C modf() function
C modf() function - Floating-point remainder value function
Syntax:
double modf(double x, double *iptr)
The modf() function is used to break down the floating-point value x into fractional and integral parts.
Parameters:
Name | Description | Required /Optional |
---|---|---|
x | Floating-point values. | Required |
iptr | Floating-point values. | Required |
Return value from modf()
- Returns the signed fractional portion of x.
Example: modf() function
The following example shows the usage of modf() function.
#include <math.h>
#include <stdio.h>
int main(void)
{
double x, y, d;
x = -14.876;
y = modf(x, &d);
printf("x = %lf\n", x);
printf("Integral part = %lf\n", d);
printf("Fractional part = %lf\n", y);
x = 7654.887;
y = modf(x, &d);
printf("\nx = %lf\n", x);
printf("Integral part = %lf\n", d);
printf("Fractional part = %lf\n", y);
}
Output:
x = -14.876000 Integral part = -14.000000 Fractional part = -0.876000 x = 7654.887000 Integral part = 7654.000000 Fractional part = 0.887000
C Programming Code Editor:
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C Programming: Tips of the Day
What's the point of const pointers?
const is a tool which you should use in pursuit of a very important C++ concept:
Find bugs at compile-time, rather than run-time, by getting the compiler to enforce what you mean.
Even though it does not change the functionality, adding const generates a compiler error when you're doing things you didn't mean to do. Imagine the following typo:
void foo(int* ptr) { ptr = 0;// oops, I meant *ptr = 0 }
If you use int* const, this would generate a compiler error because you're changing the value to ptr. Adding restrictions via syntax is a good thing in general. Just don't take it too far -- the example you gave is a case where most people don't bother using const.
Ref : https://bit.ly/33Cdn3Q
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