Having trouble understanding pointer operations

Question

I can't exactly understand what the language does when I write

*(t++)

*t++

When t is a pointer?

Answer 1

These two expressions

*(t++)

*t++

are equivalent due to the operator precedence.

So the postfix operator ++ has a higher priority than the unary operator *.

The result of the postfix operator ++ is the value of its operand before incrementing.

From the C Standard (6.5.2.4 Postfix increment and decrement operators)

2 The result of the postfix ++ operator is the value of the operand. As a side effect, the value of the operand object is incremented (that is, the value 1 of the appropriate type is added to it)....

Take into account that due to the pointer arithmetic if you have a pointer like this

T *p;

where T is some type then after the operation p++ the final value of the pointer itself is incremented by the value sizeof( T ). For the type char sizeof( char ) is always equal to 1.

Consider the following demonstrative program.

#include <stdio.h>

int main(void) 
{
    char *t = "AB";

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

    return 0;
}

Its output is

A
B

You can substitute this statement

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

for this statement

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

and you will get the same result.

In fact this expression

*(t++)

is also equivalent to the expression

t++[0]