Question 1

Which of the following is not user defined data type? 1 : struct book
{
    char name[10];
    float price;
    int pages;
}; 2 : long int l = 2.35; 3 : enum day {Sun, Mon, Tue, Wed};

Accepted Answer

C data types classification are Primary data types int char float double void Secondary data types (or) User-defined data type Array Pointer Structure Union Enum So, clearly long int l = 2.35; is not User-defined data type. (i.e.long int l = 2.35; is the answer.)

Question 2

Is there any difference between following declarations? 1 : extern int fun(); 2 : int fun();

Accepted Answer

extern int fun(); declaration in C is to indicate the existence of a global function and it is defined externally to the current module or in another file. int fun(); declaration in C is to indicate the existence of a function inside the current module or in the same file.

Question 3

When we mention the prototype of a function?

Accepted Answer

A function prototype in C or C++ is a declaration of a function that omits the function body but does specify the function's name, argument types and return type. While a function definition specifies what a function does, a function prototype can be thought of as specifying its interface.

Question 4

In the following program where is the variable a getting defined and where it is getting declared? #include<stdio.h>
int main()
{
    extern int a;
    printf("%d
", a);
    return 0;
}
int a=20;

Accepted Answer

- During declaration we tell the datatype of the Variable. - During definition the value is initialized.

Question 5

Which of the following statements should be used to obtain a remainder after dividing 3.14 by 2.1?

Accepted Answer

fmod(x,y) - Calculates x modulo y, the remainder of x/y. This function is the same as the modulus operator. But fmod() performs floating point divisions. Example: #include #include int main () { printf ("fmod of 3.14/2.1 is %lf ", fmod (3.14,2.1) ); return 0; } Output: fmod of 3.14/2.1 is 1.040000

Question 6

Is the following statement a declaration or definition? extern int i;

Accepted Answer

Declaring is the way a programmer tells the compiler to expect a particular type, be it a variable, class/struct/union type, a function type (prototype) or a particular object instance. (ie. extern int i) Declaration never reserves any space for the variable or instance in the program's memory; it simply a "hint" to the compiler that a use of the variable or instance is expected in the program. This hinting is technically called "forward reference".

Question 7

How would you round off a value from 1.66 to 2.0?

Accepted Answer

/* Example for ceil() and floor() functions: */ #include #include int main() { printf(" Result : %f" , ceil(1.44) ); printf(" Result : %f" , ceil(1.66) ); printf(" Result : %f" , floor(1.44) ); printf(" Result : %f" , floor(1.66) ); return 0; } // Output: // Result : 2.000000 // Result : 2.000000 // Result : 1.000000 // Result : 1.000000

Question 8

What are the types of linkages?

Accepted Answer

External Linkage-> means global, non-static variables and functions. Internal Linkage-> means static variables and functions with file scope. None Linkage-> means Local variables.

Question 9

Identify which of the following are declarations 1 : extern int x; 2 : float square ( float x ) { ... } 3 : double pow(double, double);

Accepted Answer

extern int x; - is an external variable declaration. double pow(double, double); - is a function prototype declaration. Therefore, 1 and 3 are declarations. 2 is definition.

Question 10

By default a real number is treated as a

Accepted Answer

In computing, 'real number' often refers to non-complex floating-point numbers. It include both rational numbers, such as 42 and 3/4, and irrational numbers such as pi = 3.14159265... When the accuracy of the floating point number is insufficient, we can use the double to define the number. The double is same as float but with longer precision and takes double space (8 bytes) than float. To extend the precision further we can use long double which occupies 10 bytes of memory space.

Question 11

Which of the following special symbol allowed in a variable name?

Accepted Answer

Variable names in C are made up of letters (upper and lower case) and digits. The underscore character ("_") is also permitted. Names must not begin with a digit. Examples of valid (but not very descriptive) C variable names: => foo => Bar => BAZ => foo_bar => _foo42 => _ => QuUx

Question 12

What will be the output of the program in 16 bit platform (Turbo C under DOS)? #include<stdio.h>
int main()
{
    extern int i;
    i = 20;
    printf("%d
", sizeof(i));
    return 0;
}

Accepted Answer

Linker Error : Undefined symbol 'i' The statement extern int i specifies to the compiler that the memory for 'i' is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name 'i' is available in any other program with memory space allocated for it. Hence a linker error has occurred.

Question 13

What is the output of the program given below? #include<stdio.h>
int main()
{
    enum status { pass, fail, atkt};
    enum status stud1, stud2, stud3;
    stud1 = pass;
    stud2 = atkt;
    stud3 = fail;
    printf("%d, %d, %d
", stud1, stud2, st…

Accepted Answer

enum takes the format like {0,1,2..) so pass=0, fail=1, atkt=2 stud1 = pass (value is 0) stud2 = atkt (value is 2) stud3 = fail (value is 1) Hence it prints 0, 2, 1

Question 14

What is the output of the program? #include<stdio.h>
int main()
{
    extern int a;
    printf("%d
", a);
    return 0;
}
int a=20;

Accepted Answer

extern int a; indicates that the variable a is defined elsewhere, usually in a separate source code module. printf("%d
", a); it prints the value of local variable int a = 20. Because, whenever there is a conflict between local variable and global variable, local variable gets the highest priority. So it prints 20.

Question 15

What will be the output of the program? #include<stdio.h>
int X=40;
int main()
{
    int X=20;
    printf("%d
", X);
    return 0;
}

Accepted Answer

Whenever there is conflict between a local variable and global variable, the local variable gets priority.

Question 16

What is the output of the program #include<stdio.h>
int main()
{
    extern int fun(float);
    int a;
    a = fun(3.14);
    printf("%d
", a);
    return 0;
}
int fun(int aa)
{
	return (int)++aa;
}

Accepted Answer

2 Errors 1. Type mismatch in redeclaration of fun 2. Type mismatch in parameter aa

Question 17

In the following program how long will the for loop get executed? #include<stdio.h>
int main()
{
    int i=5;
    for(;scanf("%s", &i); printf("%d
", i));
    return 0;
}

Accepted Answer

During the for loop execution scanf() ask input and then printf() prints that given input. This process will be continued repeatedly because, scanf() returns the number of input given, the condition is always true(user gives a input means it reurns '1'). Hence this for loop would get executed infinite times.

Question 18

What is the output of the program #include<stdio.h>
int main()
{
    int a[5] = {2, 3};
    printf("%d, %d, %d
", a[2], a[3], a[4]);
    return 0;
}

Accepted Answer

When an automatic array is partially initialized, the remaining elements are initialized to 0.

Question 19

What is the output of the program in Turbo C (in DOS 16-bit OS)? #include<stdio.h>
int main()
{
    char *s1;
    char far *s2;
    char huge *s3;
    printf("%d, %d, %d
", sizeof(s1), sizeof(s2), sizeof(s3));
    return 0;
}

Accepted Answer

Any pointer size is 2 bytes. (only 16-bit offset) So, char *s1 = 2 bytes. So, char far *s2; = 4 bytes. So, char huge *s3; = 4 bytes. A far, huge pointer has two parts: a 16-bit segment value and a 16-bit offset value. Since C is a compiler dependent language, it may give different output in other platforms. The above program works fine in Windows (TurboC), but error in Linux (GCC Compiler).

Question 20

What is the output of the program #include<stdio.h>
int main()
{
    int x = 10, y = 20, z = 5, i;
    i = x < y < z;
    printf("%d
", i);
    return 0;
}

Accepted Answer

Since x < y turns to be TRUE it is replaced by 1. Then 1 < z is compared and to be TRUE. The 1 is assigned to i.

Declarations and Initializations Questions