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Reference parameters can indeed modify the value in the calling function because they pass the address of the variable, allowing direct modification of the original variable.
Whether reference parameters can modify the value in the calling function
Introduction
In programming, there are two main ways to pass parameters: pass by value and pass by reference. Reference parameters refer to passing a pointer to the address of a variable, allowing the passed parameters to be modified from outside the function. This article will explore whether reference parameters can modify the value in the calling function.
Pass by value vs. Pass by reference
Practical case
The following is a C program that demonstrates how passing by reference modifies the value in the calling function:
#include <iostream> using namespace std; void swap(int& a, int& b) { int temp = a; a = b; b = temp; } int main() { int x = 5; int y = 10; cout << "Before swap: x = " << x << ", y = " << y << endl; swap(x, y); cout << "After swap: x = " << x << ", y = " << y << endl; return 0; }
Output:
Before swap: x = 5, y = 10 After swap: x = 10, y = 5
In this example, the swap()
function receives arguments by reference and swaps their addresses within the function. Therefore, for the call to function main()
, the values of the original variables x
and y
are modified.
Conclusion
Reference parameters allow the passed parameters to be modified from outside the function, but the original variables are only affected when passed by reference. Any modification of parameters passed by value will only affect the copy inside the function. Understanding the difference between passing by value and passing by reference is crucial to correctly understanding the behavior of function parameters.
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