I'm trying to printf unsigned short in hex, e.g. 0XFFFF. The problem is that I want a lowercase x not X. I have no idea how to do this.
fprintf(c,"%#06X,\n\t", pixel[i]));
Change the %X specifier to %x and you should get lower case:
fprintf(c, "%#06x\n", pixel[i]);
Outputs the number as 0xffff. If pixel is an array of unsigned shorts and you want the format to match the data, you can do:
fprintf(c, "%#06hx\n", pixel[i]);
If you really want 0xFFFF, then that's trivial:
fprintf(c, "0x%04hX\n", pixel[i]);
Related
I would like to print the unsigned char in this format:
0xff. How can I do so? Thanks.
printf("0x%02X", value & 0xFF);
if you want to values to always be 2 characters. also "& 0xFF" will avoid some values to be printed with a lot of "F" in the front (happens sometimes)
%x is the basic conversion specifier for printing an unsigned int (that's what an unsigned char is converted to when passed to a variadic function, such as printf()) in hexadecimal format with lowercase letters. If you want leading zeroes and at least two digits, use %02x. If, in addition, you want the 0x prefix as well, prepend the # modifier. So your format string will look like
"%#02x"
printf("0x%x\n", variable);
Visit printf documentation page for more details
%x is the format specifier you are looking for.
printf("0x%x\n",your_char);
For chars I use this
printf("0x%hhx\n",your_char);
I have a simple program :
#include <stdio.h>
int main()
{
long i = 16843009;
printf ("%02x \n" ,i);
}
I am using %02x format specifier to get 2 char output,
However, the output I am getting is:
1010101
while I am expecting it to be :01010101 .
%02x means print at least 2 digits, prepend it with 0's if there's less. In your case it's 7 digits, so you get no extra 0 in front.
Also, %x is for int, but you have a long. Try %08lx instead.
%x is a format specifier that format and output the hex value. If you are providing int or long value, it will convert it to hex value.
%02x means if your provided value is less than two digits then 0 will be prepended.
You provided value 16843009 and it has been converted to 1010101 which a hex value.
Your string is wider than your format width of 2. So there's no padding to be done.
You are actually getting the correct value out.
The way your x86 (compatible) processor stores data like this, is in Little Endian order, meaning that, the MSB is last in your output.
So, given your output:
10101010
the last two hex values 10 are the Most Significant Byte (2 hex digits = 1 byte = 8 bits (for (possibly unnecessary) clarification).
So, by reversing the memory storage order of the bytes, your value is actually: 01010101.
Hope that clears it up!
NSLog function accepts printf format specifiers.
My question is about %x specifier.
Does this print hex codes as sequence on memory? Or does it have it's own printing sequence style?
unsigned int a = 0x000000FF;
NSLog(#"%x", a);
Results of above code on little or big endian processors are equal or different?
And how about NSString's -initWithFormat method? Does it follows this rule equally?
%x always prints the most significant digits first. Doesn't matter what kind of processor it is running on.
MASSIVE EDIT:
I have a long int variable that I need to convert to a signed 24bit hexadecimal string without the "0x" at the start. The string must be 6 characters followed by a string terminator '\0', so leading zeros need to be added.
Examples:
[-1 -> FFFFFF] ---
[1 -> 000001] ---
[71 -> 000047]
Answer
This seems to do the trick:
long int number = 37;
char string[7];
snprintf (string, 7, "%lX", number);
Because you only want six digits, you are probably going to have to do some masking to make sure that the number is as you require. Something like this:
sprintf(buffer, "%06lx", (unsigned long)val & 0xFFFFFFUL);
Be aware that you are mapping all long integers into a small range of representations. You may want to check the number is in a specific range before printing it (E.g. -2^23 < x < 2^23 - 1)
Look at sprintf. The %lx specifier does what you want.
Use itoa. It takes the desired base as an argument.
Or on second thought, no. Use sprintf, which is standard-compliant.
In the title you say you want a signed hex string, but all your examples are unsigned hex strings. Assuming the examples are what you want, the easiest way is
sprintf(buffer, "%06X", (int)value & 0xffffff);
hi i am reading a binary file using c as shown here link text
so that all the information read from binary file is stored in "char *buffer".
i have the format standard where it says that one of the lines should be
format: unsigned char, size: 1 byte
i am doing the following:
printf("%x\n", buffer[N]);
but what should i do when the format says:
format: unsigned short, size: 2 bytes
if i do it as follows, would this be correct:
printf("%d%d\n", buffer[N], buffer[N+1]);
if not can you show me the correct way?
Also can you tell me if the following are correct way while printing:
char %c
unsigned long %ul
unsigned short %d
unsigned char %x
double %f
long %ld
all of the data in binary file is in little-endian format! thanks a lot in advance!
Try printf("%d", (short)(buffer[N] + buffer[N+1]<<8)). Now notice that I had to assume that the byte order in the buffer had the least significant byte of the two-byte short stored at the lower address.
I could likely have written *(short *)(&buffer[N]), but that assumes that N has the right alignment to hold a short on your platform, and that the buffer and the platform agree on byte order.
This is actually just the tip of a very large iceberg of a topic. There are many subtle issues lurking, and some really unsubtle ones when you wander into floating point values.