Small C Question

finnpark

Hi Forum,

Small question for someone into c programming. Some code:

for( count = (N - 2 * ofs); count > 0L; count-- )
        facc += *(p++);

My question relates to the line:

facc += *(p++);

p points at a an array called 'D'.

Is the above line equal to:

facc=facc+D[p];
p=p+1;

Is that the basic function of the line. The p++ is confusing me. Does it not increment straight away? Or does it increment after facc is assigned it's value?

Many thanks for any help.

kenmc

Your intrepretation is correct. It does not increment "p" straight away as it's a "post incrementor", same as the "count--" in the line above it. if you wanted to increment p before hand you would use a "pre-incrementor" i.e. ++p;
This sort of stuff is incredibly common in C, especially using pointers (arrays are pointers too don't forget). See how much more elegent the
"facc += *(p++);"
is relative to the
"facc = facc + D[p];
p=p+1;"

edit: In fact, if you were to replace the facc+=*(p++); line with yours, you would in fact break the code, unless you surrounded your code with braces "{....}" as the for loop would only increment the first line (facc = facc+D[p]).
Ken

Syth

kenmc wrote:

"facc += *(p++);"
is relative to the
"facc = facc + D[p];
p=p+1;"

Kenmc is correct about the p++ vs ++p, however the above is incorrect.

p points to the i-th element of D, thus p = *D + i, ie the start of D plus the index. So *p is not equal to D[p].

kenmc

Doh yeah how the hell did I miss that??? I knew what he was trying to say though so i read between the lines!
Anyway - point proven - facc += *(p++); is nicer.
so*(p+i) = D; or since initially i = 0, *p = D[0], then *p++ = D[1] the first time, D[2] the second time etc.

finnpark

Cheers:) .

Thanks again Ken for info and thanks Syth for pointing out slight flaw:eek: .;)

Am getting there now:cool: .

Thanks Again.