assigning 128 to char variable in c

Question

The output comes to be the 32-bit 2\'s complement of 128 that is 4294967168. How?

#include 
int main()
{
    char a;
    a=128;
    if(a==-128         


        

Answer 1

First of all, as I hope you understand, the code you've posted is full of errors, and you would not want to depend on its output. If you were trying to perform any of these manipulations in a real program, you would want to do so in some other, more well-defined, more portable way.

So I assume you're asking only out of curiosity, and I answer in the same spirit.

Type char on your machine is probably a signed 8-bit quantity. So its range is from -128 to +127. So +128 won't fit.

When you try to jam the value +128 into a signed 8-bit quantity, you probably end up with the value -128 instead. And that seems to be what's happening for you, based on the fact that your if statement is evidently succeeding.

So next we try to take the value -128 and print it as if it was an unsigned int, which on your machine is evidently an 32-bit type. It can hold numbers in the range 0 to 4294967295, which obviously does not include -128. But unsigned integers typically behave pretty nicely modulo their range, so if we add 4294967296 to -128 we get 4294967168, which is precisely the number you saw.

Now that we've worked through this, let's resolve in future not to jam numbers that won't fit into char variables, or to print signed quantities with the %u format specifier.