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+When you pass a value into a C function as an argument it creates a copy of the
+value that is stored outside of the function (the parameter value). This value
+is modified by the function.
+
+Changes to that value within function scope are not reflected in the main scope
+
+- the values that were passed in as arguments..
+
+Say the function returns the value it modifies. The modified version will only
+be accessible from the function, the original main scope value will be the same
+as it was when the value entered the function.
+
+This is why [pointers](./Pointers_in_C.md) are used in cases where you want a
+function to change the value of a variable in main scope.
+
+## Demonstration
+
+```c
+void swap (int a, int b) {
+    int t = a;
+    a = b;
+    b = t;
+    printf("a = %d, b = %d", a, b);
+}
+```
+
+With this form of the function we get:
+
+```c
+int main(){
+    int a = 21;
+    int b = 17;
+    swap(a,b);
+    printf("a = %d, b = %d", a, b);
+}
+
+// a = 17, b = 21
+// a = 21, b = 17
+```
+
+Refactoring the function to use pointers gives us:
+
+```c
+void swap (int *pointer_a, int *pointer_b) {
+    int t = *pointer_a;
+    *pointer_a = *pointer_b;
+    *pointer_b = t;
+}
+```
+
+Now that we are using pointers, we can pass in the main scope variables:
+
+```c
+swap(&a, &b);
+```
+
+Which will cause the main function to return what we expect:
+
+```txt
+a = 17, b = 21
+```
+
+While the non-pointerized version of the function enshrines **call by value**,
+the pointerized version enshrines **call by reference**.