Comparing unsigned char and EOF

Question

when the following code is compiled it goes into an infinite loop:

int main()
{
    unsigned char  ch;
    FILE *fp;
    fp = fopen(\"abc\",\"r\");
    if(fp         


        

Answer 1

There are several implicit conversions going on. They aren't really relevant to the specific warning, but I included them in this answer to show what the compiler really does with that expression.

• ch in your example is of type unsigned char.
• EOF is guaranteed to be of type int (C99 7.19.1).

So the expression is equivalent to

(unsigned char)ch != (int)EOF

The integer promotion rules in C will implicitly convert the unsigned char to unsigned int:

(unsigned int)ch != (int)EOF

Then the balancing rules (aka the usual arithmetic conversions) in C will implicitly convert the int to unsigned int, because each operand must have the same type:

(unsigned int)ch != (unsigned int)EOF

On your compiler EOF is likely -1:

(unsigned int)ch != (unsigned int)-1

which, assuming 32-bit CPU, is the same as

(unsigned int)ch != 0xFFFFFFFFu

A character can never have such a high value, hence the warning.