Why doesn't `std::cout.precision()` display trailing zeros in floating-point numbers in C ?

Correct Use of std::cout.precision() to Display Trailing Zeros

When working with floating-point numbers in C , the std::cout.precision() method is crucial for controlling the number of decimal places displayed. However, in certain cases, users encounter unexpected results, such as the absence of trailing zeros.

Consider the following code:

int main() {
  int a = 5;
  int b = 10;
  std::cout.precision(4);
  std::cout << (float)a / (float)b << "\n";
  return 0;
}

This code attempts to print the result of dividing two integers (5 and 10) with a precision of 4 decimal places. However, the output is "0.5" instead of the expected "0.5000".

The reason for this behavior lies in the fact that the numbers 'a' and 'b' are initially stored as integers. When we perform the division, the result is also an integer by default. To display floating-point numbers correctly, we need to explicitly cast them to floats, as seen in the line:

std::cout << (float)a / (float)b << "\n";

However, even after casting to floats, the absence of trailing zeros persists. This is where the std::fixed manipulator comes into play. To ensure that trailing zeros are displayed, we must pass the std::fixed manipulator to std::cout. This manipulator rounds the floating-point value and displays it in fixed-point notation.

The corrected code is:

int main() {
  int a = 5;
  int b = 10;
  std::cout << std::fixed;
  std::cout.precision(4);
  std::cout << (float)a / (float)b << "\n";
  return 0;
}

With the inclusion of std::fixed, the output now correctly displays "0.5000", adhering to the specified precision of 4 decimal places.

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