What is the output of this C code?
int main()
{
printf("before continue ");
continue;
printf("after continue\n");
}
Answer: D
Correct answer: Compile time error
continue must appear inside a loop. In a function body with no loop around it there is no next iteration to jump to, so the compiler rejects the program - neither printf ever runs.
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Predict the output of the below program:
#include <stdio.h>
int main()
{
int i = 3;
switch(i)
{
printf("Outside ");
case 1: printf("Geeks");
break;
case 2: printf("Quiz");
break;
defau1t: printf("GeeksQuiz");
}
return 0;
}
Answer: C
Correct answer: Nothing gets printed
Two things happen here. The printf("Outside ") sits inside the switch body but before any label, so control never reaches it - a switch jumps straight to a matching label. And defau1t is spelt with the digit one, so it is an ordinary goto label, not the default case. With i equal to 3 nothing matches and there is no real default, so the switch does nothing at all.
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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;
}
Answer: C
Correct answer: Error: L value required for b
Assignment binds more loosely than the conditional operator, so the line parses as
(a >= 5 ? b = 100 : b) = 200;
and the left-hand side is now a conditional expression, which is a value rather than an object - so it cannot be assigned to. Written as b = a >= 5 ? 100 : 200; it would compile and give 100.
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What will be the output of the program?
#include <stdio.h>
int main()
{
int x = 10, y = 20;
if(!(!x) && x)
printf("x = %d\n", x);
else
printf("y = %d\n", y);
return 0;
}
Answer: C
Correct answer: x = 10
!x is 0 since x is non-zero, and !(!x) is therefore 1 - the double negation normalises any non-zero value to 1. The condition 1 && 10 is true, so the first branch runs and prints the value of x, which the logical operators never altered: 10.
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What will be the output of the program?
#include <stdio.h>
int main()
{
char ch;
if(ch = printf(""))
printf("It matters\n");
else
printf("It doesn't matters\n");
return 0;
}
Answer: B
Correct answer: It doesn't matters
printf("") writes nothing and therefore returns 0 - the count of characters written. That 0 is assigned to ch, and the value of the assignment is also 0, which is false. So the else branch runs.
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