Swap two numbers using pointers
Write a C program to Swap two numbers using pointers. Here’s simple program to Swap two numbers using pointers in C Programming Language.
What are Pointers?
A pointer is a variable whose value is the address of another variable, i.e., direct address of the memory location. Like any variable or constant, you must declare a pointer before using it to store any variable address.
The general form of a pointer variable declaration is −
- type *var-name;
Here, type is the pointer’s base type; it must be a valid C data type and var-name is the name of the pointer variable.
The asterisk * used to declare a pointer is the same asterisk used for multiplication. However, in this statement the asterisk is being used to designate a variable as a pointer.
The unary or monadic operator & gives the “address of a variable’”.
The indirection or dereference operator * gives the “contents of an object pointed to by a pointer”.
Below is the source code for C program to Swap two numbers using pointers which is successfully compiled and run on Windows System to produce desired output as shown below :
SOURCE CODE : :
/* C program to Swap two numbers using pointers */ #include<stdio.h> #include<conio.h> int main() { int num1,num2; int *a,*b,*temp; printf("Enter 1st number :: "); scanf("%d",&num1); printf("\nEnter 2nd number :: "); scanf("%d",&num2); a=&num1; b=&num2; printf("\nBefore Swapping ::\n\n\ta = %d\tb = %d\n",*a,*b); temp=a; a=b; b=temp; printf("\nAfter Swapping ::\n\n\ta = %d\tb = %d\n",*a,*b); return 0; }
Output : :
/* C program to Swap two numbers using pointers */ Enter 1st number :: 5 Enter 2nd number :: 9 Before Swapping :: a = 5 b = 9 After Swapping :: a = 9 b = 5 Process returned 0
Above is the source code for C program to Swap two numbers using pointers which is successfully compiled and run on Windows System.The Output of the program is shown above .
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The above program is wrong.
Correct logic is: