The following code ‘for(;;)’ represents an infinite loop. Which statement exits the loop itself?
Answer: A
Correct answer: break
break is the statement that ends the loop: control passes to the statement after it and the program carries on. exit(0) and abort() would also stop the loop running, but only by ending the whole program, which is a different thing from leaving the loop.
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Which keyword is used to come out of a loop only for that iteration?
Answer: B
Correct answer: continue
continue abandons the current iteration and jumps to the next one - to the update expression in a for, or to the condition test in a while. break leaves the loop altogether, and return leaves the function.
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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;
}
Answer: D
Correct answer: No Error
case 1*2+4 is a constant expression, which the compiler evaluates to 6 - that is allowed, and 6 does not clash with case 1. A default label is optional. So the program compiles and, with i equal to 1, prints Case1.
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#include <stdio.h>
int main()
{
int i;
for (i = 1; i != 10; i += 2)
printf(" GeeksQuiz ");
return 0;
}
Answer: B
Correct answer: GeeksQuiz GeeksQuiz GeeksQuiz .... infinite times
i starts at 1 and rises by 2, so it takes only the odd values 1, 3, 5, 7, 9, 11, … and never equals 10. The test i != 10 is therefore never false. This is why < is safer than != in a loop condition whose variable moves by more than one.
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What will be the output of the program?
#include <stdio.h>
int main()
{
int i=4;
switch(i)
{
default:
printf("This is default\n");
case 1:
printf("This is case 1\n");
break;
case 2:
printf("This is case 2\n");
break;
case 3:
printf("This is case 3\n");
}
return 0;
}
Answer: A
Correct answer: This is default This is case 1
i is 4, which matches no case, so control jumps to default - and default need not be last. Having no break, execution then falls through into case 1, whose break finally ends the switch. So two lines are printed.
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What will be the output of the following C program segment?
char inchar = 'A';
switch (inchar)
{
case 'A' :
printf ("choice A \n") ;
case 'B' :
printf ("choice B ") ;
case 'C' :
case 'D' :
case 'E' :
default:
printf ("No Choice") ;
}
Answer: C
Correct answer: Choice A Choice B No Choice
There is no break after the first printf, so execution falls through every following label. Matching case 'A' prints “choice A”, then “choice B”, then - passing through the empty labels C, D and E - the default prints “No Choice”. A case label marks an entry point, not a block.
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Which of the following for loops covers the same range of values of i as for (i = 0; i < n; i++)?
Answer: D
Correct answer: for (i = n-1; i>-1; i--)
for (i = 0; i < n; i++) visits 0 to n-1. Running the same range backwards means starting at n-1 and continuing while i is still at least 0 - that is, while i > -1. The alternatives either start at n, which is out of range, or stop at i > 0, which misses the element at index 0.
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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;
}
Answer: A
Correct answer: There should be a condition in the while loop
Unlike for, a while requires a controlling expression - while() is a syntax error. The equivalent infinite loop is written while(1), or for(;;), where an omitted condition is defined to be true.
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A labelled statement consists of an identifier followed by
Answer: B
Correct answer: :
A label is an identifier followed by a colon - here: - marking a point that goto can jump to. The same colon form is used by case and default labels inside a switch. Labels have function scope, so a goto can only reach one in the same function.
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goto can be used to jump from main into another function.
Answer: B
Correct answer: FALSE
goto can only reach a label in the same function; labels have function scope and are invisible outside it. Jumping between functions would leave the stack in an inconsistent state. Where a non-local jump really is needed, setjmp and longjmp provide a disciplined version of it.
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