**Code A returns the correct conversion: 6.55957.**

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

    float convert(float currencyA)
    {
    float currencyB = 0;
    currencyB = 6.55957 * currencyA;
    return currencyB;
    }

    int main(int argc, const char *argv[])
    {

    float amount = 0;

    printf("How much\n");
    scanf("%f", &amount);


    printf("You get %f in currencyB", convert(amount));

    return 0;
    }

**Code B returns an incorrect conversion: 0.051247.**

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

double convert(double currencyA)
{
double currencyB = 0;
currencyB = 6.55957 * currencyA;
return currencyB;
}

int main(int argc, const char *argv[])
{

double amount = 0;

printf("How much\n");
scanf("%f", &amount);


printf("You get %f in currencyB", convert(amount));

return 0;
}

If I remove printf and scanf, and assign 1 as the value to the "amount" variable the result is correct.

I suspect scanf is causing the error. If so, why would that be?

Thank you for reading and feel free to ask for any additional info you require.

有帮助吗?

解决方案

The correct format specifier for double is %lf, not %f.

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