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Home/ Questions/Q 7794511
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Editorial Team
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Editorial Team
Asked: June 1, 20262026-06-01T22:47:51+00:00 2026-06-01T22:47:51+00:00

I have a function that involves floats and ints. When I run the code,

  • 0

I have a function that involves floats and ints. When I run the code, the result comes out wrong. When I enter 4.3 for the input_voltage, I get 3803 for the decimal and EDB for HEX. The correct answer should be 3808 and EE0. Can anyone tell me what might be going on?

#include <stdio.h>

int digital_encoding(float voltage);

int main()
{
    float input_voltage;
    int valid, ch;

    do{
        printf("Please enter the input voltage between 0 and 5 volts: \n");
        valid = scanf("%f", &input_voltage);
        if(input_voltage <= 0){
            printf("Enter a number larger than 0! \n");
            valid = 0;
        }
        if(input_voltage >= 5){
            printf("Enter a number less than 5! \n");
            valid = 0;
        }
    }while(valid != 1);

    digital_encoding(input_voltage);
}
int digital_encoding(float voltage)
{
    int dig_encode;
    dig_encode = ((voltage + 5)*(4095/10));
    printf("The digital encoding equals %d in decimal, and %X in HEX.\n", dig_encode, dig_encode);
    return dig_encode;
}
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-01T22:47:54+00:00Added an answer on June 1, 2026 at 10:47 pm

    The problem lies here :

    int digital_encoding(float voltage)
    {
        **int** dig_encode;
        dig_encode = ((voltage + 5)*(4095/10));
        printf("The digital encoding equals %d in decimal, and %X in HEX.\n", dig_encode,    dig_encode);
        return dig_encode;
    }
    

    Use this instead

    int digital_encoding(float voltage)
    {
        float dig_encode;
        dig_encode = ((voltage + 5)*(4095.00/10));
        printf("The digital encoding equals %d in decimal, and %X in HEX.\n", (int)dig_encode, (int)dig_encode);
        return (int)dig_encode;
    }
    
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