How do I put two increment statements in a C++ 'for' loop?

Question

I would like to increment two variables in a for-loop condition instead of one.

So something like:

for (int i = 0; i != 5; ++i and ++j)          


        

Answer 1

for (int i = 0; i != 5; ++i, ++j) 
    do_something(i, j);

Answer 2

Use Maths. If the two operations mathematically depend on the loop iteration, why not do the math?

int i, j;//That have some meaningful values in them?
for( int counter = 0; counter < count_max; ++counter )
    do_something (counter+i, counter+j);

Or, more specifically referring to the OP's example:

for(int i = 0; i != 5; ++i)
    do_something(i, j+i);

Especially if you're passing into a function by value, then you should get something that does exactly what you want.

Answer 3

Try not to do it!

From http://www.research.att.com/~bs/JSF-AV-rules.pdf:

AV Rule 199
The increment expression in a for loop will perform no action other than to change a single loop parameter to the next value for the loop.

Rationale: Readability.

Answer 4

I agree with squelart. Incrementing two variables is bug prone, especially if you only test for one of them.

This is the readable way to do this:

int j = 0;
for(int i = 0; i < 5; ++i) {
    do_something(i, j);
    ++j;
}

For loops are meant for cases where your loop runs on one increasing/decreasing variable. For any other variable, change it in the loop.

If you need j to be tied to i, why not leave the original variable as is and add i?

for(int i = 0; i < 5; ++i) {
    do_something(i,a+i);
}

If your logic is more complex (for example, you need to actually monitor more than one variable), I'd use a while loop.

Answer 5

int main(){
    int i=0;
    int a=0;
    for(i;i<5;i++,a++){
        printf("%d %d\n",a,i);
    } 
}

Answer 6

A common idiom is to use the comma operator which evaluates both operands, and returns the second operand. Thus:

for(int i = 0; i != 5; ++i,++j) 
    do_something(i,j);

But is it really a comma operator?

Now having wrote that, a commenter suggested it was actually some special syntactic sugar in the for statement, and not a comma operator at all. I checked that in GCC as follows:

int i=0;
int a=5;
int x=0;

for(i; i<5; x=i++,a++){
    printf("i=%d a=%d x=%d\n",i,a,x);
}

I was expecting x to pick up the original value of a, so it should have displayed 5,6,7.. for x. What I got was this

i=0 a=5 x=0
i=1 a=6 x=0
i=2 a=7 x=1
i=3 a=8 x=2
i=4 a=9 x=3

However, if I bracketed the expression to force the parser into really seeing a comma operator, I get this

int main(){
    int i=0;
    int a=5;
    int x=0;

    for(i=0; i<5; x=(i++,a++)){
        printf("i=%d a=%d x=%d\n",i,a,x);
    }
}

i=0 a=5 x=0
i=1 a=6 x=5
i=2 a=7 x=6
i=3 a=8 x=7
i=4 a=9 x=8

Initially I thought that this showed it wasn't behaving as a comma operator at all, but as it turns out, this is simply a precedence issue - the comma operator has the lowest possible precedence, so the expression x=i++,a++ is effectively parsed as (x=i++),a++

Thanks for all the comments, it was an interesting learning experience, and I've been using C for many years!