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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Break takes control out of a switch. Does continue, used inside a switch, take control back to the beginning of that switch?
Answer: B
Correct answer: No
break does end a switch, but continue has no meaning inside one: it asks for the next iteration, and a switch has no iterations. A continue written inside a switch applies to the loop enclosing the switch, and if there is no such loop the program will not compile.
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What is the output?
#include <stdio.h>
int main()
{
int n;
for (n = 9; n!=0; n--)
printf("n = %d", n--);
return 0;
}
Answer: C
Correct answer: Infinite Loop
n is decremented twice on every pass - once by n-- inside the printf and once by the loop’s own n--. Starting from 9 it therefore takes only odd values: 9, 7, 5, 3, 1, -1, -3, … and never equals 0, so the test n != 0 never fails. It runs until n underflows, which is undefined behaviour.
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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;
}
Answer: B
Correct answer: Error: Constant expression required at line case P:
A case label must be a compile-time constant, because the compiler builds the jump table before the program runs. P is a variable, so case P: is rejected - even though its value happens to be known here. A const int would not help either in C; only a literal, an enumeration constant or a constant expression will do.
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