c pass int by reference

Passing an Integer by Reference in C

In C, it is not directly possible to pass an integer by reference as in some other programming languages. However, this can be achieved by passing a pointer to the integer. Here are the steps to achieve this:

  1. Declare a Function that Takes a Pointer as an Argument:
  2. Declare a function that takes a pointer to an integer as an argument. This pointer will allow the function to modify the original integer.

Example:void modifyInt(int *ptr) { *ptr = 100; // Modifying the value at the address pointed to by ptr }

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  1. Pass the Address of the Integer:
  2. When calling the function, pass the address of the integer using the address-of operator (&).

Example:int num = 50; modifyInt(&num); // Passing the address of 'num' to the function

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  1. Accessing and Modifying the Integer:
  2. Inside the function, use the dereference operator (*) to access and modify the value at the address pointed to by the pointer.

Example:void modifyInt(int *ptr) { *ptr = 100; // Modifying the value at the address pointed to by ptr }

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By following these steps, it is possible to achieve pass-by-reference behavior for integers in C by using pointers.