A short integer is at least 16 bits wide and a long integer is at least 32 bits wide.
A.
True
B.
False
Answer: Option A
Explanation:
The basic C compiler is 16 bit compiler, below are the size of it's data types
The size of short int is 2 bytes wide(16 bits).
The size of long int is 4 bytes wide(32 bits).
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.
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;
}
A float occupies 4 bytes. If the hexadecimal equivalent of these 4 bytes are A, B, C and D, then when this float is stored in memory in which of the following order do these bytes gets stored?
A.
ABCD
B.
DCBA
C.
0xABCD
D.
Depends on big endian or little endian architecture
#include<stdio.h>
int sumdig(int);
int main()
{
int a, b;
a = sumdig(123);
b = sumdig(123);
printf("%d, %d\n", a, b);
return 0;
}
int sumdig(int n)
{
int s, d;
if(n!=0)
{
d = n%10;
n = n/10;
s = d+sumdig(n);
}
else
return 0;
return s;
}
The macro function SQUARE(x) x*x calculate the square of the given number 'x'. (Eg: 102)
Step 1: float s=10, u=30, t=2, a; Here the variable s, u, t, a are declared as an floating point type and the variable s, u, t are initialized to 10, 30, 2.
Step 2: a = 2*(s-u*t)/SQUARE(t); becomes,
=> a = 2 * (10 - 30 * 2) / t * t; Here SQUARE(t) is replaced by macro to t*t .
=> a = 2 * (10 - 30 * 2) / 2 * 2;
=> a = 2 * (10 - 60) / 2 * 2;
=> a = 2 * (-50) / 2 * 2 ;
=> a = 2 * (-25) * 2 ;
=> a = (-50) * 2 ;
=> a = -100;
Step 3: printf("Result=%f", a); It prints the value of variable 'a'.
Step 1: char str1[20] = "Hello", str2[20] = " World"; The variable str1 and str2 is declared as an array of characters and initialized with value "Hello" and " World" respectively.
Step 1: void swap(char *, char *); This prototype tells the compiler that the function swap accept two strings as arguments and it does not return anything.
Step 2: char *pstr[2] = {"Hello", "IndiaBIX"}; The variable pstr is declared as an pointer to the array of strings. It is initialized to
pstr[0] = "Hello", pstr[1] = "IndiaBIX"
Step 3: swap(pstr[0], pstr[1]); The swap function is called by "call by value". Hence it does not affect the output of the program.
If the swap function is "called by reference" it will affect the variable pstr.
Step 4: printf("%s\n%s", pstr[0], pstr[1]); It prints the value of pstr[0] and pstr[1].
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