Why doesn't this program print '4'?

Question

Shouldn't ptrj value be 4 after the execution of *ptrj++?

int j=3,*ptrj = NULL;
ptrj = &j;
*ptrj++;
printf("%i",*ptrj);

it prints 0, it also says that for *ptrj++ the result is unused — , Jun 10 '19 at 08:19
And this is why `*ptr++` is bad style, no matter how common. People keep getting confused over this, the story of C for the past 40 years. Good programmers write `ptr++` on a line of its own, bad programmers write "idiomatic" goo like `*dst++ = *src++`. — Lundin, Jun 10 '19 at 08:25
I don't understand the close votes here. This is perfectly adequate to reproduce the problem, and it's NOT a simple typographical error. — klutt, Jun 10 '19 at 08:26

klutt · Accepted Answer · 2020-02-09T12:30:13.533

*ptrj++ is the same as *(ptrj++). What you expect is instead (*ptrj)++. You should look up operator precedence to learn more about which operators that acts before others. To understand what ptrj++ does, you should read about pointer arithmetic. But here is a quick explanation:

*(ptrj++) returns the value that ptrj points to (3), and THEN increments ptrj to point to the next value.
(*ptrj)++ returns the value that ptrj points to (3), and THEN increments the value that ptrj points at from 3 to 4.

This means that what you're printing is the value at address &j + 1, which is the value that lies right after the variable j in memory. and this is undefined behavior. As Sourav pointed out, you would get a warning that points you to this if you enable compiler warnings.

The only difference between *ptrj++ and ptrj++ is what it is returning. And since you don't use the return value, your code is equivalent to:

int j=3,*ptrj = NULL;
ptrj = &j;
ptrj++;
printf("%i",*ptrj);

OK, you were mentioning about the printf argument, I see that now. — Sourav Ghosh, Jun 10 '19 at 08:29

Sourav Ghosh · Answer 2 · 2019-06-10T08:28:43.813

If you compile the program with warnings enabled, you'll see

source_file.c:9:5: warning: value computed is not used [-Wunused-value]
     *ptrj++;
     ^

That means, the value computation is useless.

In other words, according to the operator precedence *ptrj++; is same as *(ptrj++);, and as per the post-increment operator property, the value of the operation is the value of the operand, and the value is increased as the side-effect.

Quoting C11, chapter

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). [....]

So, this is same as

 *ptr;
  ptr++;

If you want to increment that value at the address, you need to enforce the operator precedence by using explicit parenthesis, like

(*ptrj)++;   // first get the value, then update the value.

score 0 · Answer 3 · answered Jun 10 '19 at 08:22

0

*ptrj++ is equivalent to *(ptrj++).

desired output can be achieved using (*ptrj)++.

Please refer https://www.geeksforgeeks.org/c-operator-precedence-associativity/ to get how operator works.

answered Jun 10 '19 at 08:22

Shivam Seth

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score 0 · Answer 4 · answered Jun 10 '19 at 08:35

Precedence of postfix ++ is higher than *. The expression *ptrj++ is treated as *(ptrj++) as the precedence of postfix ++ is higher than *.If you want to print 4 ( ie ptrj+1 ),You should use the following code:-

int j=3,*ptrj = NULL;
ptrj = &j;
(*ptrj)++;
printf("%i",*ptrj);
return 0;

To know more about operator precedence, refer follow link: https://en.cppreference.com/w/c/language/operator_precedence

Why doesn't this program print '4'?

4 Answers4

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