With float my program looks like this:
#include <stdio.h>
int main() {
float result;
float input;
scanf("%f", &input);
result = input / .25;
printf("%f", result);
return 0;
}Entering the value 3 produces 12.000 as output. Replacing float with double, however, consistently gives me 0.000 instead. Is there something I'm overlooking here?
#include <stdio.h>
int main() {
double result;
double input;
scanf("%lf",&input);
result=input/.25;
printf("%lf",result);
return 0;
}Here you must choose the format specifier meant for doubles, namely %lf.
The format string specifications (https://en.wikipedia.org/wiki/Scanf_format_string#Format_string_specifications) explain the available specifiers in more detail.
As explained in this question about scanning doubles (https://stackoverflow.com/questions/33269283/scanf-not-reading-in-double),
Your claim to the compiler was false: for scanning, %f promises a pointer to float. What you actually supplied was a pointer to double.
Either switch to %lf or make the input a float.
Note that printf has different format rules, accepting %f with either float or double arguments. This works because the arguments undergo promotion to doubles instead of being pointers.
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