what i did today......................????: November 2010

What is a variable?? Why do we use variables??

Variable is data holder, in which we can store some data and we can modify that data later in the program.

In C programming before we store any data in a variable we need to declare that variable.

declaring a variable:
declaring a variable means telling the compiler that what type of values you want store in that variable.

datatype variablename ; // declaring a variable

In C we have 4 types of Data Types:
1. int - for integer values
2. float - for floating points
3. double - for long floating points such as scientific values
4.char - for characters

declaring variables example:

int a; // tells the compiler that a is a integer type variable in which you want to store integer value.

float f; //

double d;

char c;

once you declare the variables you can store actual values in those variables.

As I told you variable is data holder. so once you declare any variable C compiler will allocate memory depending on the specified data type.

int a; when do you this one C compiler will allocate 4 bytes of space for this variable called a.

its like this a

Point out the correct statements are correct about the program below?

#include
int main()
{
    char ch;
    while(x=0;x<=255;x++)
        printf("ASCII value of %d character %c\n", x, x);
    return 0;
}

The code generates an infinite loop

The code prints all ASCII values and its characters

Error: x undeclared identifier

Error: while statement missing

Answer: Option D

Explanation:

There are 2 errors in this program.
1. "Undefined symbol x" error. Here x is not defined in the program.
2. Here while() statement syntax error.

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What will be the output of the program?

#include
int main()
{
    int i=-3, j=2, k=0, m;
    m = ++i && ++j || ++k;
    printf("%d, %d, %d, %d\n", i, j, k, m);
    return 0;
}

1, 2, 0, 1

-3, 2, 0, 1

-2, 3, 0, 1

2, 3, 1, 1

Answer: Option C

Explanation:

Step 1: int i=-3, j=2, k=0, m; here variable i, j, k, m are declared as an integer type and variable i, j, k are initialized to -3, 2, 0 respectively.

Step 2: m = ++i && ++j || ++k;
becomes m = (-2 && 3) || ++k;
becomes m = TRUE || ++k;.
(++k) is not executed because (-2 && 3) alone return TRUE.
Hence this statement becomes TRUE. So it returns '1'(one). Hence m=1.

Step 3: printf("%d, %d, %d, %d\n", i, j, k, m); In the previous step the value of i,j are increemented by '1'(one).

Hence the output is "-2, 3, 0, 1".

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What will be the output of the program?

#include
int fun(int, int);
typedef int (*pf) (int, int);
int proc(pf, int, int);

int main()
{
    printf("%d\n", proc(fun, 6, 6));
    return 0;
}
int fun(int a, int b)
{
   return (a==b);
}
int proc(pf p, int a, int b)
{
   return ((*p)(a, b));
}

-1

Answer: Option B

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What will be the output of the program?

#include
#define FUN(arg) do\
                 {\
                    if(arg)\
                        printf("IndiaBIX...", "\n");\
                  }while(--i)

int main()
{
    int i=2;
    FUN(i<3);
    return 0;
}

IndiaBIX...
IndiaBIX...
IndiaBIX

IndiaBIX... IndiaBIX...

Error: cannot use control instructions in macro

No output

Answer: Option B

Explanation:

The macro FUN(arg) prints the statement "IndiaBIX..." untill the while condition is satisfied.

Step 1: int i=2; The variable i is declared as an integer type and initialized to 2.

Step 2: FUN(i<3); becomes,

do
{
    if(2 < 3)
    printf("IndiaBIX...", "\n");
}while(--2)

After the 2 while loops the value of i becomes '0'(zero). Hence the while loop breaks.

Hence the output of the program is "IndiaBIX... IndiaBIX..."

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What will be the output of the program?

#include
#define MAX(a, b) (a > b ? a : b)

int main()
{
    int x;
    x = MAX(3+2, 2+7);
    printf("%d\n", x);
    return 0;
}

Answer: Option B

Explanation:

The macro MAX(a, b) (a > b ? a : b) returns the biggest value of the given two numbers.

Step 1 : int x; The variable x is declared as an integer type.

Step 2 : x = MAX(3+2, 2+7); becomes,

=> x = (3+2 > 2+7 ? 3+2 : 2+7)

=> x = (5 > 9 ? 5 : 9)

=> x = 9

Step 3 : printf("%d\n", x); It prints the value of variable x.

Hence the output of the program is 9.

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What will be the output of the program?

#include
#define MAX(a, b, c) (a>b ? a>c ? a : c: b>c ? b : c)

int main()
{
    int x;
    x = MAX(3+2, 2+7, 3+7);
    printf("%d\n", x);
    return 0;
}

3+7

Answer: Option C

Explanation:

The macro MAX(a, b, c) (a>b ? a>c ? a : c: b>c ? b : c) returns the biggest of given three numbers.

Step 1: int x; The variable x is declared as an integer type.

Step 2: x = MAX(3+2, 2+7, 3+7); becomes,

=> x = (3+2 >2+7 ? 3+2 > 3+7 ? 3+2 : 3+7: 2+7 > 3+7 ? 2+7 : 3+7)

=> x = (5 >9 ? (5 > 10 ? 5 : 10): (9 > 10 ? 9 : 10) )

=> x = (5 >9 ? (10): (10) )

=> x = 10

Step 3: printf("%d\n", x); It prints the value of 'x'.

Hence the output of the program is "10".

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What will be the output of the program ?

#include

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;
}
change(int *b, int n)
{
    int i;
    for(i=0; i1) = *(b+i)+5;
}

7, 9, 11, 13, 15

2, 15, 6, 8, 10

2 4 6 8 10

3, 1, -1, -3, -5

Answer: Option B

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In the following program add a statement in the function fact(). such that the factorial gets stored in j.

#include
void fact(int*);

int main()
{
    int i=5, j;
    fact(&i);
    printf("%d\n", i);
    return 0;
}
void fact(int *j)
{
    static int s=1;
    if(*j!=0)
    {
        s = s**j;
        *j = *j-1;
        fact(j);
        /* Add a statement here */
    }
}

j=s;

*j=s;

*j=&s;

&j=s;

Answer: Option B

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Is the NULL pointer same as an uninitialised pointer?

Yes

Answer: Option B

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10.

What will be the output of the program if the array begins at address 65486?

#include

int main()
{
    int arr[] = {12, 14, 15, 23, 45};
    printf("%u, %u\n", arr, &arr);
    return 0;
}

65486, 65488

65486, 65486

65486, 65490

65486, 65487

Answer: Option B

Explanation:

Step 1: int arr[] = {12, 14, 15, 23, 45}; The variable arr is declared as an integer array and initialized.

Step 2: printf("%u, %u\n", arr, &arr); Here,

The base address of the array is 65486.

=> arr, &arr is pointing to the base address of the array arr.

Hence the output of the program is 65486, 65486

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11.

What will be the output of the program ?

#include
#include

int main()
{
    char str1[20] = "Hello", str2[20] = " World";
    printf("%s\n", strcpy(str2, strcat(str1, str2)));
    return 0;
}

Hello

World

Hello World

WorldHello

Answer: Option C

Explanation:

Step 1: char str1[20] = "Hello", str2[20] = " World"; The variable str1 andstr2 is declared as an array of characters and initialized with value "Hello" and " World" respectively.

Step 2: printf("%s\n", strcpy(str2, strcat(str1, str2)));

=> strcat(str1, str2)) it append the string str2 to str1. The result will be stored instr1. Therefore str1 contains "Hello World".

=> strcpy(str2, "Hello World") it copies the "Hello World" to the variable str2.

Hence it prints "Hello World".

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12.

For the following statements will arr[3] and ptr[3] fetch the same character?
char arr[] = "IndiaBIX";
char *ptr = "IndiaBIX";

Yes

Answer: Option A

Explanation:

Yes, both the statements prints the same character 'i'.

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13.

It is not possible to create an array of pointer to structures.

True

False

Answer: Option B

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14.

Can I increase the size of statically allocated array?

Yes

Answer: Option B

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15.

What will be the output of the program?

#include
#include
void fun1(int num, ....);
void fun2(int num, ....);

int main()
{
    fun1(1, "Apple", "Boys", "Cats", "Dogs");
    fun2(2, 12, 13, 14);
    return 0;
}
void fun1(int num, ...)
{
    char *str;
    va_list ptr;
    va_start(ptr, num);
    str = va_arg(ptr, char *);
    printf("%s", str);
}
void fun2(int num, ...)
{
    va_list ptr;
    va_start(ptr, num);
    num = va_arg(ptr, int);
    printf("%d", num);
}

Dogs 12

Cats 14

Boys 13

Apple 12

Answer: Option D

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16.

Point out the error in the following program.

#include
#include
void display(char *s, ...);
void show(char *t, ...);

int main()
{
    display("Hello", 4, 12, 13, 14, 44);
    return 0;
}
void display(char *s, ...)
{
    show(s, ...);
}
void show(char *t, ...)
{
    int a;
    va_list ptr;
    va_start(ptr, s);
    a = va_arg(ptr, int);
    printf("%f", a);
}

Error: invalid function display() call

Error: invalid function show() call

No error

Error: Rvalue required for t

Answer: Option B

Explanation:

The call to show() is improper. This is not the way to pass variable argument list to a function.

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17.

In a function that receives variable number of arguments the fixed arguments passed to the function can be at the end of argument list.

True

False

Answer: Option B

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18.

What will be the output of the program?

#include

int main()
{
    char huge *near *far *ptr1;
    char near *far *huge *ptr2;
    char far *huge *near *ptr3;
    printf("%d, %d, %d\n", sizeof(ptr1), sizeof(*ptr2), sizeof(**ptr3));
    return 0;
}

4, 4, 4

2, 4, 4

4, 4, 2

2, 4, 8

Answer: Option A

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19.

What will be the output of the program?

#include

int main()
{
    char huge *near *ptr1;
    char huge *far *ptr2;
    char huge *huge *ptr3;
    printf("%d, %d, %d\n", sizeof(ptr1), sizeof(ptr2), sizeof(ptr3));
    return 0;
}

4, 4, 8

2, 4, 4

4, 4, 2

2, 4, 8

Answer: Option B

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20. ftell() returns the current position of the pointer in a file stream.

True

False

Answer: Option A

Explanation:

The ftell() function shall obtain the current value of the file-position indicator for the stream pointed to by stream.

Example:


#include 

int main(void)
{
    FILE *stream;
    stream = fopen("MYFILE.TXT", "w+");
    fprintf(stream, "This is a test");
    printf("The file pointer is at byte %ld\n", ftell(stream));
    fclose(stream);
    return 0;
}

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what i did today......................????

Friday, November 19, 2010

VARIABLES & CONSTANTS IN C .....!!

Monday, November 8, 2010

C QUESTION AND ANSWERS: 3