format specifier for short integer

Question

I don't use correctly the format specifiers in C. A few lines of code:

int main()
{
        char dest[]="stack";
        unsigned short val = 500;
        char c = 'a';

        char* final = (char*) malloc(strlen(dest) + 6);

        snprintf(final, strlen(dest)+6, "%c%c%hd%c%c%s", c, c, val, c, c, dest); 

        printf("%s\n", final);
        return 0;
}

What I want is to copy at

final [0] = a random char final [1] = a random char final [2] and final [3] = the short array final [4] = another char ....

My problem is that i want to copy the two bytes of the short int to 2 bytes of the final array.

thanks.

Answer 1

I'm confused - the problem is that you are saying strlen(dest)+6 which limits the length of the final string to 10 chars (plus a null terminator). If you say strlen(dest)+8 then there will be enough space for the full string.

Update

Even though a short may only be 2 bytes in size, when it is printed as a string each character will take up a byte. So that means it can require up to 5 bytes of space to write a short to a string, if you are writing a number above 10000.

Now, if you write the short to a string as a hexadecimal number using the %x format specifier, it will take up no more than 2 bytes.

Answer 2

You need to allocate space for 13 characters - not 11. Don't forget the terminating NULL.

Answer 3

When formatted the number (500) takes up three spaces, not one. So your snsprintf should give the final length as strlen(dest)+5+3. Then also fix your malloc call to adjust. If you want to compute the strlen of the number, do that with a call like this strlen(itoa(val)). Also, cant forget the NULL at the end of dest, but I think strlen takes this into account, but I'm not for sure.

Answer 4

Simple answer is you only allocated enough space for the strlen(dest) + 6 characters when in all reality it looks like you're going to have 8 extra characters... since you have 2 chars + 3 chars in your number + 2 chars after + dest (5 chars) = 13 char when you allocated 11 chars.

Answer 5

Unsigned shorts can take up to 5 characters, right? (0 - 65535)

Seems like you'd need to allocate 5 characters for your unsigned short to cover all of the values.

Which would point to using this:

char* final = (char*) malloc(strlen(dest) + 10);

Answer 6

You lose one byte because you think the short variable takes 2 byte. But it takes three: one for each digit character ('5', '0', '0'). Also you need a '\0' terminator (+1 byte).

==> You need strlen(dest) + 8

Answer 7

Use 8 instead of 6 on:

char* final = (char*) malloc(strlen(dest) + 6);

and

snprintf(final, strlen(dest)+6, "%c%c%hd%c%c%s", c, c, val, c, c, dest);

Answer 8

Seems like the primary misunderstanding is that a "2-byte" short can't be represented on-screen as 2 1-byte characters.

First, leave enough room:

char* final = (char*) malloc(strlen(dest) + 9);

The entire range of possible values for a 1-byte character are not printable. If you want to display this on screen and be readable, you'll have to encode the 2-byte short as 4 hex bytes, such as:

## as hex, 4 characters
snprintf(final, sizeof(final), "%c%c%4x%c%c%s", c, c, val, c, c, dest);

If you are writing this to a file, that's OK, and you might try the following:

## print raw bytes, upper byte, then lower byte.
snprintf(final, sizeof(final), "%c%c%c%c%c%c%s", c, c, ((val<<8)&0xFF), ((val>>8)&0xFF), c, c, dest); 

But that won't make sense to a human looking at it, and is sensitive to endianness. I'd strongly recommend against it.