why wont my printf float display? (beginner)

Question

this is the code i wrote for my school program my question is why wont the printf show the float value it comes up as inf ( which i believe is infinity?) i know my values previously are all int's which if why i put float in () anyways help is appreciated

cheerzx

#include <stdio.h>
#include <conio.h>

int main()
{
    int c1, c2, c3, parallelCap;
    float seriesCap;

    printf("please enter the value of the first resistor:");
    scanf(" %d", &c1);
    printf("please enter the value of the second rstr:");
    scanf(" %d", &c2);
    printf("please enter the value of the third resistor:");
    scanf(" %d", &c3);
    parallelCap = c1 + c2 + c3;
    seriesCap = (float)1/(1/c1 + 1/c2 + 1/c3);
    printf("the parallel capacitance is: %d the series 
    capacitance is: %f/n ", parallelCap, seriesCap);
getch();
return 0;
}

All the divisions in `1/c1 + 1/c2 + 1/c3` are done using integer arithmetic, so they're all `0`. — Barmar, Feb 02 '19 at 02:41

score 3 · Accepted Answer · answered Feb 02 '19 at 02:50

3

You are getting inf because each of 1/c1 + 1/c2 + 1/c3 evaluates to 0 when c1, c2 and c2 are greater than 1 and so you end up dividing 1 by 0. This is because it works as integer division. You can either cast each one of them separately to float or use 1.0 as numerator like this

seriesCap = 1.0/(1.0/c1 + 1.0/c2 + 1.0/c3);

answered Feb 02 '19 at 02:50

xashru

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accepting the answer would suffice <3 – xashru Feb 02 '19 at 03:14
check mark ? sorry im new – babySteps Feb 02 '19 at 03:54
Don't sweat it! It's perfectible fine :) – xashru Feb 02 '19 at 03:57

why wont my printf float display? (beginner)

1 Answers1