# Online C Programming Test - C Programming Test 7

Instruction:

• Total number of questions : 20.
• Time alloted : 20 minutes.
• Each question carry 1 mark, no negative marks.
• DO NOT refresh the page.
• All the best :-).

1.

A short integer is at least 16 bits wide and a long integer is at least 32 bits wide.

A.
 True
B.
 False

Explanation:

The basic C compiler is 16 bit compiler, below are the size of it's data types
The size of short int is 2 bytes wide(16 bits).
The size of long int is 4 bytes wide(32 bits).

2.

Which of the following correctly shows the hierarchy of arithmetic operations in C?

A.
 / + * -
B.
 * - / +
C.
 + - / *
D.
 / * + -

Explanation:

Simply called as BODMAS (Bracket of Division, Multiplication, Addition and Subtraction).

How Do I Remember ? BODMAS !

• B - Brackets first
• O - Orders (ie Powers and Square Roots, etc.)
• DM - Division and Multiplication (left-to-right)
• AS - Addition and Subtraction (left-to-right)

3.

In which order do the following gets evaluated
 1 Relational 2 Arithmetic 3 Logical 4 Assignment

A.
 2134
B.
 1234
C.
 4321
D.
 3214

Explanation:

2. Arithmetic operators: *, /, %, +, -
1. Relational operators: >, <, >=, <=, ==, !=
3. Logical operators : !, &&, ||
4. Assignment operators: =

4.

Associativity has no role to play unless the precedence of operator is same.

A.
 True
B.
 False

Explanation:

Associativity is only needed when the operators in an expression have the same precedence. Usually + and - have the same precedence.

Consider the expression 7 - 4 + 2. The result could be either (7 - 4) + 2 = 5 or 7 - (4 + 2) = 1. The former result corresponds to the case when + and - are left-associative, the latter to when + and - are right-associative.

Usually the addition, subtraction, multiplication, and division operators are left-associative, while the exponentiation, assignment and conditional operators are right-associative. To prevent cases where operands would be associated with two operators, or no operator at all, operators with the same precedence must have the same associativity.

5.

If the binary eauivalent of 5.375 in normalised form is 0100 0000 1010 1100 0000 0000 0000 0000, what will be the output of the program (on intel machine)?

``````#include<stdio.h>
#include<math.h>
int main()
{
float a=5.375;
char *p;
int i;
p = (char*)&a;
for(i=0; i<=3; i++)
printf("%02x\n", (unsigned char)p[i]);
return 0;
}
``````

A.
 40 AC 00 00
B.
 04 CA 00 00
C.
 00 00 AC 40
D.
 00 00 CA 04

6.

A float occupies 4 bytes. If the hexadecimal equivalent of these 4 bytes are A, B, C and D, then when this float is stored in memory in which of the following order do these bytes gets stored?

A.
 ABCD
B.
 DCBA
C.
 0xABCD
D.
 Depends on big endian or little endian architecture

7.

What will be the output of the program?

``````#include<stdio.h>
int sumdig(int);
int main()
{
int a, b;
a = sumdig(123);
b = sumdig(123);
printf("%d, %d\n", a, b);
return 0;
}
int sumdig(int n)
{
int s, d;
if(n!=0)
{
d = n%10;
n = n/10;
s = d+sumdig(n);
}
else
return 0;
return s;
}
``````

A.
 4, 4
B.
 3, 3
C.
 6, 6
D.
 12, 12

8.

Point out the error in the program

``````
f(int a, int b)
{
int a;
a = 20;
return a;
}
``````

A.
 Missing parenthesis in return statement
B.
 The function should be defined as int f(int a, int b)
C.
 Redeclaration of a
D.
 None of above

Explanation:

f(int a, int b) The variable a is declared in the function argument statement.

int a; Here again we are declaring the variable a. Hence it shows the error "Redeclaration of a"

9.

What will be the output of the program?

``````#include<stdio.h>
#define SQUARE(x) x*x

int main()
{
float s=10, u=30, t=2, a;
a = 2*(s-u*t)/SQUARE(t);
printf("Result = %f", a);
return 0;
}
``````

A.
 Result = -100.000000
B.
 Result = -25.000000
C.
 Result = 0.000000
D.
 Result = 100.000000

Explanation:

The macro function SQUARE(x) x*x calculate the square of the given number 'x'. (Eg: 102)

Step 1: float s=10, u=30, t=2, a; Here the variable s, u, t, a are declared as an floating point type and the variable s, u, t are initialized to 10, 30, 2.

Step 2: a = 2*(s-u*t)/SQUARE(t); becomes,

=> a = 2 * (10 - 30 * 2) / t * t; Here SQUARE(t) is replaced by macro to t*t .

=> a = 2 * (10 - 30 * 2) / 2 * 2;

=> a = 2 * (10 - 60) / 2 * 2;

=> a = 2 * (-50) / 2 * 2 ;

=> a = 2 * (-25) * 2 ;

=> a = (-50) * 2 ;

=> a = -100;

Step 3: printf("Result=%f", a); It prints the value of variable 'a'.

Hence the output of the program is -100

10.

Preprocessor directive #undef can be used only on a macro that has been #define earlier

A.
 True
B.
 False

Explanation:

True, #undef can be used only on a macro that has been #define earlier

Example: #define PI 3.14

We can undefine PI macro by #undef PI

11.

Which statement will you add to the following program to ensure that the program outputs "IndiaBIX" on execution?

``````#include<stdio.h>

int main()
{
char s[] = "IndiaBIX";
char t[25];
char *ps, *pt;
ps = s;
pt = t;
while(*ps)
*pt++ = *ps++;

/* Add a statement here */
printf("%s\n", t);
return 0;
}
``````

A.
 *pt='';
B.
 pt='\0';
C.
 pt='\n';
D.
 *pt='\0';

12.

What will be the output of the program ?

``````#include<stdio.h>
#include<string.h>

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

A.
 Hello
B.
 World
C.
 Hello World
D.
 WorldHello

Explanation:

Step 1: char str1[20] = "Hello", str2[20] = " World"; The variable str1 and str2 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 in str1. Therefore str1 contains "Hello World".

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

Hence it prints "Hello World".

13.

What will be the output of the program ?

``````#include<stdio.h>

int main()
{
char p[] = "%d\n";
p[1] = 'c';
printf(p, 65);
return 0;
}
``````

A.
 A
B.
 a
C.
 c
D.
 65

Explanation:

Step 1: char p[] = "%d\n"; The variable p is declared as an array of characters and initialized with string "%d".

Step 2: p[1] = 'c'; Here, we overwrite the second element of array p by 'c'. So array p becomes "%c".

Step 3: printf(p, 65); becomes printf("%c", 65);

Therefore it prints the ASCII value of 65. The output is 'A'.

14.

What will be the output of the program ?

``````#include<stdio.h>
void swap(char *, char *);

int main()
{
char *pstr[2] = {"Hello", "IndiaBIX"};
swap(pstr[0], pstr[1]);
printf("%s\n%s", pstr[0], pstr[1]);
return 0;
}
void swap(char *t1, char *t2)
{
char *t;
t=t1;
t1=t2;
t2=t;
}
``````

A.
 IndiaBIXHello
B.
C.
 HelloIndiaBIX
D.
 IelloHndiaBIX

Explanation:

Step 1: void swap(char *, char *); This prototype tells the compiler that the function swap accept two strings as arguments and it does not return anything.

Step 2: char *pstr[2] = {"Hello", "IndiaBIX"}; The variable pstr is declared as an pointer to the array of strings. It is initialized to

pstr[0] = "Hello", pstr[1] = "IndiaBIX"

Step 3: swap(pstr[0], pstr[1]); The swap function is called by "call by value". Hence it does not affect the output of the program.

If the swap function is "called by reference" it will affect the variable pstr.

Step 4: printf("%s\n%s", pstr[0], pstr[1]); It prints the value of pstr[0] and pstr[1].

Hence the output of the program is

Hello
IndiaBIX

15.

What will be the output of the program ?

``````#include<stdio.h>

int main()
{
union var
{
int a, b;
};
union var v;
v.a=10;
v.b=20;
printf("%d\n", v.a);
return 0;
}
``````

A.
 10
B.
 20
C.
 30
D.
 0

16.

Nested unions are allowed

A.
 True
B.
 False

17.

Can we have an array of bit fields?

A.
 Yes
B.
 No

18.

To scan a and b given below, which of the following scanf() statement will you use?

``````#include<stdio.h>

float a;
double b;
``````

A.
 scanf("%f %f", &a, &b);
B.
 scanf("%Lf %Lf", &a, &b);
C.
 scanf("%f %Lf", &a, &b);
D.
 scanf("%f %lf", &a, &b);

Explanation:

To scan a float value, %f is used as format specifier.

To scan a double value, %lf is used as format specifier.

Therefore, the answer is scanf("%f %lf", &a, &b);

19.

What will be the output of the program?

``````#define P printf("%d\n", -1^~0);
#define M(P) int main()\
{\
P\
return 0;\
}
M(P)
``````

A.
 1
B.
 0
C.
 -1
D.
 2

20.

What will be the output of the program?

``````#include<stdio.h>

int main()
{
unsigned int res;
res = (64 >>(2+1-2)) & (~(1<<2));
printf("%d\n", res);
return 0;
}
``````

A.
 32
B.
 64
C.
 0
D.
 128