Point out the compile time error in the program given below.
#include<stdio.h>
int main()
{
int *x;
*x=100;
return 0;
}
A.
Error: invalid assignment for x
B.
Error: suspicious pointer conversion
C.
No error
D.
None of above
Answer: Option C
Explanation:
While reading the code there is no error, but upon running the program having an unitialised variable can cause the program to crash (Null pointer assignment).
#include<stdio.h>
#include<string.h>
int main()
{
char str1[5], str2[5];
int i;
gets(str1);
gets(str2);
i = strcmp(str1, str2);
printf("%d\n", i);
return 0;
}
A.
Unpredictable integer value
B.
0
C.
-1
D.
Error
Answer: Option A
Explanation:
gets() gets collects a string of characters terminated by a new line from the standard input stream stdin.
The gets(str1) read the input string from user and store in variable str1.
The gets(str2) read the input string from user and store in variable str2.
The code i = strcmp(str1, str2); The strcmp not only returns -1, 0 and +1, but also other negative or positive values. So the value of i is "unpredictable integer value".
printf("%d\n", i); It prints the value of variable i.
#include<stdio.h>
int main()
{
int arr[1]={10};
printf("%d\n", 0[arr]);
return 0;
}
A.
1
B.
10
C.
0
D.
6
Answer: Option B
Explanation:
Step 1: int arr[1]={10}; The variable arr[1] is declared as an integer array with size '2' and it's first element is initialized to value '10'(means arr[0]=10)
Step 2: printf("%d\n", 0[arr]); It prints the first element value of the variable arr.
Assunming, integer is 2 byte, What will be the output of the program?
#include<stdio.h>
int main()
{
printf("%x\n", -1>>1);
return 0;
}
A.
ffff
B.
0fff
C.
0000
D.
fff0
Answer: Option A
Explanation:
Negative numbers are treated with 2's complement method.
1's complement: Inverting the bits ( all 1s to 0s and all 0s to 1s)
2's complement: Adding 1 to the result of 1's complement.
Binary of 1(2byte) : 0000 0000 0000 0001
Representing -1:
1s complement of 1(2byte) : 1111 1111 1111 1110
Adding 1 to 1's comp. result : 1111 1111 1111 1111
Right shift 1bit(-1>>1): 1111 1111 1111 1111 (carry out 1)
Hexadecimal : f f f f
(Filled with 1s in the left side in the above step)
Note:
1. Fill with 1s in the left side for right shift for negative numbers.
2. Fill with 0s in the right side for left shift for negative numbers.
3. Fill with 0s in the left side for right shift for positive numbers.
4. Fill with 0s in the right side for left shift for positive numbers.
#include<stdio.h>
int main()
{
extern int a;
printf("%d\n", a);
return 0;
}
int a=20;
A.
20
B.
0
C.
Garbage Value
D.
Error
Answer: Option A
Explanation:
extern int a; indicates that the variable a is defined elsewhere, usually in a separate source code module.
printf("%d\n", a); it prints the value of local variable int a = 20. Because, whenever there is a conflict between local variable and global variable, local variable gets the highest priority. So it prints 20.
If a function contains two return statements successively, the compiler will generate warnings. Yes/No ?
A.
Yes
B.
No
Answer: Option A
Explanation:
Yes. If a function contains two return statements successively, the compiler will generate "Unreachable code" warnings.
Example:
#include<stdio.h>
int mul(int, int); /* Function prototype */
int main()
{
int a = 4, b = 3, c;
c = mul(a, b);
printf("c = %d\n", c);
return 0;
}
int mul(int a, int b)
{
return (a * b);
return (a - b); /* Warning: Unreachable code */
}
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;
}
A.
1 ... 65535
B.
Expression syntax error
C.
No output
D.
0, 1, 2, 3, 4, 5
Answer: Option A
Explanation:
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)
#include<stdio.h>
int main()
{
int i=0;
for(; i<=5; i++);
printf("%d,", i);
return 0;
}
A.
0, 1, 2, 3, 4, 5
B.
5
C.
1, 2, 3, 4
D.
6
Answer: Option D
Explanation:
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'.
Loading Test...
