Loops, one rung at a time

#include <stdio.h>
int main(void)
{
    int n;
    printf("Enter n: ");
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        printf("%d ", i);
    }
    printf("\n");
    return 0;
}

The printf("\n") after the loop is what ends the line; inside the loop there is no \n, which is why everything stays on one line.

#include <stdio.h>
int main(void)
{
    for (int i = 1; i <= 10; i++) {
        printf("7 x %d = %d\n", i, 7 * i);
    }
    return 0;
}

Debugging Exercise 1 on the sheet is this program with < 10 instead of <= 10. Type this one and you will never make that slip.

#include <stdio.h>
int main(void)
{
    int n;
    printf("Enter n: ");
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        printf("*");
    }
    printf("\n");
    return 0;
}

Count the stars in your output against n. Off by one here is the single most common loop error; the fix is always in the pair (start value, comparison).

#include <stdio.h>
int main(void)
{
    int n, sum = 0;
    printf("Enter n: ");
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        sum += i;
    }
    printf("Sum of 1 to %d is %d\n", n, sum);
    return 0;
}

Debugging Exercise 3 on the sheet is this program with int sum; left uninitialised. Check: 5050 is the famous answer for 100.

#include <stdio.h>
int main(void)
{
    int n;
    long long fact = 1;
    printf("Enter n: ");
    scanf("%d", &n);
    for (int i = 2; i <= n; i++) {
        fact *= i;
    }
    printf("%d! = %lld\n", n, fact);
    return 0;
}

long long and %lld because 13! already overflows an int. Try 12 (479001600) and 13 with a plain int to see garbage appear.

#include <stdio.h>
int main(void)
{
    int n, x, sum = 0;
    printf("How many numbers? ");
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        printf("Number %d: ", i);
        scanf("%d", &x);
        sum += x;
    }
    printf("Average: %.2f\n", (double) sum / n);
    return 0;
}

The cast (double) sum / n is essential: 105 / 4 with two ints is 26, not 26.25. Integer division is the trap of this rung.

#include <stdio.h>
int main(void)
{
    int n, digits = 0;
    printf("Enter n: ");
    scanf("%d", &n);
    int temp = n;
    while (temp != 0) {
        digits++;
        temp /= 10;
    }
    printf("%d has %d digits\n", n, digits);
    return 0;
}

Work on a copy so n survives for the final message. Try 0: the loop body never runs and the program says "0 digits" — the zero-iteration case the sheet's §19 tells you to test.

#include <stdio.h>
int main(void)
{
    int n, sum = 0;
    printf("Enter n: ");
    scanf("%d", &n);
    int temp = n;
    while (temp != 0) {
        sum += temp % 10;
        temp /= 10;
    }
    printf("Digit sum of %d is %d\n", n, sum);
    return 0;
}

Problems 7, 8 and 9 are the same six-line loop with a different body. The sheet's Problem 1 asks for all three at once — do them one at a time here first.

#include <stdio.h>
int main(void)
{
    int n, reversed = 0;
    printf("Enter n: ");
    scanf("%d", &n);
    int temp = n;
    while (temp != 0) {
        reversed = reversed * 10 + temp % 10;
        temp /= 10;
    }
    printf("%d reversed is %d\n", n, reversed);
    return 0;
}

Try 1200: the answer is 21, not 0021 — leading zeros do not exist in an int. Say that out loud if asked; it is correct, not a bug.

#include <stdio.h>
int main(void)
{
    int x, count = 0, sum = 0;
    printf("Enter numbers, 0 to stop:\n");
    scanf("%d", &x);
    while (x != 0) {
        count++;
        sum += x;
        scanf("%d", &x);
    }
    printf("You entered %d numbers with sum %d\n", count, sum);
    return 0;
}

Read once before the loop, then again at the end of every pass — "read, test, use, read". Put the scanf at the top of the body instead and the 0 gets counted.

#include <stdio.h>
int main(void)
{
    int choice;
    do {
        printf("1. Say hello  2. Say bye  0. Quit\n");
        printf("Choice: ");
        scanf("%d", &choice);
        if (choice == 1)
            printf("Hello!\n");
        else if (choice == 2)
            printf("Bye!\n");
        else if (choice != 0)
            printf("Not a valid choice.\n");
    } while (choice != 0);
    printf("Quitting.\n");
    return 0;
}

This is the skeleton of the sheet's Problem 4 (PIN validator) and Problem 7 (pattern menu). The invalid-choice branch is the habit the sheet's §19 calls "test the bad input".

#include <stdio.h>
int main(void)
{
    int rows, cols;
    printf("Enter rows and columns: ");
    scanf("%d %d", &rows, &cols);
    for (int r = 1; r <= rows; r++) {
        for (int c = 1; c <= cols; c++) {
            printf("*");
        }
        printf("\n");
    }
    return 0;
}

Where the \n goes decides everything. Inside the inner loop: one star per line. Outside both loops: one long line. Between them: a rectangle.

#include <stdio.h>
int main(void)
{
    int n;
    printf("Enter n: ");
    scanf("%d", &n);
    for (int r = 1; r <= n; r++) {
        for (int c = 1; c <= r; c++) {
            printf("*");
        }
        printf("\n");
    }
    return 0;
}

Debugging Exercise 4 on the sheet is the rectangle mistakenly left as j <= n. Change n to r and the triangle appears — that is the "change exactly one variable" the sheet asks for.

#include <stdio.h>
int main(void)
{
    int n;
    printf("Enter n: ");
    scanf("%d", &n);
    for (int r = 1; r <= n; r++) {
        for (int c = 1; c <= n; c++) {
            printf("%4d", r * c);
        }
        printf("\n");
    }
    return 0;
}

This is the sheet's Problem 7 pattern menu without the switch: rows outside, columns inside, newline between. After this rung the sheet's patterns are yours.