#include<stdio.h>
int fun(int i)
{
i++;
return i;
}
int main()
{
int fun(int);
int i=3;
fun(i=fun(fun(i)));
printf("%d\n", i);
return 0;
}
A.
5
B.
4
C.
Error
D.
Garbage value
Answer: Option A
Explanation:
Step 1: int fun(int); This is prototype of function fun(). It tells the compiler that the function fun() accept one integer parameter and returns an integer value.
Step 2: int i=3; The variable i is declared as an integer type and initialized to value 3.
Step 3: fun(i=fun(fun(i)));. The function fun(i) increements the value of i by 1(one) and return it.
Lets go step by step,
=> fun(i) becomes fun(3) is called and it returns 4.
=> i = fun(fun(i)) becomes i = fun(4) is called and it returns 5 and stored in variable i.(i=5)
=> fun(i=fun(fun(i))); becomes fun(5); is called and it return 6 and nowhere the return value is stored.
Step 4: printf("%d\n", i); It prints the value of variable i.(5)
A function prototype in C or C++ is a declaration of a function that omits the function body but does specify the function's name, argument types and return type.
While a function definition specifies what a function does, a function prototype can be thought of as specifying its interface.
Which of the following statements should be used to obtain a remainder after dividing 3.14 by 2.1 ?
A.
rem = 3.14 % 2.1;
B.
rem = modf(3.14, 2.1);
C.
rem = fmod(3.14, 2.1);
D.
Remainder cannot be obtain in floating point division.
Answer: Option C
Explanation:
fmod(x,y) - Calculates x modulo y, the remainder of x/y.
This function is the same as the modulus operator. But fmod() performs floating point divisions.
Example:
#include <stdio.h>
#include <math.h>
int main ()
{
printf ("fmod of 3.14/2.1 is %lf\n", fmod (3.14,2.1) );
return 0;
}
Does this mentioning array name gives the base address in all the contexts?
A.
Yes
B.
No
Answer: Option B
Explanation:
No, Mentioning the array name in C or C++ gives the base address in all contexts except one.
Syntactically, the compiler treats the array name as a pointer to the first element. You can reference elements using array syntax, a[n], or using pointer syntax, *(a+n), and you can even mix the usages within an expression.
When you pass an array name as a function argument, you are passing the "value of the pointer", which means that you are implicitly passing the array by reference, even though all parameters in functions are "call by value".
Associativity has no role to play unless the precedence of operator is same.
A.
True
B.
False
Answer: Option A
Explanation:
Associativity is only needed when the operators in an expression have the same precedence. Usually + and - have the same precedence.
Consider the expression 7 - 4 + 2. The result could be either (7 - 4) + 2 = 5 or 7 - (4 + 2) = 1. The former result corresponds to the case when + and - are left-associative, the latter to when + and - are right-associative.
Usually the addition, subtraction, multiplication, and division operators are left-associative, while the exponentiation, assignment and conditional operators are right-associative. To prevent cases where operands would be associated with two operators, or no operator at all, operators with the same precedence must have the same associativity.
#include<stdio.h>
int main()
{
int k, num=30;
k = (num>5 ? (num <=10 ? 100 : 200): 500);
printf("%d\n", num);
return 0;
}
A.
200
B.
30
C.
100
D.
500
Answer: Option B
Explanation:
Step 1: int k, num=30; here variable k and num are declared as an integer type and variable num is initialized to '30'. Step 2: k = (num>5 ? (num <=10 ? 100 : 200): 500); This statement does not affect the output of the program. Because we are going to print the variable num in the next statement. So, we skip this statement. Step 3: printf("%d\n", num); It prints the value of variable num '30' Step 3: Hence the output of the program is '30'
Does there any function exist to convert the int or float to a string?
A.
Yes
B.
No
Answer: Option A
Explanation:
1. itoa() converts an integer to a string.
2. ltoa() converts a long to a string.
3. ultoa() converts an unsigned long to a string.
4. sprintf() sends formatted output to a string, so it can be used to convert any type of values to string type.
If the binary eauivalent of 5.375 in normalised form is 0100 0000 1010 1100 0000 0000 0000 0000, what will be the output of the program (on intel machine)?
#include<stdio.h>
#include<math.h>
int main()
{
float a=5.375;
char *p;
int i;
p = (char*)&a;
for(i=0; i<=3; i++)
printf("%02x\n", (unsigned char)p[i]);
return 0;
}
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