
can represent all values of the original type, the value is converted to anint
; otherwise, it is converted to anunsigned int
.
This results in passing an int
to a format specifier %u
, which expects an unsigned int
.
To avoid undefined behavior in your program, add explicit type casts as follows:
unsigned char ch = (unsigned char)212;
printf("%u", (unsigned int)ch);
char
up to the implementation. See this question for more details.
There are two bugs in this code. First, in most C implementations with signed char
, there is a problem in char ch = 212
because 212 does not fit in an 8-bit signed char
, and the C standard does not fully define the behavior (it requires the implementation to define the behavior). It should instead be:
unsigned char ch = 212;
Second, in printf("%u",ch)
, ch
will be promoted to an int
in normal C implementations. However, the %u
specifier expects an unsigned int
, and the C standard does not define behavior when the wrong type is passed. It should instead be:
printf("%u", (unsigned) ch);
In case you cannot change the declaration for whatever reason, you can do:
char ch = 212;
printf("%d", (unsigned char) ch);
The range of char is 127 to -128. If you assign 212, ch stores -44 (212-128-128) not 212.So if you try to print a negative number as unsigned you get (MAX value of unsigned int)-abs(number) which in this case is 4294967252
So if you want to store 212 as it is in ch the only thing you can do is declare ch as
unsigned char ch;
now the range of ch is 0 to 255.
Because char
is by default signed
declared that means the range of the variable is
-127 to +127>
your value is overflowed. To get the desired value you have to declared the unsigned
modifier. the modifier's (unsigned
) range is:
0 to 255
to get the the range of any data type follow the process 2^bit
example char
is 8 bit length to get its range just 2 ^(power) 8
.
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