How to divide float by integer when both are variables?

问题

I am writing a program in C. I have two variables one of which is integer and the other one is float. I want to divide the float by the integer and want to get result in float. I am doing the following:

int a;
float b=0,c=0;
scanf("%d",&a);

Here I am doing some computations on b so its value is increased randomly i.e 33 etc.

c = b/(float)a;
printf("c = %2.f\n", c);

The problem is I am getting 'c' as a rounded number (integer) rather than a float value. How can I get 'c' as a float value.

回答1:

Your problem is here:

printf("c = %d\n", c);

%d formats c as integer. Use %f instead.

Or std::cout << "c = " << c << std::endl if you prefer.

回答2:

For those here coming from a Google search because of the question's name:

The result of dividing a float by an integer is a float, this is clean and safe. Example:

#include <iostream>

int main()
{
    float y = 5.0f;
    int x = 4;
    std::cout << y/x << std::endl;
}

The output is as expected: 1.25

回答3:

When printing on a screen try this (this worked for me):

printf("c = %f\n", c);

来源：https://stackoverflow.com/questions/21307390/how-to-divide-float-by-integer-when-both-are-variables