Simple Character Interpretation In C

Question

Here is my code

 #include

 void main()
 {
     char ch = 129;
     printf(\"%d\", ch);
 }

I get the output as -127. What d

Answer 1

The type char can be either signed or unsigned, it's up to the compiler. Most compilers have it as `signed.

In your case, the compiler silently converts the integer 129 to its signed variant, and puts it in an 8-bit field, which yields -127.