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- Question
Point out the error, if any in the program.
#include<stdio.h> int main() { int i = 1; switch(i) { printf("This is c program."); case 1: printf("Case1"); break; case 2: printf("Case2"); break; } return 0; }
Options- A. Error: No default specified
- B. Error: Invalid printf statement after switch statement
- C. No Error and prints "Case1"
- D. None of above
- Correct Answer
- No Error and prints "Case1"
Explanationswitch(i) becomes switch(1), then the case 1: block is get executed. Hence it prints "Case1".printf("This is c program."); is ignored by the compiler.
Hence there is no error and prints "Case1".
Control Instructions problems
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- 1. Point out the error, if any in the while loop.
#include<stdio.h> int main() { int i=1; while() { printf("%d\n", i++); if(i>10) break; } return 0; }
Options- A. There should be a condition in the while loop
- B. There should be at least a semicolon in the while
- C. The while loop should be replaced with for loop.
- D. No error Discuss
Correct Answer: There should be a condition in the while loop
Explanation:
The while() loop must have conditional expression or it shows "Expression syntax" error.Example: while(i > 10){ ... }
- 2. Point out the error, if any in the program.
#include<stdio.h> int main() { int P = 10; switch(P) { case 10: printf("Case 1"); case 20: printf("Case 2"); break; case P: printf("Case 2"); break; } return 0; }
Options- A. Error: No default value is specified
- B. Error: Constant expression required at line case P:
- C. Error: There is no break statement in each case.
- D. No error will be reported. Discuss
Correct Answer: Error: Constant expression required at line case P:
Explanation:
The compiler will report the error "Constant expression required" in the line case P: . Because, variable names cannot be used with case statements.The case statements will accept only constant expression.
- 3. Point out the error, if any in the for loop.
#include<stdio.h> int main() { int i=1; for(;;) { printf("%d\n", i++); if(i>10) break; } return 0; }
Options- A. There should be a condition in the for loop
- B. The two semicolons should be dropped
- C. The for loop should be replaced with while loop.
- D. No error Discuss
Correct Answer: No error
Explanation:
Step 1: for(;;) this statement will genereate infinite loop.
Step 2: printf("%d\n", i++); this statement will print the value of variable i and increement i by 1(one).
Step 3: if(i>10) here, if the variable i value is greater than 10, then the for loop breaks.
Hence the output of the program is
1
2
3
4
5
6
7
8
9
10
- 4. What will be the output of the program?
#include<stdio.h> int main() { int x=1, y=1; for(; y; printf("%d %d\n", x, y)) { y = x++ <= 5; } printf("\n"); return 0; }
Options- A. 2 1
3 1
4 1
5 1
6 1
7 0 - B. 2 1
3 1
4 1
5 1
6 1 - C. 2 1
3 1
4 1
5 1 - D. 2 2
3 3
4 4
5 5
Discuss
Correct Answer: 2 1
3 1
4 1
5 1
6 1
7 0
- 5. What will be the output of the program?
#include<stdio.h> int main() { int i=3; switch(i) { case 1: printf("Hello\n"); case 2: printf("Hi\n"); case 3: continue; default: printf("Bye\n"); } return 0; }
Options- A. Error: Misplaced continue
- B. Bye
- C. No output
- D. Hello Hi Discuss
Correct Answer: Error: Misplaced continue
Explanation:
The keyword continue cannot be used in switch case. It must be used in for or while or do while loop. If there is any looping statement in switch case then we can use continue.- 6. Point out the error, if any in the while loop.
#include<stdio.h> int main() { void fun(); int i = 1; while(i <= 5) { printf("%d\n", i); if(i>2) goto here; } return 0; } void fun() { here: printf("It works"); }
Options- A. No Error: prints "It works"
- B. Error: fun() cannot be accessed
- C. Error: goto cannot takeover control to other function
- D. No error Discuss
Correct Answer: Error: goto cannot takeover control to other function
Explanation:
A label is used as the target of a goto statement, and that label must be within the same function as the goto statement.Syntax: goto <identifier> ;
Control is unconditionally transferred to the location of a local label specified by <identifier>.
Example:
#include <stdio.h> int main() { int i=1; while(i>0) { printf("%d", i++); if(i==5) goto mylabel; } mylabel: return 0; }Output: 1,2,3,4
- 7. Which of the following errors would be reported by the compiler on compiling the program given below?
#include<stdio.h> int main() { int a = 5; switch(a) { case 1: printf("First"); case 2: printf("Second"); case 3 + 2: printf("Third"); case 5: printf("Final"); break; } return 0; }
Options- A. There is no break statement in each case.
- B. Expression as in case 3 + 2 is not allowed.
- C. Duplicate case case 5:
- D. No error will be reported. Discuss
Correct Answer: Duplicate case case 5:
Explanation:
Because, case 3 + 2: and case 5: have the same constant value 5.- 8. Point out the error, if any in the program.
#include<stdio.h> int main() { int a = 10, b; a >=5? b=100: b=200; printf("%d\n", b); return 0; }
Options- A. 100
- B. 200
- C. Error: L value required for b
- D. Garbage value Discuss
Correct Answer: Error: L value required for b
Explanation:
Variable b is not assigned.It should be like:
b = a >= 5 ? 100 : 200;- 9. Point out the error, if any in the program.
#include<stdio.h> int main() { int i = 1; switch(i) { case 1: printf("Case1"); break; case 1*2+4: printf("Case2"); break; } return 0; }
Options- A. Error: in case 1*2+4 statement
- B. Error: No default specified
- C. Error: in switch statement
- D. No Error Discuss
Correct Answer: No Error
Explanation:
Constant expression are accepted in switchIt prints "Case1"
- 10. Point out the error, if any in the program.
#include<stdio.h> int main() { int a = 10; switch(a) { } printf("This is c program."); return 0; }
Options- A. Error: No case statement specified
- B. Error: No default specified
- C. No Error
- D. Error: infinite loop occurs Discuss
Correct Answer: No Error
Explanation:
There can exists a switch statement, which has no case.
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- 1. Point out the error, if any in the while loop.