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()
{
void v = 0;
printf("%d", v);
return 0;
}
#include<stdio.h>
int main()
{
int i=1;
while()
{
printf("%d\n", i++);
if(i>10)
break;
}
return 0;
}
The while() loop must have conditional expression or it shows "Expression syntax" error.
Example: while(i > 10){ ... }
#include<stdio.h>
int addmult(int ii, int jj)
{
int kk, ll;
kk = ii + jj;
ll = ii * jj;
return (kk, ll);
}
int main()
{
int i=3, j=4, k, l;
k = addmult(i, j);
l = addmult(i, j);
printf("%d, %d\n", k, l);
return 0;
}Step 1: int i=3, j=4, k, l; The variables i, j, k, l are declared as an integer type and variable i, j are initialized to 3, 4 respectively.
The function addmult(i, j); accept 2 integer parameters.
Step 2: k = addmult(i, j); becomes k = addmult(3, 4)
In the function addmult(). The variable kk, ll are declared as an integer type int kk, ll;
kk = ii + jj; becomes kk = 3 + 4 Now the kk value is '7'.
ll = ii * jj; becomes ll = 3 * 4 Now the ll value is '12'.
return (kk, ll); It returns the value of variable ll only.
The value 12 is stored in variable 'k'.
Step 3: l = addmult(i, j); becomes l = addmult(3, 4)
kk = ii + jj; becomes kk = 3 + 4 Now the kk value is '7'.
ll = ii * jj; becomes ll = 3 * 4 Now the ll value is '12'.
return (kk, ll); It returns the value of variable ll only.
The value 12 is stored in variable 'l'.
Step 4: printf("%d, %d\n", k, l); It prints the value of k and l
Hence the output is "12, 12".
#include<stdio.h>
int main()
{
int a=10;
void f();
a = f();
printf("%d\n", a);
return 0;
}
void f()
{
printf("Hi");
}
The function void f() is not visible to the compiler while going through main() function. So we have to declare this prototype void f(); before to main() function. This kind of error will not occur in modern compilers.
No, Mentioning the array name in C or C++ gives the base address in all contexts except one.
Syntactically, the compiler treats the array name as a pointer to the first element. You can reference elements using array syntax, a[n], or using pointer syntax, *(a+n), and you can even mix the usages within an expression.
When you pass an array name as a function argument, you are passing the "value of the pointer", which means that you are implicitly passing the array by reference, even though all parameters in functions are "call by value".
#include<stdio.h>
#include<string.h>
int main()
{
char str1[5], str2[5];
int i;
gets(str1);
gets(str2);
i = strcmp(str1, str2);
printf("%d\n", i);
return 0;
}
gets() gets collects a string of characters terminated by a new line from the standard input stream stdin.
The gets(str1) read the input string from user and store in variable str1.
The gets(str2) read the input string from user and store in variable str2.
The code i = strcmp(str1, str2); The strcmp not only returns -1, 0 and +1, but also other negative or positive values. So the value of i is "unpredictable integer value".
printf("%d\n", i); It prints the value of variable i.
#include<stdio.h>
int main()
{
char str[10] = "India";
str[6] = "BIX";
printf("%s\n", str);
return 0;
}
str[6] = "BIX"; - Nonportable pointer conversion.
#include<stdio.h>
int main()
{
char str = "IndiaBIX";
printf("%s\n", str);
return 0;
}
The line char str = "IndiaBIX"; generates "Non portable pointer conversion" error.
To eliminate the error, we have to change the above line to
char *str = "IndiaBIX"; (or) char str[] = "IndiaBIX";
Then it prints "IndiaBIX".
#include<stdio.h>
int main()
{
struct emp
{
char name[25];
int age;
float bs;
};
struct emp e;
e.name = "Suresh";
e.age = 25;
printf("%s %d\n", e.name, e.age);
return 0;
}
We cannot assign a string to a struct variable like e.name = "Suresh"; in C.
We have to use strcpy(char *dest, const char *source) function to assign a string.
Ex: strcpy(e.name, "Suresh");
Declaration: char *fgets(char *s, int n, FILE *stream);
fgets reads characters from stream into the string s. It stops when it reads either n - 1 characters or a newline character, whichever comes first.
Therefore, the string str contain "I am a boy\n\0"
FILE *fp;
fp = fopen("source.txt", "rb");
The file source.txt will be opened in the binary mode.
#include<stdio.h>
int main()
{
char ch;
int i;
scanf("%c", &i);
scanf("%d", &ch);
printf("%c %d", ch, i);
return 0;
}
cmd> sample
/* sample.c */
#include<stdio.h>
int main(int argc, char **argv)
{
printf("%s\n", argv[argc-1]);
return 0;
}
#include<stdio.h>
#define MAX 128
int main()
{
const int max=128;
char array[max];
char string[MAX];
array[0] = string[0] = 'A';
printf("%c %c\n", array[0], string[0]);
return 0;
}
Step 1: A macro named MAX is defined with value 128
Step 2: const int max=128; The constant variable max is declared as an integer data type and it is initialized with value 128.
Step 3: char array[max]; This statement reports an error "constant expression required". Because, we cannot use variable to define the size of array.
To avoid this error, we have to declare the size of an array as static. Eg. char array[10]; or use macro char array[MAX];
Note: The above program will print A A as output in Unix platform.
#include<stdio.h>
#include<stdlib.h>
int main()
{
static char *p = (char *)malloc(10);
return 0;
}