Small C Question

finnpark · 14-11-2005 12:34PM #1

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 · 14-11-2005 02:29PM

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 · 14-11-2005 05:39PM

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 · 14-11-2005 06:41PM

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 · 14-11-2005 08:47PM

Cheers:) .

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

Am getting there now:cool: .

Thanks Again.

Small C Question

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