Solution

The C code is an example of recursion.

If n is odd then returns n, else returns (n-1).

Hence in the end as it is post-increment, the 19 will be returned and after that in memory i’s value will be incremented to 20.

Solution

Address of arr[40][50] = Base address + 40*100*element_size + 50*element_size

= 0 + 4000*1 + 50*1

= 4050

Based on row major if row major then the result will be 4050

Solution

rec(989, 11) =>

989%11 * rec(89, 11)

10 * [89%11 * rec(89/11, 11)]

10 * [ 1 * rec(8, 11)]

10 * [ 1 * [ 8%11 * rec (8/11, 11) ] ]

10 * [ 1 - [ 8 * rec(0,11) ] ]

10 * [ 1 * [ 8 * 1 ] ]

10 * [ 1 * 8 ]

10 * [8]

80

Solution

Array subscripting is commutative in C.

Hence, str[3] == 3[str]

Solution

For this above program, you must know what this is - '>>='. This is a bitwise right shift assignment operator. Let me first tell what the bitwise right shift operator is. The bitwise right shift operator is a binary operator that requires operands. The first operator is the value to be shifted and the second operand specifies the number of bits to be shifted.

Shifting binary numbers is just like sifting decimal numbers. If we have an eight-digit decimal number and we shift it one place to the right, then the rightmost digit is lost and one zero digit is added on the left. So basically, we are dividing the number by 10. The below table shows the shift operator values for division:

SHIFT VALUE DIVIDES BY (2^m) SHIFT OPERATOR

1 2 >>1

2 4 >>2

.... ..... .....

n 2^n >>n

So, in this program, 'i' has been initialized to 1024. The for loop will run until the value of 'i' is not equal to zero. And in the last statement, i>>=1 will mean that i=i>>1 i.e divide the value of 'i' by 2 (check in the above table for shift operator) and then save it in 'i' itself. So, the for loop will be executed 11 times i.e "GeeksQuiz" will be printed 11 times.

STEP # i

1. 1024

2. 512

3. 256

4. 128

5. 64

6. 32

7. 16

8. 8

9. 4

10. 2

11. 1

After this, the control wont enter the body of the for loop because the value of 'i' becomes zero.

Solution

array and &array both point to base address of array[5], hence A == B.

But “array+1” and “&array+1” shouldn’t point to the same address.

Basically, “array” is a “pointer to the first element of array” but “&array” is a “pointer to whole array of 5 int”. Since “array” is pointer to int, addition of 1 resulted in an address with increment of 4 (assuming int size in your machine is 4 bytes). Since “&array” is pointer to an array of 5 ints, addition of 1 resulted in an address with increment of 4 x 5 = 20 .

Output of the code:

0x7ffc0e692650 0x7ffc0e692650

0x7ffc0e692654 0x7ffc0e692664

Solution

In C, Post-increment works as follows:

x=i++;

the first value of i is assigned to x and then the value of i is incremented.

Solution

// In C anything other that 0 i.e non-zero considered as 1 in && operation

1 && 0 - 2.9 =>

1&& (0 - 2.9) // By precedence rule

1 && (- 2.9)

1

switch_check(1)

case 1 will print “H”, as there is no break statement, everything after H will print from switch case.

Hence HIMALAYA!

Solution

As i is static, in the second and further calls of recursive functions the value of i will not be reinitiated to 5 again, whereas it will update.

In if condition, i-- is post-increment. Hence the decrement will happen after testing if condition.

Solution

Bitwise & has higher precedence than bitwise OR

Decimal => Binary

110 => 0110 1110

110 => 0110 1110

& ====================

110 & 110 => 0110 1110

1 => 0000 0001

| ======================

110 & 110 | 1 => 0110 1111

Final decimal value => 111

Solution

To check the step by step handrun of the code, please check the following print values in sequence

void ptrFun(void);

main()

{

static int a=1; // LINE 1

printf("In main 1 : %d \n",a);

ptrFun();

printf("In main 4 : %d \n",a);

a+=1;

printf("In main 5 : %d \n",a);

ptrFun();

printf("In main 6 : %d \n",a);

}

void ptrFun()

{

printf("----START---- \n");

static int a=2; // LINE 2

printf("In fun 2 : %d \n",a);

int b=1;

a+=++b;

printf("In fun 3 : %d \n",a);

printf("-----END------- \n");

}

Output:

In main 1 : 1

----START----

In fun 2 : 2

In fun 3 : 4

-----END-------

In main 4 : 1

In main 5 : 2

----START----

In fun 2 : 4

In fun 3 : 6

-----END-------

In main 6 : 2

If the static variable is declared in global scope and then changed/assigned in various functions the recently modified value will be considered as value of the variable.

If the static variable is declared in global scope and redeclared with the same name again in local scope as static variable then in the data section we will have 2 copies of the same variable separately. Compiler might change the name of one of the static variables internally.

Now a third scenario, where we have the same static variable declared in 2 local functions with the same name.

A function-local static variable is something different than a global static variable.

Since there can be as many function-local statics with the same name as you like (provided they are all in different scopes), the compiler might have to change their names internally (incorporating the function's name or the line number or whatever), so that the linker can tell them apart.

This 3rd scenario is happening in the above code and the compiler is treating these 2 a’s as 2 seperate static variables.

Solution

There are 3 blocks in the code

int main()

{ // BLOCK1

int q = 63, r = 29, s = 30;

{ // BLOCK2

int q = 68;

float r = 25;

{ // BLOCK3

int s = 10;

printf(" %d, %f, %d \n", q, r, s);

}

}

return 0;

}

Scope of the variables declared in each block is limited to that block itself and not more than that.

If a variable is initialised multiple times in the given program then the value printed will be from the nearest block it is acquired in.

Eg. s=30 in block 1, s=10 in block 3; now when it is printed in block 3 the value printed will be 10 and not 30.

Similarly depending on recent scope the value of the variable will be printed.

As r=25 is float it will be printed as 25.000000

Solution

Specifying the number of columns in 2-D array is mandatory, so, it will give compile time error, At line int arr[][] = {{4,5,6},{6,7,8}};

Solution

Output of the code:

const.c: In function ‘main’:

const.c:5:7: error: assignment of read-only variable ‘i’

i = 5;

^

Whenever we use a const qualifier with variable name, it becomes a read-only variable and get stored in the .rodata segment. Any attempt to modify this read-only variable will then result in a compilation error: “assignment of read-only variable”

Solution

strlen function from string.h will count the length of string assigned in a character array. here m, a,n,g, o, '\n' => these 6 characters stored din arr and hence strlen prints 6

,

size_t existed at least since C89 but the respective format specifier %zu (specifically the length modifier z) was added to the standard only since C99.

Alternatively you can use:

printf("%lu\n", (unsigned long)n);