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POCan I increase int i by more than one using the i++ syntax?
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copied!<p>int fkt(int &i) { return i++; }</p> <pre><code>int main() { int i = 5; printf("%d ", fkt(i)); printf("%d ", fkt(i)); printf("%d ", fkt(i)); } </code></pre> <p>prints '5 6 7 '. Say I want to print '5 7 9 ' like this, is it possible to do it in a similar way without a temporary variable in fkt()? (A temporary variable would marginally decrease efficiency, right?) I.e., something like </p> <pre><code>return i+=2 </code></pre> <p>or </p> <pre><code>return i, i+=2; </code></pre> <p>which both first increases i and then return it, what is not what I need.</p> <p>Thanks</p> <p>EDIT: The main reason, I do it in a function and not outside is because fkt will be a function pointer. The original function will do something else with i. I just feel that using {int temp = i; i+=2; return temp;} does not seem as nice as {return i++;}. </p> <p>I don't care about printf, this is just for illustration of the use of the result.</p> <p>EDIT2: Wow, that appears to be more of a chat here than a traditional board :) Thanks for all the answers. My fkt is actually this. Depending on some condition, I will define get_it as either get_it_1, get_it_2, or get_it_4:</p> <pre><code>unsigned int (*get_it)(char*&); unsigned int get_it_1(char* &s) {return *((unsigned char*) s++);} unsigned int get_it_2(char* &s) {unsigned int tmp = *((unsigned short int*) s); s += 2; return tmp;} unsigned int get_it_4(char* &s) {unsigned int tmp = *((unsigned int*) s); s += 4; return tmp;} </code></pre> <p>For get_it_1, it is so simple... I'll try to give more background in the future...</p>

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