Online C Programming Test - C Programming Test 10
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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.
#include<stdio.h>
int main()
{
char ch;
while(x=0;x<=255;x++)
printf("ASCII value of %d character %c\n", x, x);
return 0;
}
There are 2 errors in this program.
1. "Undefined symbol x" error. Here x is not defined in the program.
2. Here while() statement syntax error.
Option A: assignment statements are always return in paranthesis in the case of conditional operator. It should be a>b? (c=30):(c=40);
Option B: it is syntatically wrong.
Option D: syntatically wrong, it should be return(a>b ? a:b);
Option C: it uses nested conditional operator, this is logic for finding greatest number out of three numbers.
#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>
#include<math.h>
int main()
{
float n=1.54;
printf("%f, %f\n", ceil(n), floor(n));
return 0;
}
ceil(x) round up the given value. It finds the smallest integer not < x.
floor(x) round down the given value. It finds the smallest integer not > x.
printf("%f, %f\n", ceil(n), floor(n)); In this line ceil(1.54) round up the 1.54 to 2 and floor(1.54) round down the 1.54 to 1.
In the printf("%f, %f\n", ceil(n), floor(n)); statement, the format specifier "%f %f" tells output to be float value. Hence it prints 2.000000 and 1.000000.
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>
int main()
{
int i;
#if A
printf("Enter any number:");
scanf("%d", &i);
#elif B
printf("The number is odd");
return 0;
}
The conditional macro #if must have an #endif. In this program there is no #endif statement written.
#include<stdio.h>
int main()
{
float arr[] = {12.4, 2.3, 4.5, 6.7};
printf("%d\n", sizeof(arr)/sizeof(arr[0]));
return 0;
}
The sizeof function return the given variable. Example: float a=10; sizeof(a) is 4 bytes
Step 1: float arr[] = {12.4, 2.3, 4.5, 6.7}; The variable arr is declared as an floating point array and it is initialized with the values.
Step 2: printf("%d\n", sizeof(arr)/sizeof(arr[0]));
The variable arr has 4 elements. The size of the float variable is 4 bytes.
Hence 4 elements x 4 bytes = 16 bytes
sizeof(arr[0]) is 4 bytes
Hence 16/4 is 4 bytes
Hence the output of the program is '4'.
int fun(int arr[]);
int fun(int arr[2]);
No, both the statements are same. It is the prototype for the function fun() that accepts one integer array as an parameter and returns an integer value.
#include<stdio.h>
#include<string.h>
int main()
{
char *str1 = "India";
char *str2 = "BIX";
char *str3;
str3 = strcat(str1, str2);
printf("%s %s\n", str3, str1);
return 0;
}
It prints 'IndiaBIX IndiaBIX' in TurboC (in 16 bit platform).
It may cause a 'segmentation fault error' in GCC (32 bit platform).
#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' |
#include<stdio.h>
int get();
int main()
{
const int x = get();
printf("%d", x);
return 0;
}
int get()
{
return 20;
}
Step 1: int get(); This is the function prototype for the funtion get(), it tells the compiler returns an integer value and accept no parameters.
Step 2: const int x = get(); The constant variable x is declared as an integer data type and initialized with the value "20".
The function get() returns the value "20".
Step 3: printf("%d", x); It prints the value of the variable x.
Hence the output of the program is "20".
#include<stdio.h>
#include<stdlib.h>
int fun(const union employee *e);
union employee
{
char name[15];
int age;
float salary;
};
const union employee e1;
int main()
{
strcpy(e1.name, "A");
fun(&e1);
printf("%s %d %f", e1.name, e1.age, e1.salary);
return 0;
}
int fun(const union employee *e)
{
strcpy((*e).name, "B");
return 0;
}
#include<stdio.h>
#include<stdlib.h>
int main()
{
struct ex
{
int i;
float j;
char *s
};
struct ex *p;
p = (struct ex *)malloc(sizeof(struct ex));
p->s = (char*)malloc(20);
return 0;
}
void *cmp();#include<stdio.h>
int main()
{
char huge *near *far *ptr1;
char near *far *huge *ptr2;
char far *huge *near *ptr3;
printf("%d, %d, %d\n", sizeof(**ptr1), sizeof(ptr2), sizeof(*ptr3));
return 0;
}
#include<stdio.h>
void display(int (*ff)());
int main()
{
int show();
int (*f)();
f = show;
display(f);
return 0;
}
void display(int (*ff)())
{
(*ff)();
}
int show()
{
printf("IndiaBIX");
}
#include<stdio.h>
#include<string.h>
int main()
{
char str1[] = "IndiaBIX.com";
char str2[20];
strncpy(str2, str1, 8);
printf("%s", str2);
return 0;
}
No. It will print something like 'IndiaBIX(some garbage values here)' .
Because after copying the first 8 characters of source string into target string strncpy() doesn't terminate the target string with a '\0'. So it may print some garbage values along with IndiaBIX.