What will be the output of the program?

#include<stdio.h>

int main()
{
    int i, a[] = {2, 4, 6, 8, 10};
    change(a, 5);
    for(i=0; i<=4; i++)
        printf("%d, ", a[i]);
    return 0;
}
void change(int *b, int n)
{
    int i;
    for(i=0; i<n; i++)
        *(b+1) = *(b+i)+5;
}

Here fputc('A', fp1); stores 'A' in the file1.c then fputc('B', fp2); overwrites the contents of the file1.c with value 'B'. Because the fp1 and fp2 opens the file1.c in write mode.

Hence the file1.c contents is 'B'.

FALSE

Example: #define symbol replacement

When the preprocessor encounters #define directive, it replaces any occurrence of symbol in the rest of the code by replacement. This replacement can be an statement or expression or a block or simple text.

Step 1: int i=3, j=4, k, l; The variables i, j, k, l are declared as an integer type and variable i, j are initialized to 3, 4 respectively.

The function addmult(i, j); accept 2 integer parameters.

Step 2: k = addmult(i, j); becomes k = addmult(3, 4)

In the function addmult(). The variable kk, ll are declared as an integer type int kk, ll;

kk = ii + jj; becomes kk = 3 + 4 Now the kk value is '7'.

ll = ii * jj; becomes ll = 3 * 4 Now the ll value is '12'.

return (kk, ll); It returns the value of variable ll only.

The value 12 is stored in variable 'k'.

Step 3: l = addmult(i, j); becomes l = addmult(3, 4)

kk = ii + jj; becomes kk = 3 + 4 Now the kk value is '7'.

ll = ii * jj; becomes ll = 3 * 4 Now the ll value is '12'.

return (kk, ll); It returns the value of variable ll only.

The value 12 is stored in variable 'l'.

Step 4: printf("%d, %d\n", k, l); It prints the value of k and l

Hence the output is "12, 12".

for(i=1; i<=5; i++) Here the for loop runs 5 times.

Loop 1:
scanf("%c", &c); Here we give 'a' as input.
printf("%c", c); prints the character 'a' which is given in the previous "scanf()" statement.
ungetc(c, stdin); "ungetc()" function pushes character 'a' back into input stream.

Loop 2:
Here the scanf("%c", &c); get the input from "stdin" because of "ungetc" function.
printf("%c", c); Now variable c = 'a'. So it prints the character 'a'.
ungetc(c, stdin); "ungetc()" function pushes character 'a' back into input stream.

This above process will be repeated in Loop 3, Loop 4, Loop 5.

Step 1: int check(int); This prototype tells the compiler that the function check() accepts one integer parameter and returns an integer value.

Step 2: int l=45, c; The variable i and c are declared as an integer type and i is initialized to 45.

The function check(i) return 100 if the given value of variable i is >=(greater than or equal to) 45, else it will return 10.

Step 3: c = check(i); becomes c = check(45); The function check() return 100 and it get stored in the variable c.(c = 100)

Step 4: printf("%d\n", c); It prints the value of variable c.

Hence the output of the program is '100'.

Step 1: for(;;) this statement will genereate infinite loop.
Step 2: printf("%d\n", i++); this statement will print the value of variable i and increement i by 1(one).
Step 3: if(i>10) here, if the variable i value is greater than 10, then the for loop breaks.

Hence the output of the program is
1
2
3
4
5
6
7
8
9
10

if(a < 0.7) here a is a float variable and 0.7 is a double constant. The float variable a is less than double constant 0.7. Hence the if condition is satisfied and it prints 'C'
Example:

#include<stdio.h>
int main()
{
    float a=0.7;
    printf("%.10f %.10f\n",0.7, a);
    return 0;
}

Output:
0.7000000000 0.6999999881

The integer value 2 is represented as 00000000 00000010 in binary system.

Negative numbers are represented in 2's complement method.

1's complement of 00000000 00000010 is 11111111 11111101 (Change all 0s to 1 and 1s to 0).

2's complement of 00000000 00000010 is 11111111 11111110 (Add 1 to 1's complement to obtain the 2's complement value).

Therefore, in binary we represent -2 as: 11111111 11111110.

After left shifting it by 2 bits we obtain: 11111111 11111000, and it is equal to "fff8" in hexadecimal system.