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- Question
What will be the output of the program?
#include<stdio.h> int main() { int i=4; switch(i) { default: printf("This is default\n"); case 1: printf("This is case 1\n"); break; case 2: printf("This is case 2\n"); break; case 3: printf("This is case 3\n"); } return 0; }
Options- A. This is default
This is case 1 - B. This is case 3
This is default - C. This is case 1
This is case 3 - D. This is default
- Correct Answer
- This is default
This is case 1
ExplanationIn the very begining of switch-case statement default statement is encountered. So, it prints "This is default".In default statement there is no break; statement is included. So it prints the case 1 statements. "This is case 1".
Then the break; statement is encountered. Hence the program exits from the switch-case block.
More questions
- 1. Bitwise & and | are unary operators
Options- A. True
- B. False Discuss
Correct Answer: False
Explanation:
Bitwise & and | are not unary operators only bitwise ! is unary operator.- 2. Point out the error in the program.
#include<stdio.h> const char *fun(); int main() { char *ptr = fun(); return 0; } const char *fun() { return "Hello"; }
Options- A. Error: Lvalue required
- B. Error: cannot convert 'const char *' to 'char *'.
- C. No error and No output
- D. None of above Discuss
Correct Answer: No error and No output
- 3. In a function that receives variable number of arguments the fixed arguments passed to the function can be at the end of argument list.
Options- A. True
- B. False Discuss
Correct Answer: False
- 4. Point out the error in the following program.
#include<stdio.h> #include<stdarg.h> int main() { void display(char *s, int num1, int num2, ...); display("Hello", 4, 2, 12.5, 13.5, 14.5, 44.0); return 0; } void display(char *s, int num1, int num2, ...) { double c; char s; va_list ptr; va_start(ptr, s); c = va_arg(ptr, double); printf("%f", c); }
Options- A. Error: invalid arguments in function display()
- B. Error: too many parameters
- C. Error: in va_start(ptr, s);
- D. No error Discuss
Correct Answer: Error: in va_start(ptr, s);
Explanation:
We should have use va_start(ptr, num2);- 5. Point out the error if any in the following program (Turbo C).
#include<stdio.h> #include<stdarg.h> void display(int num, ...); int main() { display(4, 'A', 'a', 'b', 'c'); return 0; } void display(int num, ...) { char c; int j; va_list ptr; va_start(ptr, num); for(j=1; j<=num; j++) { c = va_arg(ptr, char); printf("%c", c); } }
Options- A. Error: unknown variable ptr
- B. Error: Lvalue required for parameter
- C. No error and print A a b c
- D. No error and print 4 A a b c Discuss
Correct Answer: No error and print A a b c
- 6. Point out the error in the following program.
#include<stdio.h> #include<stdarg.h> void varfun(int n, ...); int main() { varfun(3, 7, -11, 0); return 0; } void varfun(int n, ...) { va_list ptr; int num; num = va_arg(ptr, int); printf("%d", num); }
Options- A. Error: ptr has to be set at begining
- B. Error: ptr must be type of va_list
- C. Error: invalid access to list member
- D. No error Discuss
Correct Answer: Error: ptr has to be set at begining
Explanation:
Using va_start(ptr, int);- 7. Point out the error in the following program.
#include<stdio.h> #include<stdarg.h> 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); }
Options- A. Error: invalid function display() call
- B. Error: invalid function show() call
- C. No error
- D. Error: Rvalue required for t Discuss
Correct Answer: Error: invalid function show() call
Explanation:
The call to show() is improper. This is not the way to pass variable argument list to a function.- 8. What will be the output of the program?
#include<stdio.h> #include<math.h> int main() { float i = 2.5; printf("%f, %d", floor(i), ceil(i)); return 0; }
Options- A. 2, 3
- B. 2.000000, 3
- C. 2.000000, 0
- D. 2, 0 Discuss
Correct Answer: 2.000000, 0
Explanation:
Both ceil() and floor() return the integer found as a double.floor(2.5) returns the largest integral value(round down) that is not greater than 2.5. So output is 2.000000.
ceil(2.5) returns 3, while converting the double to int it returns '0'.
So, the output is '2.000000, 0'.- 9. What will function gcvt() do?
Options- A. Convert vector to integer value
- B. Convert floating-point number to a string
- C. Convert 2D array in to 1D array.
- D. Covert multi Dimensional array to 1D array Discuss
Correct Answer: Convert floating-point number to a string
Explanation:
The gcvt() function converts a floating-point number to a string. It converts given value to a null-terminated string.#include <stdlib.h> #include <stdio.h> int main(void) { char str[25]; double num; int sig = 5; /* significant digits */ /* a regular number */ num = 9.876; gcvt(num, sig, str); printf("string = %s\n", str); /* a negative number */ num = -123.4567; gcvt(num, sig, str); printf("string = %s\n", str); /* scientific notation */ num = 0.678e5; gcvt(num, sig, str); printf("string = %s\n", str); return(0); }Output:
string = 9.876
string = -123.46
string = 67800- 10. What will be the output of the program?
#include<stdio.h> int main() { int i; char c; for(i=1; i<=5; i++) { scanf("%c", &c); /* given input is 'b' */ ungetc(c, stdout); printf("%c", c); ungetc(c, stdin); } return 0; }
Options- A. bbbb
- B. bbbbb
- C. b
- D. Error in ungetc statement. Discuss
Correct Answer: b
Explanation:
The ungetc() function pushes the character c back onto the named input stream, which must be open for reading.This character will be returned on the next call to getc or fread for that stream.
One character can be pushed back in all situations.
A second call to ungetc without a call to getc will force the previous character to be forgotten.
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