Exercise: Control Instructions

Questions for: Control Instructions

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;
}
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.
Answer: B

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.

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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;
}
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.
Answer: C

Because, case 3 + 2: and case 5: have the same constant value 5.

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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;
}
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
Answer: A

The while() loop must have conditional expression or it shows "Expression syntax" error.

Example: while(i > 10){ ... }

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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;
}
A:
Error: No default specified
B:
Error: Invalid printf statement after switch statement
C:
No Error and prints "Case1"
D:
None of above
Answer: C

switch(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".

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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;
}
A:
Error: No case statement specified
B:
Error: No default specified
C:
No Error
D:
Error: infinite loop occurs
Answer: C
There can exists a switch statement, which has no case.
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