++ Operator Misunderstanding

I have the following code

Code:

---

#include <stdio.h>

void show_str(const char *test_str);

int main(void){
   
    char* the_str = "A woodchuck could chop as much as a woodchuck could\n";
   
    show_str(the_str);
   
    return 0;
}

void show_str(const char *some_string){
    //using bracket notation
    int ndx = 0;
    while(some_string[ndx]){
        putchar(some_string[ndx++]);
    }
   
    //using increment operator
    while( (*some_string) ){
        putchar( *(some_string++) );
    }
   
}

---

doesn't the ++ operator have higher precedence, and thus *(string + 1) should always be the first element that is sent to stdout? As it stands it starts printing from (string + 0).

It's been bothering me so if anyone can explain it another way that I'll I understand it would be appreciated.

I think that you are trying to do this:

Code:

---

putchar( *(++some_string) );

---

You are incrementing the pointer after the line has completed, instead of before.

[edit]

Code:

---

banana = apple++ + mango;
//is the same as
banana = apple + mango;
apple = apple + 1;

banana = ++apple + mango;
//is the same as
apple = apple + 1;
banana = apple + mango;

---

[/edit]

Quote:

---

Originally Posted by hex_dump

doesn't the ++ operator have higher precedence... As it stands it starts printing from (string + 0).

---

It does have higher precedence, but this is the correct behaviour for the code you posted. Remember that with postfix ++ (i.e. ++ after the operand like"some_string++") its operand is first used and then incremented - the increment only happens *after* its value is taken. With prefix ++, it is first incremented, then the (newly incremented) value is used.

As a more concrete example:

Code:

---

int post_a = 2;
int post_b = post_a++;  // post_a++ yields value of post_a *before* it was incremented.
// At this point, post_a = 3, post_b = 2.

int pre_a = 2;
int pre_b = ++pre_a;  // ++pre_a yields value of pre_a *after* it gets incremented.
// At this point, pre_a = 3, pre_b = 3.

---

Take a look here: C Operator Precedence Table. Pay particular attention to 'Note 2' at the bottom.

To clarify more, yes, it has higher precedence, and your parentheses do ensure the ++ happens first, but you are misunderstanding the semantics of the ++ postfix operator.

First, it may help to understand the concept of a sequence point (link). Basically, the increment happens at the end of a sequence point. A sequence point is the end of a full expression. Your parenthetical is not a full expression. You have a case more similar to rule #4: the call to putchar is the sequence point at which the increment happens, however the original value of some_string is passed to putchar.

The result of the parenthetical expression

Code:

---

(some_string++);

---

is the current value of some_string, i.e. some_string+0. That value is used for the dereference operator *. Then, when you reach a sequence point, the increment happens. Basically what you have is equivalent to:

Code:

---

putchar( *(some_string) );
some_string += 1;

---

Click_here gave you a possible solution if you wish to skip the first char, using the ++ prefix operator.

Hope that clears it up a bit.

Excellent, thanks guys. Yes I think i was misunderstanding the semantics of the ++ operators and thank you very much for the links anduril462, much appreciated.