BE Civil Engineering (IOE, TU) Computer Programming in C (IOE, CT 401) Question Paper 2079 Nepal

Program

#include <stdio.h>

int main(void) {
    int N, i, pass = 0;
    float s[100], sum = 0.0f, mean, mx, mn;

    printf("Enter number of specimens: ");
    scanf("%d", &N);

    for (i = 0; i < N; i++) {
        printf("Strength of specimen %d (MPa): ", i + 1);
        scanf("%f", &s[i]);
        sum += s[i];
    }

    mean = sum / N;
    mx = mn = s[0];
    for (i = 1; i < N; i++) {
        if (s[i] > mx) mx = s[i];
        if (s[i] < mn) mn = s[i];
    }

    for (i = 0; i < N; i++) {
        if (s[i] >= 25.0f) {
            printf("Specimen %d: %.1f MPa -> PASS\n", i + 1, s[i]);
            pass++;
        } else {
            printf("Specimen %d: %.1f MPa -> FAIL\n", i + 1, s[i]);
        }
    }

    printf("Mean strength = %.2f MPa\n", mean);
    printf("Max = %.2f MPa, Min = %.2f MPa\n", mx, mn);
    printf("Pass percentage = %.2f %%\n", 100.0f * pass / N);
    return 0;
}

Hand trace for the given data $\{27.5, 24.0, 31.2, 22.8, 29.5\}$:

• Sum $= 27.5 + 24.0 + 31.2 + 22.8 + 29.5 = 135.0$ MPa.
• Mean $= 135.0 / 5 = 27.00$ MPa.
• Max $= 31.2$ MPa, Min $= 22.8$ MPa.
• Classification against $f_{ck}=25$ MPa:

• Passes $= 3$, so pass percentage $= 100 \times 3/5 = \mathbf{60.00\%}$.

Final answer: Mean $= \mathbf{27.00}$ MPa, Max $= \mathbf{31.2}$ MPa, Min $= \mathbf{22.8}$ MPa, Pass percentage $= \mathbf{60.00\%}$.

(a) Call by value vs call by reference

• Call by value: a copy of the argument is passed; changes inside the function do not affect the caller's variable.
• Call by reference: the address is passed (via pointers); the function can modify the caller's variable.

void byVal(int x) { x = 99; }          /* caller unchanged */
void byRef(int *x) { *x = 99; }        /* caller changed */

If a = 5; byVal(a); then a is still 5. If byRef(&a); then a becomes 99.

(b) Bisection function

#include <math.h>

double f(double x) { return x*x*x - x - 2.0; }

double bisection(double a, double b, double tol) {
    double m;
    while ((b - a) / 2.0 > tol) {
        m = (a + b) / 2.0;
        if (f(m) == 0.0) return m;
        if (f(a) * f(m) < 0.0) b = m;   /* root in [a,m] */
        else                  a = m;    /* root in [m,b] */
    }
    return (a + b) / 2.0;
}

Hand computation, $f(x) = x^3 - x - 2$, start $[1, 2]$:

• $f(1) = 1 - 1 - 2 = -2$ (negative), $f(2) = 8 - 2 - 2 = 4$ (positive). Root lies in $[1,2]$.

Iteration 1: $m_1 = (1+2)/2 = 1.5$. $f(1.5) = 3.375 - 1.5 - 2 = -0.125$ (negative). Since $f(1.5)$ and $f(2)$ differ in sign, root in $[1.5, 2]$.

Iteration 2: $m_2 = (1.5+2)/2 = 1.75$. $f(1.75) = 5.359375 - 1.75 - 2 = 1.609375$ (positive). Root in $[1.5, 1.75]$.

Iteration 3: $m_3 = (1.5+1.75)/2 = 1.625$. $f(1.625) = 4.291015625 - 1.625 - 2 = 0.666015625$ (positive). Root in $[1.5, 1.625]$.

Approximate root after 3 iterations $\approx (1.5 + 1.625)/2 = \mathbf{1.5625}$. (The true root is $\approx 1.5214$.)

Program

#include <stdio.h>

int main(void) {
    int i, j;
    double K[3][3], d[3], F[3] = {0};

    printf("Enter 3x3 stiffness matrix K:\n");
    for (i = 0; i < 3; i++)
        for (j = 0; j < 3; j++)
            scanf("%lf", &K[i][j]);

    printf("Enter displacement vector d (3 values):\n");
    for (i = 0; i < 3; i++)
        scanf("%lf", &d[i]);

    for (i = 0; i < 3; i++) {
        F[i] = 0.0;
        for (j = 0; j < 3; j++)
            F[i] += K[i][j] * d[j];
    }

    printf("Force vector F = [K]{d}:\n");
    for (i = 0; i < 3; i++)
        printf("F[%d] = %.2f\n", i, F[i]);
    return 0;
}

Hand computation $\{F\} = [K]\{d\}$:

• $F_1 = (4)(2) + (-1)(1) + (0)(3) = 8 - 1 + 0 = 7$.
• $F_2 = (-1)(2) + (4)(1) + (-1)(3) = -2 + 4 - 3 = -1$.
• $F_3 = (0)(2) + (-1)(1) + (4)(3) = 0 - 1 + 12 = 11$.

Final answer: $\{F\} = \mathbf{[\,7,\;-1,\;11\,]^{T}}$.

Program

#include <stdio.h>

struct Station {
    char name[20];
    double rl;        /* reduced level in metres */
};

int main(void) {
    struct Station st[50], *p, *best;
    int n, i;
    double sum = 0.0;

    printf("Number of stations: ");
    scanf("%d", &n);
    for (i = 0; i < n; i++) {
        printf("Name and RL of station %d: ", i + 1);
        scanf("%s %lf", st[i].name, &st[i].rl);
    }

    best = st;                       /* pointer to first element */
    for (p = st; p < st + n; p++) {
        sum += p->rl;
        if (p->rl > best->rl) best = p;
    }

    printf("Highest RL station: %s (%.2f m)\n", best->name, best->rl);
    printf("Average RL = %.2f m\n", sum / n);
    return 0;
}

Hand computation for A(1245.30), B(1248.75), C(1242.10), D(1250.60):

• Sum $= 1245.30 + 1248.75 + 1242.10 + 1250.60 = 4986.75$ m.
• Average $= 4986.75 / 4 = 1246.6875 \approx 1246.69$ m.
• Highest RL $=$ D at 1250.60 m.

Final answer: Highest-RL station is $\mathbf{D\,(1250.60\,\text{m})}$; average RL $= \mathbf{1246.69}$ m.

Program

#include <stdio.h>

int main(void) {
    FILE *fp = fopen("rain.txt", "r");
    double v, total = 0.0, mx = 0.0;
    int rainy = 0;

    if (fp == NULL) {
        printf("Error: cannot open rain.txt\n");
        return 1;
    }

    while (fscanf(fp, "%lf", &v) == 1) {
        total += v;
        if (v > 0.0) rainy++;
        if (v > mx)  mx = v;
    }
    fclose(fp);

    printf("Total rainfall = %.2f mm\n", total);
    printf("Rainy days = %d\n", rainy);
    printf("Maximum daily rainfall = %.2f mm\n", mx);
    return 0;
}

Hand computation for $\{0, 12.5, 0, 30.0, 8.0, 0, 45.5, 5.0\}$:

• Total $= 0 + 12.5 + 0 + 30.0 + 8.0 + 0 + 45.5 + 5.0 = 101.0$ mm.
• Rainy days (value $> 0$): $12.5, 30.0, 8.0, 45.5, 5.0$ $\Rightarrow$ 5 days.
• Maximum daily rainfall $= 45.5$ mm.

Final answer: Total $= \mathbf{101.00}$ mm, Rainy days $= \mathbf{5}$, Maximum $= \mathbf{45.50}$ mm.

Integer evaluation (precedence: * and / before + and -, left to right):

• b * c = 2 * 5 = 10.
• a / b = 7 / 2 = 3 (integer division truncates the fractional part).
• result = a + 10 - 3 = 7 + 10 - 3 = 14.

So with int, $\mathbf{result = 14}$.

Float evaluation: With float a=7, b=2, c=5, the division is real: a/b = 3.5.

• result = 7.0 + 10.0 - 3.5 = 13.5.

So with float, $\mathbf{result = 13.5}$.

The difference arises solely from integer division truncating $7/2$ to $3$ rather than $3.5$. Multiplication/division bind tighter than addition/subtraction, so they are computed first.

Program

#include <stdio.h>

int main(void) {
    int code;
    printf("Enter soil code (1-4): ");
    scanf("%d", &code);

    switch (code) {
        case 1: printf("Soil type: Gravel\n"); break;
        case 2: printf("Soil type: Sand\n");   break;
        case 3: printf("Soil type: Silt\n");   break;
        case 4: printf("Soil type: Clay\n");   break;
        default: printf("Error: invalid soil code\n");
    }
    return 0;
}

Explanation: Each case ends with break to prevent fall-through to the next case. The default label handles any code outside 1-4. For example, input 3 prints Soil type: Silt, and input 9 prints Error: invalid soil code.

Pointers: A pointer is a variable that stores the memory address of another variable. int *p = &x; makes p point to x; the dereference *p reads or writes the value at that address. Pointers enable call by reference, dynamic memory, and efficient array traversal.

Hand trace (this is the classic swap-without-temp):

• Initially x = 10, y = 20; p points to x, q points to y.
• *p = *p + *q $= 10 + 20 = 30$, so x = 30.
• *q = *p - *q $= 30 - 20 = 10$, so y = 10.
• *p = *p - *q $= 30 - 10 = 20$, so x = 20.

Output: x = 20, y = 10. The program swaps the two values using arithmetic through pointers, with no temporary variable.

Function and call

#include <stdio.h>

double meanRL(double rl[], int n) {
    double sum = 0.0;
    int i;
    for (i = 0; i < n; i++)
        sum += rl[i];
    return sum / n;
}

int main(void) {
    double rl[] = {100.50, 101.20, 99.80, 100.50};
    int n = 4;
    printf("Mean RL = %.2f m\n", meanRL(rl, n));
    return 0;
}

The array name rl decays to a pointer when passed, so the function operates on the caller's data; n carries the element count.

Hand computation for $\{100.50, 101.20, 99.80, 100.50\}$:

• Sum $= 100.50 + 101.20 + 99.80 + 100.50 = 402.00$ m.
• Mean $= 402.00 / 4 = 100.50$ m.

Final answer: Mean RL $= \mathbf{100.50}$ m.

Program

#include <stdio.h>

int main(void) {
    int n, i;
    unsigned long long fact = 1ULL;

    printf("Enter n: ");
    scanf("%d", &n);

    if (n < 0) {
        printf("Factorial undefined for negative numbers\n");
        return 1;
    }
    for (i = 2; i <= n; i++)
        fact *= i;

    printf("%d! = %llu arrangements\n", n, fact);
    return 0;
}

Hand computation for $n = 6$:

Step by step: $1 \times 2 = 2$; $2 \times 3 = 6$; $6 \times 4 = 24$; $24 \times 5 = 120$; $120 \times 6 = 720$.

Final answer: $6! = \mathbf{720}$ arrangements.

Program

#include <stdio.h>

int main(void) {
    int a[100], n, key, i, found = -1;

    printf("Enter n: ");
    scanf("%d", &n);
    for (i = 0; i < n; i++)
        scanf("%d", &a[i]);
    printf("Enter key: ");
    scanf("%d", &key);

    for (i = 0; i < n; i++) {
        if (a[i] == key) { found = i; break; }
    }

    if (found != -1)
        printf("Found at index %d\n", found);
    else
        printf("Not found\n");
    return 0;
}

Hand trace for key = 17 in $\{4, 9, 17, 23, 8\}$:

The key is found at index 2 after 3 comparisons.

Final answer: Output is Found at index 2; number of comparisons $= \mathbf{3}$.