Change a call-by-value parameter does not affect the argument : function parameters « Function « C++ Tutorial

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C++ Tutorial » Function » function parameters 
7.3.11.Change a call-by-value parameter does not affect the argument
#include <iostream> 
using namespace std; 
 
double f(double x)
 
int main() 

  double t = 10.0
 
  cout << "1/10.0 is " << f(t<< "\n"
 
  cout << "Value of t is still: " << t << "\n"
 
  return 0

 
double f(double x

  x = / x;
  cout << "inside f " << x << "\n"
  return x; 
}
inside f 0.1
1/10.0 is 0.1
Value of t is still: 10
7.3.function parameters
7.3.1.Passing int by value
7.3.2.Define function to accept three int parameters
7.3.3.Pass a pointer to a function.
7.3.4.Pass variable address to a function
7.3.5.Pass int array to a function
7.3.6.Declare int array parameter for a function without indicating the array length
7.3.7.Function parameter: Use int pointer to accept an array
7.3.8.Use array as function's parameter
7.3.9.Change the contents of an array using a function
7.3.10.Pass a string to a function: Invert the case of the letters within a string
7.3.11.Change a call-by-value parameter does not affect the argument
7.3.12.Demonstrate the pointer version of swap(): Exchange the values of the variables pointed to by x and y
7.3.13.Using reference parameters
7.3.14.Passing a two-dimensional array to a function
7.3.15.Passing an array to a function
7.3.16.var args has to be the last one
7.3.17.the use of ... and its support macros va_arg, va_start, and va_end
7.3.18.Handling an array parameter as a pointer
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