Online C Programming Test - C Programming Test - Random
- This is a FREE online test. Beware of scammers who ask for money to attend this test.
- Total number of questions: 20.
- Time allotted: 30 minutes.
- Each question carries 1 mark; there are no negative marks.
- DO NOT refresh the page.
- All the best!
Marks : 2/20
Test Review : View answers and explanation for this test.
#include<stdio.h>
int main()
{
struct emp
{
char name[20];
int age;
float sal;
};
struct emp e = {"Tiger"};
printf("%d, %f\n", e.age, e.sal);
return 0;
}
float a = 3.14; a = a%3; gives "Illegal use of floating point" error.
The modulus (%) operator can only be used on integer types. We have to use fmod() function in math.h for float values.
z = x + y * z / 4 % 2 - 1
#include<stdio.h>
int main()
{
float d=2.25;
printf("%e,", d);
printf("%f,", d);
printf("%g,", d);
printf("%lf", d);
return 0;
}
printf("%e,", d); Here '%e' specifies the "Scientific Notation" format. So, it prints the 2.25 as 2.250000e+000.
printf("%f,", d); Here '%f' specifies the "Decimal Floating Point" format. So, it prints the 2.25 as 2.250000.
printf("%g,", d); Here '%g' "Use the shorter of %e or %f". So, it prints the 2.25 as 2.25.
printf("%lf,", d); Here '%lf' specifies the "Long Double" format. So, it prints the 2.25 as 2.250000.
#include<stdio.h>
int fun(int(*)());
int main()
{
fun(main);
printf("Hi\n");
return 0;
}
int fun(int (*p)())
{
printf("Hello ");
return 0;
}
#include<stdio.h>
int func1(int);
int main()
{
int k=35;
k = func1(k=func1(k=func1(k)));
printf("k=%d\n", k);
return 0;
}
int func1(int k)
{
k++;
return k;
}
Step 1: int k=35; The variable k is declared as an integer type and initialized to 35.
Step 2: k = func1(k=func1(k=func1(k))); The func1(k) increement the value of k by 1 and return it. Here the func1(k) is called 3 times. Hence it increements value of k = 35 to 38. The result is stored in the variable k = 38.
Step 3: printf("k=%d\n", k); It prints the value of variable k "38".
#include<stdio.h>
#define FUN(i, j) i##j
int main()
{
int va1=10;
int va12=20;
printf("%d\n", FUN(va1, 2));
return 0;
}
The following program will make you understand about ## (macro concatenation) operator clearly.
#include<stdio.h>
#define FUN(i, j) i##j
int main()
{
int First = 10;
int Second = 20;
char FirstSecond[] = "IndiaBIX";
printf("%s\n", FUN(First, Second) );
return 0;
}
Output:
-------
IndiaBIX
The preprocessor will replace FUN(First, Second) as FirstSecond.
Therefore, the printf("%s\n", FUN(First, Second) ); statement will become as printf("%s\n", FirstSecond );
Hence it prints IndiaBIX as output.
Like the same, the line printf("%d\n", FUN(va1, 2)); given in the above question will become as printf("%d\n", va12 );.
Therefore, it prints 20 as output.
#include<stdio.h>
int main()
{
printf("India" "BIX\n");
return 0;
}
#include<stdio.h>
int main()
{
int a[3][4] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 };
printf("%u, %u, %u\n", a[0]+1, *(a[0]+1), *(*(a+0)+1));
return 0;
}
If the index of the array size is exceeded, the program will crash. Hence "option c" is the correct answer. But the modern compilers will take care of this kind of errors.
Example: Run the below program, it will crash in Windows (TurboC Compiler)
#include<stdio.h>
int main()
{
int arr[2];
arr[3]=10;
printf("%d",arr[3]);
return 0;
}
Since C is a compiler dependent language, it may give different outputs at different platforms. We have given the Turbo-C Compiler (Windows) output.
Please try the above programs in Windows (Turbo-C Compiler) and Linux (GCC Compiler), you will understand the difference better.
#include<stdio.h>
int main()
{
static int arr[] = {0, 1, 2, 3, 4};
int *p[] = {arr, arr+1, arr+2, arr+3, arr+4};
int **ptr=p;
ptr++;
printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr);
*ptr++;
printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr);
*++ptr;
printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr);
++*ptr;
printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr);
return 0;
}
#include<stdio.h>
int main()
{
struct node
{
int data;
struct node *link;
};
struct node *p, *q;
p = (struct node *) malloc(sizeof(struct node));
q = (struct node *) malloc(sizeof(struct node));
printf("%d, %d\n", sizeof(p), sizeof(q));
return 0;
}
#include<stdio.h>
int main()
{
union a
{
int i;
char ch[2];
};
union a u1 = {512};
union a u2 = {0, 2};
return 0;
}
| 1: | u2 CANNOT be initialized as shown. |
| 2: | u1 can be initialized as shown. |
| 3: | To initialize char ch[] of u2 '.' operator should be used. |
| 4: | The code causes an error 'Declaration syntax error' |
True, each line may contain zero or more characters terminated by a newline character.
cmd> sample "*.c"
/* sample.c */
#include<stdio.h>
int main(int argc, int *argv)
{
int i;
for(i=1; i<argc; i++)
printf("%s\n", argv[i]);
return 0;
}
char *scr;#include<stdio.h>
double i;
int main()
{
(int)(float)(char) i;
printf("%d", sizeof((int)(float)(char)i));
return 0;
}
In Turbo C (DOS - 16 bit platform), the output will be 2.
But in GCC (Unix/Linux - 32 bit platform), the output will be 4.