newbie question

[junky_fellow]

i have a short C program as follows:

main()
{

int i=0;
i = ++i;
printf("\ni=%d",i);
}

when i compile the program, i get the following warning.
cc: Warning: test.c, line 6: In this statement, the expression "i=++i"
modifies the variable "i" more than once without an intervening
sequence point.
This behavior is undefined.
i = ++i;
^

I don't what causes the undefined bahaviour. For me the variable i
should
be incremented by 1. Also when i run the executable the value for i
printed
is 1.

 

[Richard Bos]

i = ++i;

<http://www.eskimo.com/~scs/C-faq/q3.1.html>, and the next two.

Sheesh. This must be the single most FAQed fscking FAQ in the FAQ. And
if you'd just taken the trouble to read the last week or so of this
newsgroup, or tried Google Groups, you would've known where to find the
FAQ.

Richard

 

[Miika Karanki]

main()
{

int i=0;
i = ++i;
printf("\ni=%d",i);
}

when i compile the program, i get the following warning.
cc: Warning: test.c, line 6: In this statement, the expression "i=++i"
modifies the variable "i" more than once without an intervening
sequence point.
This behavior is undefined.
i = ++i;
^

int main()
{
int i=0;
++i; /* or i = i+1 or i+=1 (or i++) */
printf("\ni=%d", i); /* or printf("\ni=%d", ++i) without
the previous line */
return 0;
}

++i increments the value of i by one, you shouldn't assign its
result back to i.

++i is not the same as "i+1" which doesn't increment the
variable i, it just returns the result of calculation
"i plus one".