Basics Of C Language: Trivia Quiz!

1. Which of the following is not a valid variable name declaration in the C programming language?

Int _a3

Int a_3

Int 3_a

Int _3a

In the C programming language, variable names cannot start with a number. They must start with either a letter or an underscore. Therefore, the declaration "int 3_a" is not valid because it starts with a number.

Explanation

In the C programming language, variable names cannot start with a number. They must start with either a letter or an underscore. Therefore, the declaration "int 3_a" is not valid because it starts with a number.

2. What is the output of this C code? #include <stdio.h> int main() { int i = 3; int l = i / -2; int k = i % -2; printf("%d %d\n", l, k); return 0; }

Compile time error

-1 1

1 -1

Implementation defined

The code initializes the variable i with the value 3. It then performs integer division between i and -2, resulting in l being assigned the value -1. It also calculates the remainder of the division, which is assigned to k. Since the remainder of dividing 3 by -2 is 1, k is assigned the value 1. The printf statement then prints the values of l and k, resulting in the output -1 1.

Explanation

The code initializes the variable i with the value 3. It then performs integer division between i and -2, resulting in l being assigned the value -1. It also calculates the remainder of the division, which is assigned to k. Since the remainder of dividing 3 by -2 is 1, k is assigned the value 1. The printf statement then prints the values of l and k, resulting in the output -1 1.

3. What is the output of this C code? #include <stdio.h> #define square(x) x*x int main() { int a = 2; int sqr = square(a+1); printf("%d",sqr); return 0; }

3

4

5

6

#define replaces the square(x) with x*x with the value of ‘x’ provided before compiling the code.
Therefore, square(a+1) would become a+1*a+1, which is 2+1*2+1 = 2+2+1 = 5

Explanation

#define replaces the square(x) with x*x with the value of ‘x’ provided before compiling the code.
Therefore, square(a+1) would become a+1*a+1, which is 2+1*2+1 = 2+2+1 = 5

4. What is the output of this C code? #include<stdio.h> int main() { printf("%d", printf("Hello 2018! ")); return 0; }

Compile time error

Hello 2018!

12

Hello 2018! 12

printf() function returns the number of characters it has printed.

Explanation

printf() function returns the number of characters it has printed.

5. What will happen if the below program is executed? #include int main() { int main = 3; printf("%d", main); return 0; }

It will cause a compile-time error.

It will cause a run-time error.

It will run without any error and prints 3.

No output and program will terminate abnormally.

In a function a variable can take the same name as that of the function

Explanation

In a function a variable can take the same name as that of the function

6. What is 'short int' in C programming?

A basic data type

Qualifier

Modifier

‘short’ is qualifier, ‘int’ is basic data type.

In C programming, 'short int' is a data type that combines the 'short' qualifier with the 'int' basic data type. The 'short' qualifier is used to reduce the range of values that can be stored in the variable, while the 'int' data type is a basic data type used to represent integer values. Therefore, 'short int' is a combination of a qualifier and a basic data type.

Explanation

In C programming, 'short int' is a data type that combines the 'short' qualifier with the 'int' basic data type. The 'short' qualifier is used to reduce the range of values that can be stored in the variable, while the 'int' data type is a basic data type used to represent integer values. Therefore, 'short int' is a combination of a qualifier and a basic data type.

7. What is the output of this C code? #include <stdio.h> int main() { if(printf("foo")) switch(printf("bar")) while(printf("infinite")); return 0; }

Foobarinfiniteinfiniteinfinite………∞

Foobar

Foobarinfinite

Compile-time error

Since printf() return the no. of characters printed so, in if it would be 3 and ‘foo’ is printed. Similarly, in switch it would also be 3, due to ‘bar’. Also, switch will search for ‘case 3’ but, it doesn’t find any ‘case 3’, so it comes out of switch and hence program terminates.

Explanation

Since printf() return the no. of characters printed so, in if it would be 3 and ‘foo’ is printed. Similarly, in switch it would also be 3, due to ‘bar’. Also, switch will search for ‘case 3’ but, it doesn’t find any ‘case 3’, so it comes out of switch and hence program terminates.

8. What is the output of the following C code? #include int main() { int i; if(i=0,2,3) printf("Hello "); else printf("World "); printf("%d",i); return 0; }

World 3

World 0

Hello 0

Hello 3

The code first initializes the variable i without assigning any value to it. Then, it checks the condition (i=0,2,3). In this condition, the comma operator is used, which evaluates each expression separated by commas and returns the value of the last expression. In this case, the value of the condition is 3. Since the condition is true, the code executes the printf("Hello ") statement. After that, it prints the value of i, which is 0. Therefore, the output is "Hello 0".

Explanation

The code first initializes the variable i without assigning any value to it. Then, it checks the condition (i=0,2,3). In this condition, the comma operator is used, which evaluates each expression separated by commas and returns the value of the last expression. In this case, the value of the condition is 3. Since the condition is true, the code executes the printf("Hello ") statement. After that, it prints the value of i, which is 0. Therefore, the output is "Hello 0".

9. What is the output of the following C code? #include <stdio.h> int main() { int i=0,j=0; while(i<5,j<10) { i++; j++; } printf("%d %d",i,j); return 0; }

5 5

10 10

10 5

5 10

Here, ‘,’ operator in while() will act as LOGICAL OR operator i.e. (i

Explanation

Here, ‘,’ operator in while() will act as LOGICAL OR operator i.e. (i

10. Does this code compile without error(s)? // no header files included int main() { int k; { int k; for (k = 0; k < 10; k++); // there's a semicolon in the end! } return 0; }

Yes

No

Depends on the compiler

Depends on the C standard you are using

Header files are only required for inbuilt fuctions(like: printf(),scanf()).

Data-types(like:int, float etc) are already known to the compiler.

When variable ‘k’ is declared second time, it’s declared inside another block, hence no contradiction of having two variables with same name. Also, putting a ‘;’ at the end of ‘for’ loop takes it as an empty statement.

Explanation

Header files are only required for inbuilt fuctions(like: printf(),scanf()).

Data-types(like:int, float etc) are already known to the compiler.

When variable ‘k’ is declared second time, it’s declared inside another block, hence no contradiction of having two variables with same name. Also, putting a ‘;’ at the end of ‘for’ loop takes it as an empty statement.

11. What is the output of the following C code? #include int main() { signed char chr; chr = 128; printf("%d\n", chr); return 0; }

128

-128

Depends on the compiler

Compile time error as char can't be assigned an int value

When we exceed the maximum limit of char, it starts taking values from the beginning i.e. -128.

Explanation

When we exceed the maximum limit of char, it starts taking values from the beginning i.e. -128.

12. What will be the output of this C code? #include <stdio.h> int main() { float f1 = 0.1; if (f1 == 0.1) printf("equal\n"); else printf("not equal\n"); return 0; }

Equal

Not equal

Output depends upon compiler

Compile-time error

Here, 0.1 is taken as double (by default) by the compiler. So. a float value is not equal to a double value.

Explanation

Here, 0.1 is taken as double (by default) by the compiler. So. a float value is not equal to a double value.

13. What is the output of this C code? #include int main() { int i; scanf("%d", &i); // inputting 071 printf("%d", i); return 0; }

071

71

57

56

When we declare or assign an int variable starting with 0, the compiler considers it as Hexadecimal value. But when, printf() prints, it prints it in its decimal equivalent. ‘071’ in Hexadecimal = ‘57’ in decimal But if we take input by scanf() it ignore the first digit '0'.

Explanation

When we declare or assign an int variable starting with 0, the compiler considers it as Hexadecimal value. But when, printf() prints, it prints it in its decimal equivalent. ‘071’ in Hexadecimal = ‘57’ in decimal But if we take input by scanf() it ignore the first digit '0'.

14. What will be the value of d in the following program? #include int main() { int a = 10, b = 5, c = 5; int d; d = b + (c == a); printf("%d", d); }

Syntax Error

1

5

10

In the given program, the value of d is determined by the expression "b + (c == a)". Here, (c == a) is a comparison expression which evaluates to false because c is not equal to a. In C language, false is represented by 0 and true is represented by 1. So, the expression becomes "b + 0" which is equal to b. Therefore, the value of d will be 5, as the value of b is 5.

Explanation

In the given program, the value of d is determined by the expression "b + (c == a)". Here, (c == a) is a comparison expression which evaluates to false because c is not equal to a. In C language, false is represented by 0 and true is represented by 1. So, the expression becomes "b + 0" which is equal to b. Therefore, the value of d will be 5, as the value of b is 5.

15. What is the Output of this C code? #include <stdio.h> int main() { int a = 10; if (a == a--) printf("TRUE 1\t"); a = 10; if (a == --a) printf("TRUE 2\t"); }

TRUE 1

TRUE 2

TRUE 1 TRUE 2

Compiler Dependent

The output of this C code is Compiler Dependent. This is because the behavior of the code depends on how the compiler evaluates the expressions and the order in which the side effects of the expressions are applied. In the first if statement, the expression "a == a--" compares the value of a with the value of a before it is decremented. Some compilers may evaluate the expression as true because the value of a is copied before it is decremented, while others may evaluate it as false because the value of a is decremented before the comparison. In the second if statement, the expression "a == --a" compares the value of a with the value of a after it is decremented. Again, the behavior may vary depending on the compiler.

Explanation

The output of this C code is Compiler Dependent. This is because the behavior of the code depends on how the compiler evaluates the expressions and the order in which the side effects of the expressions are applied. In the first if statement, the expression "a == a--" compares the value of a with the value of a before it is decremented. Some compilers may evaluate the expression as true because the value of a is copied before it is decremented, while others may evaluate it as false because the value of a is decremented before the comparison. In the second if statement, the expression "a == --a" compares the value of a with the value of a after it is decremented. Again, the behavior may vary depending on the compiler.