#include<conio.h> main() { int a,b; scanf("%d%d",&a,&b); printf("Before swapping A= %d B=%d",a,b); swap(a,b); getch(); } void swap(int a,int b) { int c; c=a; a=b; b=c; printf("After swapping A= %d B=%d",a,b); }
You have to pass the address of the variables.void swap (int *pa, int *pb){...}
The only way to swap two values using call by value semantics is to pass pointer variables by value. A pointer is a variable that stores an address. Passing a pointer by value copies the address, the value of the pointer, not the pointer itself. By passing the addresses of the two values to be swapped, you are effectively passing those values by reference. Both C and C++ use pass by value semantics by default, however C++ also has a reference data type to support native pass by reference semantics. By contrast, Java uses pass by reference semantics by default. In C, to swap two variables using pass by value: void swap (int* p, int* q) { int t = *p; *p = *q; *q = t; } In C++, to swap two variables using pass by reference: void swap (int& p, int& q) { std::swap (p, q); } Note that C++ is more efficient because std::swap uses move semantics; there is no temporary variable required to move variables. With copy semantics, a temporary is required. However, with primitive data types, there is a way to swap values without using a temporary, using a chain of exclusive-or assignments: void swap (int* p, int* q) { *p^=*q^=*p^=*q; }
You cannot swap values via pass by value since the values would be copies of the original values, therefore only the copies would be swapped. You have to use pass by value, either by reference or by pointer. References are easier to implement: void swap(int& x,int& y){x^=y^=x^=y;} Using pointers: void swap(int* const x,int* const y){ assert(x && y); // null pointers not permitted! (*x)^=(*y)^=(*x)^=(*y);}
You cant you have to use php or javascript.
Arrays are reference type. array values are always pass by reference.
You can write a C++ fib pro using arrays but the problem is the prog becomes very complicated since u need to pass the next adding value in an array.....
exactly: leave this function, and (optionally) pass the following value to the caller.
it is used to check or verify how much current pass through the circuit using voltage source.
A palindrome is any value that reads the same backwards as it does forwards. The value may be a string of characters, a multi-digit numerical value, or an array. To determine if an object is a palindromic, point to the first and last elements of the object and work towards the middle element(s). If the values at each point are the same and the pointers meet or pass each other, the object is palindromic. If the values are not the same at any point, the object is non-palindromic. If the object is a numeric value, you must determine the individual digits for the comparison. If the object is a string, skip white-space values.
The most important use of pointers comes when we pass value by reference to any function. You do not need to create a second memory location as in pass by value. You can mofify the original variable by using its address.
In principle, values are always passed by reference (a copy of the value provided is copied onto the stack, the receiving method accesses this).In the case of objects, this value is the memory address of an object. In other words, the object itself is not copied - only a copy of the address is made. Therefore, the receiving method works with the SAME object, and any changes will be reflected back to the calling program. The result is that, for all practical purposes, this works as if the object is passed by reference.
#include<iostream> void swap(int* x, int* x){ (*x)^=(*y)^=(*x)^=(*y); } int main() { int a, b; a=10; b=20; std::cout<<"Before swap: a="<<a<<", b="<<b<<std::endl; swap(&a,&b); std::cout<<"After swap: a="<<a<<", b="<<b<<std::endl; return(0); }