The continue statement cannot be used with
Answer: D
Correct answer: switch
continue means “begin the next iteration”, so it belongs to the three loops. A switch has no iterations, so a continue written inside one is either applied to an enclosing loop or, if there is none, rejected. break, by contrast, works in all four.
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The output of the code below is
#include <stdio.h>
int x;
void main()
{
if (x)
printf("hi");
else
printf("how are u");
}
Answer: B
Correct answer: how are u
x is declared at file scope, so it has static storage duration and is zero-initialised - not left holding junk, as an uninitialised local would be. The condition is therefore false and the else branch runs.
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What will be the output of the program?
#include <stdio.h>
int main()
{
int a = 500, b = 100, c;
if(!a >= 400)
b = 300;
c = 200;
printf("b = %d c = %d\n", b, c);
return 0;
}
Answer: D
Correct answer: b = 100 c = 200
! binds far tighter than >=, so the condition is (!a) >= 400, not !(a >= 400). With a non-zero, !a is 0, and \(0 \ge 400\) is false - so b keeps its initial 100. The assignment c = 200 is a separate statement, outside the if altogether, so it always runs.
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By default, the data type of a constant without a decimal point is int, whereas the one with a decimal point is a double.
Answer: A
Correct answer: Yes
An unsuffixed integer literal such as 10 has type int, and one with a decimal point such as 3.14 has type double - not float, which catches many people out. Suffixes override this: 10L, 10U, 3.14f.
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When 1 is entered, The output of the code below is?
#include <stdio.h>
void main()
{
int ch;
printf("enter a value btw 1 to 2:");
scanf("%d", &ch);
switch (ch)
{
case 1:
printf("1\n");
default:
printf("2\n");
}
}
Answer: C
Correct answer: 1 2
With no break after case 1, execution falls through into the default label - a label marks an entry point, not the boundary of a block. So both printf calls run, giving 1 and then 2 on the next line.
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Predict the output of the below program:
#include <stdio.h>
#define EVEN 0
#define ODD 1
int main()
{
int i = 3;
switch (i & 1)
{
case EVEN: printf("Even");
break;
case ODD: printf("Odd");
break;
default: printf("Default");
}
return 0;
}
Answer: B
Correct answer: Odd
i & 1 isolates the lowest bit, which is 1 for an odd number and 0 for an even one. With i equal to 3 the expression is 1, matching ODD, so Odd is printed. This bitwise test is a common alternative to i % 2.
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What will be the output of the program, if a short int is 2 bytes wide?
#include <stdio.h>
int main()
{
short int i = 0;
for(i<=5 && i>=-1; ++i; i>0)
printf("%u,", i);
return 0;
}
Answer: A
Correct answer: 1 ... 65535
The three clauses are not doing what they look like. The initialisation i<=5 && i>=-1 merely computes a value and throws it away; the condition is ++i, so the loop runs while the incremented value is non-zero; and the update i>0 changes nothing. So i simply counts up and is printed with %u, which reinterprets the negative half of the range as large positive numbers - ending when the value wraps back to 0. (Signed overflow is formally undefined, so this is a description of what happens in practice, not a guarantee.)
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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");
}
Answer: C
Correct answer: Error: goto cannot takeover control to other function
A goto can only jump to a label within the same function. The label here: is inside fun(), so it is not visible in main and the compiler reports an undefined label. Transferring control between functions is what a call, or setjmp/longjmp, is for.
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The output of the code below is
#include <stdio.h>
void main()
{
int x = 0;
if (x == 0)
printf("hi");
else
printf("how are u");
printf("hello");
}
Answer: D
Correct answer: hihello
The condition is true, so the if branch prints hi. The second printf is outside the if-else altogether - there are no braces, so the else takes only the single statement that follows it - and therefore runs whichever branch was taken. The output is hihello, with no separator since neither string has a newline.
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The modulus operator cannot be used with a long double.
Answer: A
Correct answer: True
The % operator requires integer operands, so it cannot be applied to float, double or long double. For a floating-point remainder the library provides fmodl() in <math.h>.
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