Online C Programming Test - Engineering Mechanics Test 2
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- 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.
typedef enum error { warning, test, exception } err;
A typedef gives a new name to an existing data type.
So err is a new name for enum error.
6.68 is double.
6.68L is long double constant.
6.68f is float constant.
6.68LF is not allowed in c.
Yes, In all the global variable declarations, you need to use the keyword extern.
#include<stdio.h>
int main()
{
int j=1;
while(j <= 255)
{
printf("%c %d\n", j, j);
j++;
}
return 0;
}
The while(j <= 255) loop will get executed 255 times. The size short int(2 byte wide) does not affect the while() loop.
#include<stdio.h>
int main()
{
int k, num = 30;
k = (num < 10) ? 100 : 200;
printf("%d\n", num);
return 0;
}
| 1: | for loop works faster than a while loop. |
| 2: | All things that can be done using a for loop can also be done using a while loop. |
| 3: | for(;;); implements an infinite loop. |
| 4: | for loop can be used if we want statements in a loop get executed at least once. |
#include<stdio.h>
int main()
{
float a=0.7;
if(a < 0.7)
printf("C\n");
else
printf("C++\n");
return 0;
}
if(a < 0.7) here a is a float variable and 0.7 is a double constant. The float variable a is less than double constant 0.7. Hence the if condition is satisfied and it prints 'C'
Example:
#include<stdio.h>
int main()
{
float a=0.7;
printf("%.10f %.10f\n",0.7, a);
return 0;
}
Output:
0.7000000000 0.6999999881
#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>
int i;
int fun();
int main()
{
while(i)
{
fun();
main();
}
printf("Hello\n");
return 0;
}
int fun()
{
printf("Hi");
}
Step 1: int i; The variable i is declared as an integer type.
Step 1: int fun(); This prototype tells the compiler that the function fun() does not accept any arguments and it returns an integer value.
Step 1: while(i) The value of i is not initialized so this while condition is failed. So, it does not execute the while block.
Step 1: printf("Hello\n"); It prints "Hello".
Hence the output of the program is "Hello".
True, A function can be called either call by value or call by reference.
Example:
Call by value means c = sub(a, b); here value of a and b are passed.
Call by reference means c = sub(&a, &b); here address of a and b are passed.
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 */
}
Output:
c = 12
#include<stdio.h>
#define MESS junk
int main()
{
printf("MESS\n");
return 0;
}
printf("MESS\n"); It prints the text "MESS". There is no macro calling inside the printf statement occured.
True, the header file contains classes, function prototypes, structure declaration, macros.
True, these macros are used for conditional operation.
#if <constant-expression>
#elif <constant-expression>
#endif
#include<stdio.h>
#define X (4+Y)
#define Y (X+3)
int main()
{
printf("%d\n", 4*X+2);
return 0;
}
#include<stdio.h>
int main()
{
int a[2][3][4] = { {1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 2},
{2, 1, 4, 7, 6, 7, 8, 9, 0, 0, 0, 0} };
printf("%u, %u, %u, %d\n", a, *a, **a, ***a);
return 0;
}
#include<stdio.h>
int main()
{
void *vp;
char ch=74, *cp="JACK";
int j=65;
vp=&ch;
printf("%c", *(char*)vp);
vp=&j;
printf("%c", *(int*)vp);
vp=cp;
printf("%s", (char*)vp+2);
return 0;
}
#include<stdio.h>
#include<string.h>
int main()
{
int i, n;
char *x="Alice";
n = strlen(x);
*x = x[n];
for(i=0; i<=n; i++)
{
printf("%s ", x);
x++;
}
printf("\n", x);
return 0;
}
If you compile and execute this program in windows platform with Turbo C, it will give "lice ice ce e".
It may give different output in other platforms (depends upon compiler and machine). The online C compiler given in this site will give the Option C as output (it runs on Linux platform).
#include<stdio.h>
int main()
{
int arr[3][3] = {1, 2, 3, 4};
printf("%d\n", *(*(*(arr))));
return 0;
}