increment operator at left hand side of equation.

[Nishu]

Hi all,

Is this valid in as per C standard ? --> *(pt++) = temp;

int *ptr;
int temp = 0;
*(pt++) = temp;

Would there be any confusion to compiler about which operation should
it do first? increment or assignment ?

Thanks,
Nishu

 

[Nishu]

*(pt++) = temp;

Can I treat the above statement as invalid in accordance to below
faq ?

http://c-faq.com/expr/evalorder2.html

Thanks,
Nishu

 

[Nishu]

The result of pt++ will always be the value of pt before the increment, and the
value of the subexpression *pt++ will be computed before the assignment. However
the sequencing of the assignment of pt+1 to pt relative to assignment of temp to
*pt is undefined; since the value of the subexpression (temp) doesn't depend on
the value of (pt), and vice versa, this is a safe assignment.

Thank you. I understood you point. I also read http://c-faq.com/expr/confused.html.
Nishu

 

[Peter Nilsson]

Is this valid in as per C standard ? -->  *(pt++) = temp;

int *ptr;
int temp = 0;
*(pt++) = temp;

No, for two reasons. Firstly, pt is not declared or defined
anywhere. If you meant ptr, then still no since ptr has no
initial value. You're incrementing an undetermined value.

Would there be any confusion to compiler about which
operation should it do first? increment or assignment ?

The general case you're hinting at is fine, although the
following would be problematic...

int temp = 0;
int *ptr = &temp;
*(ptr++) = temp++;

....because temp is modified twice between sequence points.