* Vs ++ precedence in C

Question

I am not able to understand the output of the following C code :

#include
main()
{
   char * something = \"something\";
   printf(\"%c\", *someth         


        

Answer 1

always use the clockwise rule clockwise rule

printf("%c\n", *something++);

according to the rule you will first encounter * so get the value then ++ means increment

in the 3rd case printf("%c\n", *something++);

so according to the image increment the value ++ and then get the value *

Answer 2

See http://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B#Operator_precedence for details

printf("%c", *something++);

Gets the char at *something and then increments it ('s')

printf("%c", *something);

Just get the char (now the second, due to the increment in the last statement ('o')

printf("%c", *++something);

increment and then get the char of the new position ( 'm' )

printf("%c", *something++);

Gets the char at *something and then increments it ('m')

Answer 3

// main entrypoint
int main(int argc, char *argv[])
{
    char * something = "something";

    // increment the value of something one type-width (char), then
    //  return the previous value it pointed to, to be used as the 
    //  input for printf.
    // result: print 's', something now points to 'o'.
    printf("%c", *something++);

    // print the characer at the address contained in pointer something
    // result: print 'o'
    printf("%c", *something);

    // increment the address value in pointer something by one type-width
    //  the type is char, so increase the address value by one byte. then
    //  print the character at the resulting address in pointer something.
    // result: something now points at 'm', print 'm'
    printf("%c", *++something);

    // increment the value of something one type-width (char), then
    //  return the previous value it pointed to, to be used as the 
    //  input for printf.
    // result: print 's', something now points to 'o'.
    printf("%c", *something++);
}

Result:

somm

Answer 4

It is pretty simple.

char * something = "something";

Assignment of pointer.

printf("%c\n", *something++);//equivalent to *(something++)

Pointer is incremented but the value before increment is dereferenced ans it is post-increment.

printf("%c\n", *something);//equivalent to *(something)

Pointer is now pointing to 'o' after increment in the previous statement.

printf("%c\n", *++something);//equivalent to *(++something)

Pointer is incremented to point to 'm' and dereferenced after incrementing the pointer as this is pre-increment.

printf("%c\n", *something++);//equivalent to *(something++)

Same as the first answer. Also notice '\n' at the end of every string in printf. It makes the output buffer flush and makes the line print. Always use a \n at the end of your printf.

You may want to look at this question as well.