why it is giving output as 0.0000?

Question

#include <stdio.h>
int main()
{
    signed char ch;
    ch = 128;
    printf("%f", ch);
    return 0;
}

Can anyone explain why it is printing 0.000 every time??

I tried %f as a format specifier for signed char value.

Answer 1

If you want to print a value, that you have in an integer variable, as if it were floating point, you have to write an explicit cast.

#include <stdio.h>
int main()
{
    signed char ch;
    ch = 128;
    printf("%f\n", (double)ch);
    return 0;
}

In a call to most functions that take a double or float argument, the compiler would insert this cast for you, but it doesn't do it for printf because printf's argument list isn't declared. See the older question I just linked this one to, for more detail.

Since you know the value is an integer, you can also write out the fractional part yourself:

#include <stdio.h>
int main()
{
    signed char ch;
    ch = 128;
    printf("%d.000000\n", ch);
    return 0;
}

will print the same thing as the first example.

Answer 2

To print the value of signed char you need to use %d format. %f id to use float and double types.

#include <stdio.h>
int main()
{
    signed char ch;
    ch = 128;
    printf("%d\n", ch);
    return 0;
}

https://godbolt.org/z/5816GvPWd

It will output -128 as you set the last bit of 8-bit integer (assuming CHAR_BIT == 8) and the system uses two's complement integers

---

Assignment of 128 to 8-bit signed char is implementation-defined as 128 cannot be accommodated in 8 bit integer. Most implementations will behave as I wrote above - ie print -128