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

Whether a plain char is signed or unsigned, is implementation-defined behavior. This is a quite stupid, obscure rule in the C language. int, long etc are guaranteed to be signed, but char could be signed or unsigned, it is up to the compiler implementation.

On your particular compiler, char is apparently signed. This means, assuming that your system uses two's complement, that it can hold values of -128 to 127.

You attempt to store the value 129 in such a variable. This leads to undefined behavior, because you get an integer overflow. Strictly speaking, anything can happen when you do this. The program could print "hello world" or start shooting innocent bystanders, and still conform to ISO C. In practice, most (all?) compilers will however implement this undefined behavior as "wrap around", as described in other answers.

To sum it up, your code relies on two different behaviors that aren't well defined by the standard. Understanding how the result of such unpredictable code ends up in a certain way has limited value. The important thing here is to recognize that the code is obscure, and learn how to write it in a way that isn't obscure.

The code could for example be rewritten as:

unsigned char ch = 129;

Or even better:

#include 
...
uint8_t ch = 129;

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As a rule of thumb, make sure to follow these rules in MISRA-C:2004:

6.1 The plain char type shall be used only for the storage and use of character values.

6.2 signed and unsigned char type shall be used only for the storage and use of numeric values.