Printing declared char value in C

Question

I understand that character variable holds from (signed)-128 to 127 and (unsigned)0 to 255

char x;
x = 128;

printf(\"%d\\n\", x); 

But how doe

Answer 1

printf is a variadic function, only providing an exact type for the first argument.
That means the default promotions are applied to the following arguments, so all integers of rank less than int are promoted to int or unsigned int, and all floating values of rank smaller double are promoted to double.

If your implementation has CHAR_BIT of 8, and simple char is signed and you have an obliging 2s-complement implementation, you thus get

128 (literal) to -128 (char/signed char) to -128 (int) printed as int => -128

If all the listed condition but obliging 2s complement implementation are fulfilled, you get a signal or some implementation-defined value.

Otherwise you get output of 128, because 128 fits in char / unsigned char.

Standard quote for case 2 (Thanks to Matt for unearthing the right reference):

6.3.1.3 Signed and unsigned integers

1 When a value with integer type is converted to another integer type other than _Bool, if the value can be represented by the new type, it is unchanged.
2 Otherwise, if the new type is unsigned, the value is converted by repeatedly adding or subtracting one more than the maximum value that can be represented in the new type until the value is in the range of the new type.60)
3 Otherwise, the new type is signed and the value cannot be represented in it; either the result is implementation-defined or an implementation-defined signal is raised.