Below is a simple program.

int main()
{
    int x, b = 3; //size_t x; size_t b = 3;
    
    char const *c = "moving_left"; // "moving_right"
    int k, border;
    
    if (!strcmp(c,"moving_left"))
    {
        k = 1;     // moving left
        border = 0;
        x = 5;
        
    }
    else
    {
        k=-1;     // moving right
        border = 7;
        x = 1;
    }
    

    while(true)
    {
        if (k*x<b*k)
        {
            cerr<<c<<x<<'\n';
            if (x==border){break;}
        }
        x-=k;
    }
    return 0;
}

When I'm moving left, i.e., x is decreasing, I want to print "moving_left" after x falls behind b (x<b).

And when moving right, i.e., x is increasing, I want to print "moving_right" after x crosses b (x>b).

The above program works fine when x & b are signed integers, but might fail when I declared them as unsigned integers (size_t) because of int promotion (when k = -1).

I want to run the above code in a single while loop for unsigned x & b (I like to declared them as unsigned because they are indices of unmentioned arrays). Also, in the if condition (kx<bk), I want left than sign (<).

Multiplying by negative one works only for signed integers. Is there any math trick to make the above code work for unsigned integers with said conditions?

CodePudding user response：

It is not completely clear why you want to use unsigneds and why you want so much to keep the condition (k*x<b*k) rather than something along the line of if ( (moving_left && x<b) || (!moving_left && b<x).

Anyhow if you want to use unsignds and you want to use a single condition for both cases, then you can add a level of indirection:

size_t x;
size_t b = 3;

bool move_left = true;
int border;

if (move_left) {
    x = 5;
    border = 0;
    upper = b;
    lower = x;        
} else {
    x = 1;
    border = 0;
    upper = x;
    lower = b;
}
    
while(true) {
    if (lower < upper) {
        std::cerr<<x<<'\n';
        if (x==border){break;}
    }
    x-=k;
}