Online C Programming Test - Engineering Mechanics Test 1
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- Total number of questions: 20.
- Time allotted: 30 minutes.
- Each question carries 1 mark; there are no negative marks.
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- All the best!
Marks : 2/20
Test Review : View answers and explanation for this test.
#include<stdio.h>
int X=40;
int main()
{
int X=20;
printf("%d\n", X);
return 0;
}
#include<stdio.h>
int main()
{
char *s1;
char far *s2;
char huge *s3;
printf("%d, %d, %d\n", sizeof(s1), sizeof(s2), sizeof(s3));
return 0;
}
Any pointer size is 2 bytes. (only 16-bit offset)
So, char *s1 = 2 bytes.
So, char far *s2; = 4 bytes.
So, char huge *s3; = 4 bytes.
A far, huge pointer has two parts: a 16-bit segment value and a 16-bit offset value.
Since C is a compiler dependent language, it may give different output in other platforms. The above program works fine in Windows (TurboC), but error in Linux (GCC Compiler).
| 1 : | typedef long a; extern int a c; |
| 2 : | typedef long a; extern a int c; |
| 3 : | typedef long a; extern a c; |
typedef long a;
extern int a c; while compiling this statement becomes extern int long c;. This will result in to "Declaration syntax error".
typedef long a;
extern a int c; while compiling this statement becomes extern long int c;. This will result in to "Too many types in declaration error".
typedef long a;
extern a c; while compiling this statement becomes extern long c;. This is a valid c declaration statement. It says variable c is long data type and defined in some other file or module.
So, Option C is the correct answer.
#include<stdio.h>
int main()
{
int i=4;
switch(i)
{
default:
printf("This is default\n");
case 1:
printf("This is case 1\n");
break;
case 2:
printf("This is case 2\n");
break;
case 3:
printf("This is case 3\n");
}
return 0;
}
In the very begining of switch-case statement default statement is encountered. So, it prints "This is default".
In default statement there is no break; statement is included. So it prints the case 1 statements. "This is case 1".
Then the break; statement is encountered. Hence the program exits from the switch-case block.
#include<stdio.h>
int main()
{
int i = 1;
switch(i)
{
printf("Hello\n");
case 1:
printf("Hi\n");
break;
case 2:
printf("\nBye\n");
break;
}
return 0;
}
switch(i) has the variable i it has the value '1'(one).
Then case 1: statements got executed. so, it prints "Hi". The break; statement make the program to be exited from switch-case statement.
switch-case do not execute any statements outside these blocks case and default
Hence the output is "Hi".
#include<stdio.h>
int main()
{
int x = 3;
float y = 3.0;
if(x == y)
printf("x and y are equal");
else
printf("x and y are not equal");
return 0;
}
Step 1: int x = 3; here variable x is an integer type and initialized to '3'.
Step 2: float y = 3.0; here variable y is an float type and initialized to '3.0'
Step 3: if(x == y) here we are comparing if(3 == 3.0) hence this condition is satisfied.
Hence it prints "x and y are equal".
#include<stdio.h>
int main()
{
int i = 1;
switch(i)
{
case 1:
printf("Case1");
break;
case 1*2+4:
printf("Case2");
break;
}
return 0;
}
Constant expression are accepted in switch
It prints "Case1"
#include<stdio.h>
int main()
{
int i = 0;
i++;
if(i <= 5)
{
printf("IndiaBIX\n");
exit(0);
main();
}
return 0;
}
Step 1: int i = 0; here variable i is declared as an integer type and initialized to '0'(zero).
Step 2: i++; here variable i is increemented by 1(one). Hence, i = 1
Step 3: if(i <= 5) becomes if(1 <= 5) here we are checking '1' is less than or equal to '5'. Hence the if condition is satisfied.
Step 4: printf("IndiaBIX\n"); It prints "IndiaBIX"
Step 5: exit(); terminates the program execution.
Hence the output is "IndiaBIX".
#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 fun(int, int);
typedef int (*pf) (int, int);
int proc(pf, int, int);
int main()
{
printf("%d\n", proc(fun, 6, 6));
return 0;
}
int fun(int a, int b)
{
return (a==b);
}
int proc(pf p, int a, int b)
{
return ((*p)(a, b));
}
long fun(int num)
{
int i;
long f=1;
for(i=1; i<=num; i++)
f = f * i;
return f;
}
Yes, this function calculates and return the factorial value of an given integer num.
#include<stdio.h>
int main()
{
char *str;
str = "%d\n";
str++;
str++;
printf(str-2, 300);
return 0;
}
#include<stdio.h>
int main()
{
char str[20] = "Hello";
char *const p=str;
*p='M';
printf("%s\n", str);
return 0;
}
char ****k;
char *p=0;
char *t=NULL;
#include<stdio.h>
int main()
{
int *p1, i=25;
void *p2;
p1=&i;
p2=&i;
p1=p2;
p2=p1;
return 0;
}
#include<stdio.h>
int main()
{
int arr[]={2, 3, 4, 1, 6};
printf("%u, %u, %u\n", arr, &arr[0], &arr);
return 0;
}
Step 1: int arr[]={2, 3, 4, 1, 6}; The variable arr is declared as an integer array and initialized.
Step 2: printf("%u, %u, %u\n", arr, &arr[0], &arr); Here,
The base address of the array is 1200.
=> arr, &arr is pointing to the base address of the array arr.
=> &arr[0] is pointing to the address of the first element array arr. (ie. base address)
Hence the output of the program is 1200, 1200, 1200