How to printf variable?

UserUserUserUserUser · **Joined:** Tue Jan 12, 2021 11:09 am **Posts:** 1

I am using printf from Meaty Skeleton (https://wiki.osdev.org/Meaty_Skeleton). After some changes of terminal_putchar, it works fine, when I'm writing ready string (printf("Hello\n") gives me "Hello" output), but when I use variable (printf(a), where's a is char), it gives me random symbols. How to make it say variable value? (I am using i686-elf-gcc)

klange · **Joined:** Wed Mar 30, 2011 12:31 am **Posts:** 676

Quote:

but when I use variable (printf(a), where's a is char),

It sounds like you're trying to pass a char as an argument to a function that takes a pointer. At the risk of sounding harsh, this is a very basic C thing and if you don't see how this is wrong you'll have a difficult time moving forward with OS dev.

bzt · **Joined:** Thu Oct 13, 2016 4:55 pm **Posts:** 1584

UserUserUserUserUser wrote:

when I use variable (printf(a), where's a is char), it gives me random symbols.

The first argument must be a zero terminated string which specifies the formating. Are you sure the compiler didn't gave you a warning about using "printf(a)"?

UserUserUserUserUser wrote:

How to make it say variable value?

Depends on the variable.

Code:

printf("%c\n", a);    // print char
printf("%d\n", i);    // print integer in decimal
printf("%x\n", i);    // print integer in hexadecimal
printf("%s\n", s);    // print another string
... etc.

See section "Conversion specifiers" in man printf.

klange wrote:

At the risk of sounding harsh, this is a very basic C thing and if you don't see how this is wrong you'll have a difficult time moving forward with OS dev.

I don't want to be harsh either, but klange is right.

Cheers,
bzt

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