Pointer and char array confusion

Question

#include <iostream>
using namespace std;
int main()
{
 char str[] {"TESTING"};
 char *p {str};
 cout << (p++, *++p);
 cout << *p++;
 cout << p;
 return 0;
}

It returns "SSTING"

I know maybe this post isn't exactly for stackoverflow but I can't figure out what it does, and couldn't find any documentation about it

 cout << (p++, *++p);

First time I saw round brackets with comma in cout... what's their function?

and shouldn't this line alone say "TESTING" but it seems to say only TING

cout << p;

Thank you!

the result of (a,b,c) is c. If that helps

pm100
– pm100

2016-12-06 00:21:16 +00:00
Commented Dec 6, 2016 at 0:21 — pm100
– pm100, Commented Dec 6, 2016 at 0:21

rodrigo · Accepted Answer · 2016-12-06 00:30:11Z

2

Let's go line by line:

char str[] {"TESTING"};

This line defines a variable named str of type array of 8 chars, and initializes it with the characters TESTING plus a NUL char to mark the end.

char *p {str};

This one defines a variable named p of type pointer to char and initializes it to the address of the first char of the array str (the first T). This happens because the array automatically decays into a pointer in most uses.

cout << (p++, *++p);

This line does several things. The , operator first evaluates the left-hand operator p++, that increments the pointer, now points to the E; then it evaluates the right-hand operator *++p, but that is a pre-increment operator so it increments the pointer again (it points to S). Finally the * operator accesses to the memory pointed to by p, the result is a S. And that character is printed into STDOUT.

cout << *p++;

This one is easy. The * operator accesses the char pointed to by p (the S again) and prints it in STDOUT. Then it increments the pointer, because it is a post-increment operator. Now it points to the second T.

cout << p;

And at least, this line prints the string pointed to by p until it finds a NUL character. Since p is pointing to the second T of your array it will print TING.

Putting all those outputs together you get SSTING.

answered Dec 6, 2016 at 0:30

rodrigo

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7 Comments

user5372775 Over a year ago

Thanks a lot for your detailed answer. Could you explain again the , operator please? Why it printed S between the two (E, S)

rodrigo Over a year ago

The , operator is actually simple. The expression E, S is evaluated as follows: first evaluate E and discard the result, then evaluate S and use the result as the value of the full expression. Usually the E is there for the side-effects, not for its actual value.

user5372775 Over a year ago

Evaluate means in this case... ? So basically it just prints out the second part of ,?

rodrigo Over a year ago

@user5372775: Not exactly. Evaluate means: compute the value and run the side effects. The ++ operator (both prefix and postfix) has a side effect, that is the increment of the pointer. If you do x = (3, 4); that is silly and the 3 does nothing. But if you do x = (p++, *p); then p is first incremented and then it is dereferenced. It would be just like: p++; x = *p;.

user5372775 Over a year ago

Oh and I wonder if I could ask you about these small stuff in the site's chat (if it has) because it's not worth it creating new threads

|

Yan Zhou · Accepted Answer · 2016-12-06 00:35:44Z

1

Not exactly an answer, but a breakdown of what was the code doing,

#include <iostream>

using namespace std;

int main()
{
    char str[]{"TESTING"};
    char *p{str}; // p points to: 'T'
    p++;          // p points to: 'E'
    ++p;          // p points to: 'S'
    cout << *p;   // output a single char: 'S'
    cout << *p;   // ouptut a single char: 'S'
    p++;          // p points to: 'T'
    cout << p;    // output a (char *) type pointer, AKA a C-string, "TING";

    return 0;
}

answered Dec 6, 2016 at 0:35

Yan Zhou

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