you could use a variable to increment your counter

for(int counter = 0, increment = 0; counter < 100; increment++, counter += increment){
   ...do_something...
}

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!