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been searching everywhere, but couldn't the correct answer.

the problem is pretty simple:

i have to convert ASCII integer values into char's. for example, according to ASCII table, 108 stands for 'h' char. But when i try to convert it like this:

int i = 108
char x = i

and when I printf it, it shows me 's', no matter what number i type in(94,111...).

i tried this as well:

int i = 108;
char x = i + '0'

but i get the same problem! by the way, i have no problem in converting chars into integers, so i don't get where's the problem :/ thanks in advance

Sebastian Paaske Tørholm
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Rytis Paskevicius
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    Please copy and paste actual code; these fragments could not have been compiled, and it doesn't usually make sense to try to debug something that has never been run. – sarnold Dec 09 '11 at 00:08
  • Note that [decimal value 108 is actually lowercase `l`](http://en.wikipedia.org/wiki/ASCII#ASCII_printable_characters), not `s` or `h`. Where did you find `s` or `h`? – sarnold Dec 09 '11 at 00:10
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    please show us how your printf call. – MByD Dec 09 '11 at 00:10
  • http://www.theasciicode.com.ar/ascii-printable-characters/ascii-codes-108.html – pmg Dec 09 '11 at 00:10
  • sorry, 108 stands for 'l', but it doesn't make a difference. full code looks like: int i = 108; char x = i; printf("%s\n", x); and when i compile it, it shows 's' char(and no errors), i don't know why.. – Rytis Paskevicius Dec 09 '11 at 00:18
  • @Rytis: `%s` means print as string. A character is not a string. A string is terminated by a `\0`, for instance. You'll want to use `%c` for printing a single character. – Sebastian Paaske Tørholm Dec 09 '11 at 00:20

3 Answers3

1

That is how you do it. You probably want it unsigned, though.

Maybe your printf is wrong?

The following is an example of it working:

// Print a to z.
int i;
for (i = 97; i <= 122; i++) {
    unsigned char x = i;
    printf("%c", x);
}

This prints abcdefghijklmnopqrstuvwxyz as expected. (See it at ideone)

Note, you could just as well printf("%c", i); directly; char is simply a smaller integer type.

If you're trying to do printf("%s", x);, note that this is not correct. %s means print as string, however a character is not a string.

If you do this, it'll treat the value of x as a memory address and start reading a string from there until it hits a \0. If this merely resulted in printing s, you're lucky. You're more likely to end up getting a segmentation fault doing this, as you'll end up accessing some memory that is most likely not yours. (And almost surely not what you want.)

Sebastian Paaske Tørholm
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1

Sounds to me like your printf statement is incorrect.

Doing printf("%c", c) where c has the value 108 will print the letter l... If you look at http://www.asciitable.com/ you'll see that 108 is not h ;)

Matt Lacey
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0

I'm guessing your printf statement looks like this:

printf("s", x);

..when in fact you probably meant:

printf("%s", x);

...which is still wrong; this is expecting a string of characters e.g:

char* x = "Testing";
printf("%s", x);

What you really want is this:

int i = 108;
char x = i + '0';
printf("%c", x);

...which on my system outputs £

Jon Cage
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