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12345678910111213141516171819#include <stdio.h> void calc(double *result, double a, double b) { *result = (a * a) + (2 * a * b) + (b * b); } int main() { double res; calc(&res, 3.0, 4.0); printf("%.2lf", res); return 0; }

Question

12345678910111213141516171819#include <stdio.h> void calc(double *result, double a, double b) { *result = (a * a) + (2 * a * b) + (b * b); } int main() { double res; calc(&res, 3.0, 4.0); printf("%.2lf", res); return 0; }

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Solution

The code you've posted is written in C language. It's a simple program that calculates the result of the equation (a^2) + (2ab) + (b^2) where a and b are inputs to the function calc().

Here's a step-by-step explanation:

  1. The function calc() is declared, which takes three parameters: a pointer to a double result, and two double values a and b.

  2. Inside the calc() function, the equation (a^2) + (2ab) + (b^2) is calculated and the result is stored in the memory location pointed to by result.

  3. In the main() function, a double variable res is declared.

  4. The calc() function is called with the address of res (&res), and two double values 3.0 and 4.0.

  5. The result of the calculation is stored in res.

  6. The value of res is printed to the console with a precision of 2 decimal places using printf().

  7. The program returns 0 and ends.

This problem has been solved

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