what is the output of the following program error find Set 1:

1.1 The Following code is not displaying the correct output, which is 1500. Identify the error. C programming language.

    #include<stdio.h>
    #include<conio.h>
    main()
    {
        /* This Program calculates the simple interest */
        int p, t, si;
        float r;
        p=5000;
        t=4;
        r=7.5;
        si=p*t*r/100;
        clrscr();
        printf("%f",si);
        getch();
        }

1.2 The following code is not displaying the output on the screen. Identify the error. C programming language.

    #include<stdio.h>

    #include<conio.h>
    main()

    {

        /* This program allow you to enter a number and display that number on the screen */

        int number;

        printf("Enter a number ");

        scanf(" %d ",&number );

        printf(" You have entered the number", number);

        getch();

    }

1.3 Identify the error in following code c programming language.

#include<stdio.h>
#include<conio.h>
main()
    {
        /* This program calculate the area of a circle.        */
        float area;
        int radius=4;
        area = 3.14*radius**2;
        printf("%f",area);
        getch();
    }

1.4 Identify the errors in the following code c programming languages.

#include<stdio.h>
#include<conio.h>
main()
    {
        int a=5, c=2, d=4, n=6;
        float ans;
        ans = 10.2/a+(2*a(3c+4)/a*d/(12/n);
        clrscr();
        printf("%f",ans);
        getch();
    }

1.5 Identify the errors in the following code c programming languages.

#include<stdio.h>
#include<conio.h>
main()
    {

        char x,y;

        int z;

        x=a;

        y=b;

        z=x+y;

        printf("%d",z);

        getch();

    }

1.6 what is the difference between the following two statement in c programming ?

        char *str = "Computer";

        char str[] = "Computer";

1.7 What is the output of the following statement c language ?

        enum p { a=1, b, c, d, f=50, y};

        printf("%d",y);

1.8 what is the output of the following code?

    main()

    {

        int ii=5;

        printf("%d",++ii++);

    }

1.9 What is the output of the following code?

        main()

        {

            char prnt[ ]="%d\n";

                   char prnt[ ] = 'c';

            printf(prnt,67);

        }

1.10 What is the output of the following code?

void main()

      {

            int var=5;

            printf("%d",var++*var++);

        }

Solution 1.1:

The code you provided in your previous message doesn't have any syntax errors. However, it does have a logical issue. The problem is that you are calculating the simple interest and storing it as an integer in the si variable, but you are trying to print it using the %f format specifier, which is meant for floating-point numbers.

To fix the issue, you should either change the data type of si to float or use the correct format specifier for an integer (e.g., %d) when you print si

Solution 1.2:

There is an error in the printf statement in your code. You forgot to include a format specifier for the variable number. You should use %d to indicate that you want to print an integer. Here's the corrected code:

Solution 1.3:

The error in your code is the use of the ** operator for exponentiation in the expression 3.14*radius**2. In C, you should use the * operator for multiplication, not ** for exponentiation. To calculate the area of a circle, you need to square the radius using * twice. Here's the corrected code:

Solution 1.4:

1. Missing multiplication operators (*) in the expression 2*a(3c+4) and (12/n).
2. You're using clrscr(), which is not a standard C function. If you intend to clear the screen, you should use platform-specific functions, but note that it's not portable.
3. Solution 1.5:
4. You are trying to assign the values of a and b to x and y, but a and b are not defined anywhere in your code. You need to define them as character constants by enclosing them in single quotes.

5. 1. You should enclose the format specifier %d in double quotes within the printf function.

6. Solution 1.6:
7. char *str = "Computer";This line declares a pointer to a string literal. It creates a pointer variable named str that points to the string literal "Computer" stored in read-only memory. The content of the string cannot be modified through this pointer. You can use this pointer to access the characters in the string but should not attempt to change them.

char str[] = "Computer";This line declares an array of characters and initializes it with the string "Computer." The array has a size determined by the number of characters in the string plus one for the null-terminator '\0'. You can modify the content of this array because it is not a pointer to a string literal; it's an array of characters. You can change individual characters in the array.

solution 1.7:

In the given enumeration enum p, you have explicitly assigned values to some of the enumeration constants, while others are implicitly assigned values starting from the last explicitly assigned value.

Here's how the values are assigned:

• a is explicitly assigned the value 1.
• b follows a and gets the value 2.
• c follows b and gets the value 3.
• d follows c and gets the value 4.
• f is explicitly assigned the value 50.
• y follows f and gets the next integer value in sequence, which is 51.

So, when you print y, it will output 51 because y is implicitly assigned the value 51 based on the enumeration sequence.

Solution 1.8:

This code will correctly increment ii by 1 and print the result, which is 6.

Solution 1.9:

The code you provided has a few issues, including the declaration of two different arrays named `prnt` with conflicting types. To achieve the intended output, which appears to be printing the integer `67`, you can modify the code as follows:

#include <stdio.h>

int main()

{

  char prnt[] = "%d\n"; // Use [] for a character array

printf(prnt, 67); return 0;

 }

In this corrected code, we have a single character array `prnt` containing the format string `"%d\n"`. When you use `printf` with this format string and provide `67` as an argument, it will correctly print the integer `67` followed by a newline character, resulting in the output: 67

Solution 1.10:

The code you provided contains undefined behavior due to the use of post-increment (`var++`) on the same variable within the same expression. The result is undefined because the order of evaluation of sub-expressions is not specified in this case, and it can vary between compilers. Therefore, the output of this program is undefined, and it may produce different results on different compilers or even under different optimization settings with the same compiler. You should avoid writing code that relies on undefined behavior for predictable results.