Question

This is the line of code in C.
The condition of loop here is ++i.
So how does compiler decide which condition to consider because here other two appear as conditions?

char i=0;
for(i<=5&&i>-1;++i;i>0)
    printf("%d",i);

output

1234..127-128-127....-2-1

Answer 1

The for statement works like this:

for (X; Y; Z)
{
    ...
}

translates to

X;
while (Y)
{
    ...
    Z;
}

So your code changes from:

char i=0;
for(i<=5&&i>-1;++i;i>0)
    printf("%d",i);

to:

char i = 0;
i<=5 && i>-1;    // X
while (++i)      // Y
{
    printf("%d", i);
    i > 0;       // Z
}

As you can see, lines marked with X and Z are completely useless. Therefore:

char i = 0;
while (++i)
    printf("%d", i);

This means it will print from 1 up to whenever result of ++i is zero.

If char in your compiler is signed, then the behavior is left to implementation, even though most likely it will overflow to a negative value and work its way up to zero.

If char is positive, this will print positive values up to where it overflows back to 0.