What's the case of float division returning inf and finite value?

Question

I have written this snippet of code and I cannot understand why this float division returns inf when args[2] = 200 and 5.0 when args[2] = 2000. Is this caused because I am exceeding some decimal position boundary?

#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
int main(int nargs, char **args) {

    float t = atoi(args[1]);   
    float dt = atoi(args[2])/1000;

    float nsamples = (t/dt);
    printf("%f\n", nsamples);
    return(0);
}

let's say 10 and 200 or 2000 – Giannis May 08 '19 at 13:30 — Giannis, May 08 '19 at 13:30

score 1 · Accepted Answer · answered May 08 '19 at 13:27

1

You're just getting tripped up by integer division (200 / 1000 = 0).

Change float dt = atoi(args[2])/1000; -> float dt = atoi(args[2])/1000.0f;

answered May 08 '19 at 13:27

Paul R

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2

silly mistake. thanks a lot. it worked. – Giannis May 08 '19 at 13:28
do I need that f at the end of the line? if so, what for? – Giannis May 08 '19 at 13:29
1

`1000.0` is a double, `1000.0f` is a float. See https://en.cppreference.com/w/cpp/language/floating_literal – flau May 08 '19 at 13:30
1

@GiannisGonidelis 1000.0 is also fine – Aykhan Hagverdili May 08 '19 at 13:31
3

@Ayxan: `1000.0f` and `1000.0` are equal, but this is not true for floating-point constants in general. Programmers should be aware of the difference in types and values that result—please do not recommend that people use casually `double` constants with `float` operations. – Eric Postpischil May 08 '19 at 13:33
1

@EricPostpischil your point is valid, but programmers should generally prefer `double` over `float` anyway. – Aykhan Hagverdili May 08 '19 at 13:39

What's the case of float division returning inf and finite value?

1 Answers1