CODEWITHSUNDEEP
cprintf
BuddhaPowers
BuddhaPowers

Reputation: 9

printf() in C: How do I keep my floats/double from rounding?

Been trying to get the outputs to give me the number without rounding, not sure what I'm missing

float NumberOfTrees1;

for (NumberOfTrees1 = 1000.0; NumberOfTrees1 >= 50; NumberOfYears++)
{
    NumberOfTrees1 = NumberOfTrees1 - (NumberOfTrees1 * 0.13);
    printf("Number of Remaining Trees: %0.1f\n",NumberOfTrees1);
}

My Output:

Number of Remaining Trees: 61.7
Number of Remaining Trees: 53.7
Number of Remaining Trees: 46.7

Required Output:

Number of Remaining Trees: 61
Number of Remaining Trees: 53
Number of Remaining Trees: 46

I understand that the %0.1f is what gives me the .7 but when I use %0.0f it rounds up my numbers which I don't want. Been switching things around from using int, double, long float etc to no avail.

Any help would be great and thank you in advance!!!

Upvotes: 0

Views: 848

Answers (2)

Antonin GAVREL
Antonin GAVREL

Reputation: 11259

You can cast into int in order to truncate your number:

int main(void) {
    double d =  53.7;
    float f = 46.7f;

    int a = (int)d;
    int b = (int)f;

    printf("%d %d\n", a, b);
}

output:

53 46

Or, as suggested by @user3386109, you may use floor(), but you will need to also include <math.h> header and compile with -lm

Upvotes: 0

Andy Sukowski-Bang
Andy Sukowski-Bang

Reputation: 1420

Antonin GAVRELs answer should print the desired output, but if you don't want to cast the floating point numbers into integers, you could use the double floor (double x) function from <math.h>.

#include <math.h>
#include <stdio.h>

int main(void)
{
    double x =  53.7;
    double floored = floor(x);
    printf("⌊%0.1lf⌋ = %0.1lf\n", x, floored);

    return 0;
}

The program can be compiled with -lm and prints the following when executed:

⌊53.7⌋ = 53.0

Upvotes: 1

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