BE Civil Engineering (IOE, TU) Computer Programming in C (IOE, CT 401) Question Paper 2080 Nepal

Program

#include <stdio.h>

int main(void) {
    int n, i, countAbove = 0;
    printf("Enter number of stations: ");
    scanf("%d", &n);

    float rl[100], sum = 0.0f, avg;
    float high, low;

    for (i = 0; i < n; i++) {
        printf("Enter RL of station %d (m): ", i + 1);
        scanf("%f", &rl[i]);
        sum += rl[i];
    }

    high = low = rl[0];
    for (i = 1; i < n; i++) {
        if (rl[i] > high) high = rl[i];
        if (rl[i] < low)  low  = rl[i];
    }

    avg = sum / n;

    for (i = 0; i < n; i++)
        if (rl[i] > avg) countAbove++;

    printf("Highest RL = %.2f m\n", high);
    printf("Lowest  RL = %.2f m\n", low);
    printf("Average RL = %.2f m\n", avg);
    printf("Stations above average = %d\n", countAbove);
    return 0;
}

Hand trace for the data set $\{1245.30, 1248.75, 1243.10, 1250.40, 1247.45\}$, $n = 5$.

Sum:

$$S = 1245.30 + 1248.75 + 1243.10 + 1250.40 + 1247.45 = 6235.00\ \text{m}$$

Average:

$$\text{avg} = \frac{6235.00}{5} = 1247.00\ \text{m}$$

Highest = $1250.40$ m, Lowest = $1243.10$ m.

Stations above average ($> 1247.00$): $1248.75$ and $1250.40$ and $1247.45$ — that is $3$ stations. (Note $1245.30$ and $1243.10$ are below.)

Final printed output

Highest RL = 1250.40 m
Lowest  RL = 1243.10 m
Average RL = 1247.00 m
Stations above average = 3

(a) Call by value vs call by reference

Call by value:

void inc(int x){ x = x + 1; }   // caller's variable unchanged

Call by reference:

void inc(int *x){ *x = *x + 1; } // caller's variable incremented

(b) Power function

#include <stdio.h>

double power(double base, int exp) {
    int negative = (exp < 0);
    int e = negative ? -exp : exp;
    double result = 1.0;
    for (int i = 0; i < e; i++)
        result *= base;
    if (negative)
        result = 1.0 / result;
    return result;
}

int main(void) {
    printf("1.5^4 = %.4f\n", power(1.5, 4));
    printf("2^-3  = %.4f\n", power(2.0, -3));
    return 0;
}

Hand computation

$1.5^4 = 1.5 \times 1.5 \times 1.5 \times 1.5 = 2.25 \times 2.25 = 5.0625$.

$2^{-3} = \dfrac{1}{2^3} = \dfrac{1}{8} = 0.125$.

Output

1.5^4 = 5.0625
2^-3  = 0.1250

Program

#include <stdio.h>
#define PI 3.141592653589793
#define RHO 7850.0   /* kg/m^3 */

struct Bar {
    char tag[11];
    double diameter;  /* mm */
    double length;    /* m  */
};

int main(void) {
    struct Bar b[3];
    double totalMass = 0.0;

    for (int i = 0; i < 3; i++) {
        printf("Bar %d tag, diameter(mm), length(m): ", i + 1);
        scanf("%10s %lf %lf", b[i].tag, &b[i].diameter, &b[i].length);
    }

    for (int i = 0; i < 3; i++) {
        double d = b[i].diameter / 1000.0;   /* m */
        double mass = RHO * (PI * d * d / 4.0) * b[i].length;
        printf("%s : %.2f kg\n", b[i].tag, mass);
        totalMass += mass;
    }

    printf("Total steel mass = %.2f kg\n", totalMass);
    return 0;
}

Sample calculation for $d = 20$ mm $= 0.020$ m, $L = 12$ m:

Cross-sectional area:

$$A = \frac{\pi (0.020)^2}{4} = \frac{\pi \times 0.0004}{4} = \frac{0.00125664}{4} = 3.14159\times 10^{-4}\ \text{m}^2.$$

Mass:

$$\text{mass} = 7850 \times 3.14159\times 10^{-4} \times 12 = 7850 \times 3.76991\times 10^{-3} = 29.59\ \text{kg}.$$

For a single 20 mm bar of 12 m, mass ≈ 29.59 kg. (Equivalently the standard $0.00617 d^2 L$ rule gives $0.00617 \times 400 \times 12 = 29.62$ kg, matching within rounding.)

(a) Pointer

A pointer is a variable that stores the memory address of another variable. If int x = 10; then &x gives the address of x, and a pointer int *p = &x; holds that address. The dereference operator *p accesses the value stored at that address (here 10).

int x = 10;
int *p = &x;   // p holds address of x
printf("%d", *p);  // prints 10
*p = 25;       // changes x to 25

• & gives the address of a variable.
• * (when applied to a pointer) gives the value at that address.

(b) Reverse function

void reverse(int *arr, int n) {
    int *left = arr;
    int *right = arr + (n - 1);
    while (left < right) {
        int temp = *left;
        *left = *right;
        *right = temp;
        left++;
        right--;
    }
}

Trace on $\{4, 9, 15, 22, 30\}$, $n = 5$:

The middle element (index 2 = 15) stays in place. Final reversed array: {30, 22, 15, 9, 4}.

Program

#include <stdio.h>
#include <stdlib.h>

int main(void) {
    FILE *fin = fopen("rainfall.txt", "r");
    if (fin == NULL) {
        printf("Error: cannot open rainfall.txt\n");
        return 1;
    }

    int d;
    fscanf(fin, "%d", &d);

    double value, total = 0.0, maxVal = -1.0;
    int maxDay = 0;
    for (int i = 1; i <= d; i++) {
        fscanf(fin, "%lf", &value);
        total += value;
        if (value > maxVal) { maxVal = value; maxDay = i; }
    }
    fclose(fin);

    double avg = total / d;

    FILE *fout = fopen("report.txt", "a");
    if (fout == NULL) {
        printf("Error: cannot open report.txt\n");
        return 1;
    }
    fprintf(fout, "Total=%.1f mm, Avg=%.2f mm, Wettest day=%d (%.1f mm)\n",
            total, avg, maxDay, maxVal);
    fclose(fout);

    printf("Total=%.1f mm, Avg=%.2f mm, Wettest day=%d (%.1f mm)\n",
           total, avg, maxDay, maxVal);
    return 0;
}

Hand computation for $\{0.0, 12.5, 8.0, 25.0, 4.5\}$, $d = 5$:

Total:

$$T = 0.0 + 12.5 + 8.0 + 25.0 + 4.5 = 50.0\ \text{mm}.$$

Average:

$$\text{avg} = \frac{50.0}{5} = 10.00\ \text{mm}.$$

Maximum is $25.0$ mm on day 4.

Printed / appended summary:

Total=50.0 mm, Avg=10.00 mm, Wettest day=4 (25.0 mm)

(a) Basic data types (typical sizes on a 32/64-bit system)

(b) Expression evaluation with a = 7, b = 2:

In C, relational/logical expressions yield int values 1 (true) or 0 (false).

Program

#include <stdio.h>

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

    switch (code) {
        case 1:
            printf("Soft clay: 50-100 kPa\n");
            break;
        case 2:
            printf("Stiff clay: 100-300 kPa\n");
            break;
        case 3:
            printf("Loose sand: 100-200 kPa\n");
            break;
        case 4:
            printf("Dense gravel: 300-600 kPa\n");
            break;
        default:
            printf("Invalid code\n");
    }
    return 0;
}

Example: input 4 prints Dense gravel: 300-600 kPa; input 9 prints Invalid code. The break after each case prevents fall-through to the next case.

Program

#include <stdio.h>

int main(void) {
    int A[2][2], B[2][2], C[2][2];
    int i, j;

    printf("Enter matrix A (4 values):\n");
    for (i = 0; i < 2; i++)
        for (j = 0; j < 2; j++)
            scanf("%d", &A[i][j]);

    printf("Enter matrix B (4 values):\n");
    for (i = 0; i < 2; i++)
        for (j = 0; j < 2; j++)
            scanf("%d", &B[i][j]);

    for (i = 0; i < 2; i++)
        for (j = 0; j < 2; j++)
            C[i][j] = A[i][j] + B[i][j];

    printf("Sum:\n");
    for (i = 0; i < 2; i++) {
        for (j = 0; j < 2; j++)
            printf("%4d", C[i][j]);
        printf("\n");
    }
    return 0;
}

Hand calculation — element-wise addition:

$$A + B = \begin{bmatrix} 3+1 & 5+4 \\ 7+6 & 2+8 \end{bmatrix} = \begin{bmatrix} 4 & 9 \\ 13 & 10 \end{bmatrix}.$$

Output

   4   9
  13  10

Function

int myStrlen(char str[]) {
    int count = 0;
    while (str[count] != '\0')
        count++;
    return count;
}

How a C string is stored

A C string is an array of char terminated by the null character '\0' (ASCII value 0). The functions know where the string ends by scanning until they reach '\0'. The terminator occupies one extra byte that is not counted as a visible character.

For "Bridge" the memory layout is:

Index:  0   1   2   3   4   5    6
Char:  'B' 'r' 'i' 'd' 'g' 'e' '\0'

The loop increments count for indices 0 through 5 (six characters) and stops at index 6 where '\0' is found.

Result: myStrlen("Bridge") = 6 (the '\0' is not counted, though 7 bytes are stored).

(a) #define vs const

(b) Errors and corrections

The program has three problems:

1. Missing semicolon after int radius = 5 → should be int radius = 5;
2. printf uses %d for a float value → should be %f (or cast).
3. Missing semicolon after the printf(...) statement.

Corrected program:

#include <stdio.h>
int main()
{
    int radius = 5;
    float area;
    area = 3.14 * radius * radius;   /* = 78.50 */
    printf("Area = %f", area);
    return 0;
}

With radius = 5, area $= 3.14 \times 25 = 78.50$.

Recursive function

long factorial(int n) {
    if (n == 0 || n == 1)   /* base case */
        return 1;
    return n * factorial(n - 1);   /* recursive case */
}

• Base case: factorial(0) = 1 and factorial(1) = 1 — stops the recursion.
• Recursive case: $n! = n \times (n-1)!$.

Call stack for $n = 4$

Calls going down (unwinding):

factorial(4) = 4 * factorial(3)
factorial(3) = 3 * factorial(2)
factorial(2) = 2 * factorial(1)
factorial(1) = 1            <- base case reached

Returns coming back up:

factorial(1) -> 1
factorial(2) -> 2 * 1   = 2
factorial(3) -> 3 * 2   = 6
factorial(4) -> 4 * 6   = 24

Result: $4! = 24$.