C math library: Which statement correctly obtains the remainder of dividing 3.14 by 2.1 in floating-point arithmetic?

Introduction / Context:Unlike integer division, floating-point division in C/C++ uses library functions to compute the remainder. The operator '%' is defined only for integers, so you must select the proper function when working with doubles or floats. This question checks whether you know the standard library tool for floating-point remainders.

Given Data / Assumptions:

• Operands are non-integers: 3.14 and 2.1 (double by default).
• We want the remainder r such that 3.14 = q * 2.1 + r with r having the same sign as the dividend for fmod.
• Standard C headers: / .

Concept / Approach:

• Operator '%' applies to integers only; using it with floating types is a compile-time error.
• 'fmod(x, y)' returns the floating remainder of x / y following the definition r = x - trunc(x / y) * y.
• 'modf(x, &intpart)' splits a floating-point number into fractional and integral parts; it is not a remainder function.

Step-by-Step Solution:

Verification / Alternative check:

Why Other Options Are Wrong:

• 3.14 % 2.1: Illegal for floating types.
• modf(3.14, 2.1): Wrong signature and purpose; 'modf' requires a pointer for the integer part.
• Remainder cannot be obtained: False; use 'fmod'.
• remainder(3.14, 2.1): Different function and behavior; not the canonical answer here.

Common Pitfalls:

• Confusing 'modf' with modulo operations.
• Assuming '%' works for floats in C/C++ as in some languages; it does not.

Final Answer: