C int to char 不使用 char 格式打印出来

Question

I'm trying to print out the digits of an integer using the char format specifier.

This is what I currently have

    //counter = arbitrary int
    while (counter > 0) {
            char c = (char)(counter % 10);
            printf("%c", c);
            counter/=10;
        }

This just prints out blank but if I change the format specifier to int but leave the type cast, it'll print the correct value. Something like this

   char c = (char)(counter % 10);
   printf("%d", c);

Any ideas why this isn't printing when I use the char format specifier?

Answer 1

%c is for printing a character, not a number. If you want to print a number, you should use %d . The reason it works for char is because char is implicitly promoted to int when you pass it as an argument to a variadic function.

If c contained 'A' , and you used %c to format it, you would get the letter 'A', but if you used %d , you would get its integer value, eg 65.

If you absolutely must use %c , you can add the value of '0' to the character before printing, egL

printf("%c", c + '0');

Answer 2

%c可以在此上下文中用作printf("%c",c+'0') ，其中+'0'将其转换为与c等效的整数。