Representing array using a pointer

pointer-operations_arrays.c

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+/**
+ * @file pointer-operations_arrays.c
+ * @author kunal jain ([email protected])
+ * @brief This programs shows how to represent array using
+ *        pointers
+ * @version 0.1
+ * @date 2022-03-07
+ * 
+ * @copyright Copyright (c) 2022
+ * 
+ */
+
+#include<stdio.h>
+
+void test_illegal_pointer_assignment() {
+    int* p;
+    // Without initializing pointer with a null value,
+    // below assignment - *p = 100, could be dangerous 
+    // as we are trying access and alter the value present
+    // at an address that we know nothing about.
+
+    // A pointer should always be set to null before we assign
+    // value to that pointer.
+    // That prevents any dangerous alterations we might perform
+    // on the memory space.
+    p = NULL;
+
+
+    printf("Address of p -> %d\n", &p);
+    printf("Value of p -> %d\n", p);
+    printf("Value at p -> %d\n", *p);
+
+    printf("Before assignment\n");
+
+    // On some compilers your code may crash here if you
+    // don't assign a NULL value to pointer.
+    *p = 100;
+
+    printf("After assignment\n");
+
+    printf("Address of p -> %d\n", &p);
+    printf("Value of p -> %d\n", p);
+    printf("Value at p -> %d\n", *p);
+}
+
+int main(void) {
+
+    test_illegal_pointer_assignment();
+
+    char a[] = "HelloWorld";
+    char b[] = "HelloCS240";
+
+    // print a[6] and b[5]
+    printf("Value of a[6] -> %c\n", a[6]);
+    printf("Value of b[5] -> %c\n", b[5]);
+
+    // Print values without using array name
+    // TYPE *pointer-name = array-name
+    // TYPE -> type of element of array
+    char* pa = a;
+    // pa + k -> a[k]
+    char* pb = b;
+
+    // *pa -> a[0], *(pa + 0) -> a[0]
+    // *(pa + i) -> a[i]
+    printf("Value of a[6] -> %c\n", *(pa + 6));
+    printf("Value of b[5] -> %c\n", *(pb + 5));
+
+    char* c[2];
+    c[0] = pa;
+    c[1] = pb;
+
+    // print values without pa and pb
+
+    // *(pa + i) -> a[i]
+    // *(c[0] + i) -> a[i]
+    // *c[0] -> *(pa) -> *(&a[0]) -> Value at address of a[0] -> a[0]
+    printf("Value of a[6] -> %c\n", *(c[0] + 6));
+    printf("Value of b[5] -> %c\n", *(c[1] + 5));
+
+    // Represent c using pointer
+    // TYPE *pointer-name = array-name
+    // TYPE -> type of element of array
+    char**pc = c;
+
+
+    return 0;
+}