why the (if) condition is not satisfied, but the function is working?

Question

The code is below. When the variate i*dx is equal to 140, this would not meet the condition, but why cout works? if I changed the 1.4 to 1.3 or other numbers, this weird thing does not happen.

#include <iostream>
#include <cmath>
using namespace std;

int main()
{   
    double dx = 0.01;
    float ip = 1.4;
    for(size_t i = 0; i < 200; ++i)
    {
        if(  (i*dx > ip)   &&   (abs(i*dx - ip) > 0)    &&   (i*dx < 1.6)  )
        {
            cout << "i = " << i << endl;
            cout << "i*dx = " << i*dx << endl;
            printf("printf. i*dx = %f \n", i*dx);   
            printf("printf. ip = %0.10f \n", ip);   
            break;
        }
    }
    return 0;
}

now, problem solved

To explain the duplicate: float 1.4 is probably ever so slightly larger than 1.4. — Andras Deak -- Слава Україні, Oct 18 '20 at 12:43
Try rewriting `(i*dx > 1.4)` to `(i*dx > 1.4 + 0.0000001)`. I.e. add small epsilon value. — Arty, Oct 18 '20 at 12:46
The linked duplicate explains why it happens. In your case it's not 0.1 but 1.4, but neither can be expressed as a finite binary fraction. You also have to consider what type a literal 1.4 is in C (either float or double, I don't know). In either case you have the same situation as the linked question. Expecting floating-point values to have exact values leads to surprises. You have to account for some tolerance in your code. — Andras Deak -- Слава Україні, Oct 18 '20 at 13:02
https://www.h-schmidt.net/FloatConverter/IEEE754.html shows that the closest representable value to 1.4 is 11740551*(2^-23) = 1.39999997615814208984375 — Raymond Chen, Oct 18 '20 at 13:08
Right, I messed up my directions. If `> 1.4` is unexpectedly tripped it means that the float 1.4 is _smaller_ than 1.4. That, and `i*dx` might be larger than 1.4 (I'm not any good at float arithmetic). — Andras Deak -- Слава Україні, Oct 18 '20 at 13:35

why the (if) condition is not satisfied, but the function is working?

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