Java C Control Statement Quiz 3 | Java C Control Question and Answers

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

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”.

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.

Step 1: int i = 0; here variable i is an integer type and initialized to ´0´.
Step 2: for(; i<=5; i++); variable i=0 is already assigned in previous step. The semi-colon at the end of this for loop tells, “there is no more statement is inside the loop”.

Loop 1: here i=0, the condition in for(; 0<=5; i++) loop satisfies and then i is incremented by ´1´(one)
Loop 2: here i=1, the condition in for(; 1<=5; i++) loop satisfies and then i is incremented by ´1´(one)
Loop 3: here i=2, the condition in for(; 2<=5; i++) loop satisfies and then i is incremented by ´1´(one)
Loop 4: here i=3, the condition in for(; 3<=5; i++) loop satisfies and then i is increemented by ´1´(one)
Loop 5: here i=4, the condition in for(; 4<=5; i++) loop satisfies and then i is incremented by ´1´(one)
Loop 6: here i=5, the condition in for(; 5<=5; i++) loop satisfies and then i is incremented by ´1´(one)
Loop 7: here i=6, the condition in for(; 6<=5; i++) loop fails and then i is not incremented.
Step 3: printf(“%d”, i); here the value of i is 6. Hence the output is ´6´.

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

Variable b is not assigned.
It should be like:
b = a >= 5 ? 100 : 200;

Step 1: char str[]=”C-program”; here variable str contains “C-program”.
Step 2: int a = 5; here variable a contains “5”.
Step 3: printf(a >10?”Ps “:”%s “, str); this statement can be written as

if(a > 10)
{
printf(“Ps “);
}
else
{
printf(“%s “, str);
}
Here we are checking a > 10 means 5 > 10. Hence this condition will be failed. So it prints variable str.

Hence the output is “C-program”.

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

Example: while(i > 10){ … }

Constant expression are accepted in switch

It prints “Case1”

There can exists a switch statement, which has no case.

Step 1: x=y=z=1; here the variables x ,y, z are initialized to value ´1´.

Step 2: z = ++x || ++y && ++z; becomes z = ( (++x) || (++y && ++z) ). Here ++x becomes 2. So there is no need to check the other side because ||(Logical OR) condition is satisfied.(z = (2 || ++y && ++z)). There is no need to process ++y && ++z. Hence it returns ´1´. So the value of variable z is ´1´

Step 3: printf(“x=%d, y=%d, z=%d “, x, y, z); It prints “x=2, y=1, z=1”. here x is increemented in previous step. y and z are not increemented.

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.

for(i<=5 && i>=-1; ++i; i>0) so expression i<=5 && i>=-1 initializes for loop. expression ++i is the loop condition. expression i>0 is the increment expression.

In for( i <= 5 && i >= -1; ++i; i>0) expression i<=5 && i>=-1 evaluates to one.

Loop condition always get evaluated to true. Also at this point it increases i by one.

An increment_expression i>0 has no effect on value of i.so for loop get executed till the limit of integer (ie. 65535)

Step 1: int a=0, b=1, c=3; here variable a, b, and c are declared as integer type and initialized to 0, 1, 3 respectively.

Step 2: *((a) ? &b : &a) = a ? b : c; The right side of the expression(a?b:c) becomes (0?1:3). Hence it return the value ´3´.

The left side of the expression *((a) ? &b : &a) becomes *((0) ? &b : &a). Hence this contains the address of the variable a *(&a).

Step 3: *((a) ? &b : &a) = a ? b : c; Finally this statement becomes *(&a)=3. Hence the variable a has the value ´3´.

Step 4: printf(“%d, %d, %d “, a, b, c); It prints “3, 1, 3”.

Here unsigned int size is 2 bytes. It varies from 0,1,2,3, … to 65535.

Step 1:unsigned int i = 65536; here variable i becomes ´0´(zero). because unsigned int varies from 0 to 65535.

Step 2: while(i != 0) this statement becomes while(0 != 0). Hence the while(FALSE) condition is not satisfied. So, the inside the statements of while loop will not get executed.

Hence there is no output.

Note: Don´t forget that the size of int should be 2 bytes. If you run the above program in GCC it may run infinite loop, because in Linux platform the size of the integer is 4 bytes.

Initially variables a = 500, b = 100 and c is not assigned.

Step 1: if(!a >= 400)
Step 2: if(!500 >= 400)
Step 3: if(0 >= 400)
Step 4: if(FALSE) Hence the if condition is failed.
Step 5: So, variable c is assigned to a value ´200´.
Step 6: printf(“b = %d c = %d “, b, c); It prints value of b and c.
Hence the output is “b = 100 c = 200”