Online C Programming Test - C Programming Test 7
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- Total number of questions: 20.
- Time allotted: 20 minutes.
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Marks : 2/20
Test Review : View answers and explanation for this test.
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).
| 1. | Relational |
| 2. | Arithmetic |
| 3. | Logical |
| 4. | Assignment |
2. Arithmetic operators: *, /, %, +, -
1. Relational operators: >, <, >=, <=, ==, !=
3. Logical operators : !, &&, ||
4. Assignment operators: =
Simply called as BODMAS (Bracket of Division, Multiplication, Addition and Subtraction).
How Do I Remember ? BODMAS !
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>
#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;
}
#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;
}
f(int a, int b)
{
int a;
a = 20;
return a;
}
f(int a, int b) The variable a is declared in the function argument statement.
int a; Here again we are declaring the variable a. Hence it shows the error "Redeclaration of a"
#include<stdio.h>
#define SQUARE(x) x*x
int main()
{
float s=10, u=30, t=2, a;
a = 2*(s-u*t)/SQUARE(t);
printf("Result = %f", a);
return 0;
}
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'.
Hence the output of the program is -100
True, #undef can be used only on a macro that has been #define earlier
Example: #define PI 3.14
We can undefine PI macro by #undef PI
#include<stdio.h>
int main()
{
char s[] = "IndiaBIX";
char t[25];
char *ps, *pt;
ps = s;
pt = t;
while(*ps)
*pt++ = *ps++;
/* Add a statement here */
printf("%s\n", t);
return 0;
}
#include<stdio.h>
int main()
{
char p[] = "%d\n";
p[1] = 'c';
printf(p, 65);
return 0;
}
Step 1: char p[] = "%d\n"; The variable p is declared as an array of characters and initialized with string "%d".
Step 2: p[1] = 'c'; Here, we overwrite the second element of array p by 'c'. So array p becomes "%c".
Step 3: printf(p, 65); becomes printf("%c", 65);
Therefore it prints the ASCII value of 65. The output is 'A'.
#include<stdio.h>
void swap(char *, char *);
int main()
{
char *pstr[2] = {"Hello", "IndiaBIX"};
swap(pstr[0], pstr[1]);
printf("%s\n%s", pstr[0], pstr[1]);
return 0;
}
void swap(char *t1, char *t2)
{
char *t;
t=t1;
t1=t2;
t2=t;
}
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].
Hence the output of the program is
Hello
IndiaBIX
#include<stdio.h>
#include<string.h>
int main()
{
char str1[20] = "Hello", str2[20] = " World";
printf("%s\n", strcpy(str2, strcat(str1, str2)));
return 0;
}
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 2: printf("%s\n", strcpy(str2, strcat(str1, str2)));
=> strcat(str1, str2)) it append the string str2 to str1. The result will be stored in str1. Therefore str1 contains "Hello World".
=> strcpy(str2, "Hello World") it copies the "Hello World" to the variable str2.
Hence it prints "Hello World".
#include<stdio.h>
int main()
{
union var
{
int a, b;
};
union var v;
v.a=10;
v.b=20;
printf("%d\n", v.a);
return 0;
}
#include<stdio.h>
float a;
double b;
To scan a float value, %f is used as format specifier.
To scan a double value, %lf is used as format specifier.
Therefore, the answer is scanf("%f %lf", &a, &b);
#include<stdio.h>
int main()
{
unsigned int res;
res = (64 >>(2+1-2)) & (~(1<<2));
printf("%d\n", res);
return 0;
}
#define P printf("%d\n", -1^~0);
#define M(P) int main()\
{\
P\
return 0;\
}
M(P)