Demonstrate pointers in C

LAB B03 · 01

Understand the task.

Aim: Demonstrate pointers in C.

Review the related theory in Unit 4.

LAB B03 · 02

Know what goes in and what comes out.

Two integers from -100 to 100: an original x value and a replacement value. A fixed three-element array then demonstrates pointer arithmetic.

12 25

Pointer targets x: YES
Before: x = 12, *p = 12
After: x = 25, *p = 25
Array through pointer: 5 10 15

Source programs read the documented input directly, without prompt text, so the same sample works in the embedded compiler’s STDIN field.

LAB B03 · 03

Plan the steps before writing code.

1. Read and validate the original and replacement values.
2. Assign &x to p and verify that p identifies x.
3. Read x through *p.
4. Assign the replacement through *p and display the changed x.
5. Retarget p to the array’s first element and access each element using *(p + i).

LAB B03 · 04

One complete program. Ready to run.

#include <stdio.h>
int main(void)
{
    int x, replacement, i;
    int values[3] = {5, 10, 15};
    int *p;
    if (scanf("%d%d", &x, &replacement) != 2 || x < -100 || x > 100 || replacement < -100 || replacement > 100) {
        puts("Enter the original and replacement values from -100 to 100.");
        return 1;
    }
    p = &x;
    printf("Pointer targets x: %s\n", p == &x ? "YES" : "NO");
    printf("Before: x = %d, *p = %d\n", x, *p);
    *p = replacement;
    printf("After: x = %d, *p = %d\n", x, *p);
    p = values;
    printf("Array through pointer:");
    for (i = 0; i < 3; i++) printf(" %d", *(p + i));
    putchar('\n');
    return 0;
}

gcc B03-pointer-demonstration.c -o program

LAB B03 · 05

Connect each statement to its purpose.

• p = &x stores an address, whereas *p = replacement changes x itself.
• The equality test demonstrates the address relationship without making a particular printed address part of the expected output.
• When p points to values, p + i moves by int elements, not by a manually chosen number of bytes.
• The loop uses only indexes 0–2, so every dereference is within the array. Unit 4 separately demonstrates portable %p printing and the one-past endpoint.

LAB B03 · 06

Follow the sample calculation.

LAB B03 · 07

Check more than one input.

Time and storage

O(1) time here because the array always has three elements; O(1) extra working space.

Start with the sample, then test a boundary and a case that changes the result. A correct sample alone does not establish that every input is handled correctly.

LAB B03 · 08

Use the real C compiler.

Choose an example and load it into the embedded C compiler. Select I/O inside the compiler, check the sample STDIN, and press Run. Change the input or edit the source to explore a different case.

12 25

Pointer targets x: YES
Before: x = 12, *p = 12
After: x = 25, *p = 25
Array through pointer: 5 10 15

#include <stdio.h>
int main(void)
{
    int x, replacement, i;
    int values[3] = {5, 10, 15};
    int *p;
    if (scanf("%d%d", &x, &replacement) != 2 || x < -100 || x > 100 || replacement < -100 || replacement > 100) {
        puts("Enter the original and replacement values from -100 to 100.");
        return 1;
    }
    p = &x;
    printf("Pointer targets x: %s\n", p == &x ? "YES" : "NO");
    printf("Before: x = %d, *p = %d\n", x, *p);
    *p = replacement;
    printf("After: x = %d, *p = %d\n", x, *p);
    p = values;
    printf("Array through pointer:");
    for (i = 0; i < 3; i++) printf(" %d", *(p + i));
    putchar('\n');
    return 0;
}

LAB B03 · 09

Explain why the program works.

LAB B03 · 10

Build the next skill.

Read the viva answers, try the extra cases, and return to the lab index to mark this exercise as practised.