Online C Programming Test - C Programming Test 4
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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()
{
int x;
for(x=-1; x<=10; x++)
{
if(x < 5)
continue;
else
break;
printf("IndiaBIX");
}
return 0;
}
#include<stdio.h>
int main()
{
int i=-3, j=2, k=0, m;
m = ++i && ++j || ++k;
printf("%d, %d, %d, %d\n", i, j, k, m);
return 0;
}
Step 1: int i=-3, j=2, k=0, m; here variable i, j, k, m are declared as an integer type and variable i, j, k are initialized to -3, 2, 0 respectively.
Step 2: m = ++i && ++j || ++k;
becomes m = (-2 && 3) || ++k;
becomes m = TRUE || ++k;.
(++k) is not executed because (-2 && 3) alone return TRUE.
Hence this statement becomes TRUE. So it returns '1'(one). Hence m=1.
Step 3: printf("%d, %d, %d, %d\n", i, j, k, m); In the previous step the value of i,j are increemented by '1'(one).
Hence the output is "-2, 3, 0, 1".
#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>
#define INFINITELOOP while(1)
int main()
{
INFINITELOOP
printf("IndiaBIX");
return 0;
}
Yes, the program prints "IndiaBIX" and runs infinitely.
The macro INFINITELOOP while(1) replaces the text 'INFINITELOOP' by 'while(1)'
In the main function, while(1) satisfies the while condition and it prints "IndiaBIX". Then it comes to while(1) and the loop runs infinitely.
#include<stdio.h>
int main()
{
int arr[] = {12, 13, 14, 15, 16};
printf("%d, %d, %d\n", sizeof(arr), sizeof(*arr), sizeof(arr[0]));
return 0;
}
#include<stdio.h>
int *check(static int, static int);
int main()
{
int *c;
c = check(10, 20);
printf("%d\n", c);
return 0;
}
int *check(static int i, static int j)
{
int *p, *q;
p = &i;
q = &j;
if(i >= 45)
return (p);
else
return (q);
}
#include<stdio.h>
void fact(int*);
int main()
{
int i=5;
fact(&i);
printf("%d\n", i);
return 0;
}
void fact(int *j)
{
static int s=1;
if(*j!=0)
{
s = s**j;
*j = *j-1;
fact(j);
/* Add a statement here */
}
}
#include<stdio.h>
int main()
{
int arr[] = {12, 14, 15, 23, 45};
printf("%u, %u\n", arr, &arr);
return 0;
}
Step 1: int arr[] = {12, 14, 15, 23, 45}; The variable arr is declared as an integer array and initialized.
Step 2: printf("%u, %u\n", arr, &arr); Here,
The base address of the array is 65486.
=> arr, &arr is pointing to the base address of the array arr.
Hence the output of the program is 65486, 65486
#include<stdio.h>
int main()
{
int i;
char a[] = "\0";
if(printf("%s", a))
printf("The string is empty\n");
else
printf("The string is not empty\n");
return 0;
}
The function printf() returns the number of charecters printed on the console.
Step 1: char a[] = "\0"; The variable a is declared as an array of characters and it initialized with "\0". It denotes that the string is empty.
Step 2: if(printf("%s", a)) The printf() statement does not print anything, so it returns '0'(zero). Hence the if condition is failed.
In the else part it prints "The string is not empty".
#include<stdio.h>
int main()
{
printf("%u %s\n", &"Hello1", &"Hello2");
return 0;
}
In printf("%u %s\n", &"Hello", &"Hello");.
The %u format specifier tells the compiler to print the memory address of the "Hello1".
The %s format specifier tells the compiler to print the string "Hello2".
Hence the output of the program is "1022 Hello2".
struct ex
{
char ch;
int i;
long int a;
};
#include<stdio.h>
int main()
{
unsigned char ch;
FILE *fp;
fp=fopen("trial", "r");
while((ch = getc(fp))!=EOF)
printf("%c", ch);
fclose(fp);
return 0;
}
#include<stdio.h>
void fun(int);
int main(int argc)
{
printf("%d ", argc);
fun(argc);
return 0;
}
void fun(int i)
{
if(i!=4)
main(++i);
}
#include<stdio.h>
int main()
{
unsigned int num;
int i;
scanf("%u", &num);
for(i=0; i<16; i++)
{
printf("%d", (num<<i & 1<<15)?1:0);
}
return 0;
}
If we give input 4, it will print 00000000 00000100 ;
If we give input 3, it will print 00000000 00000011 ;
If we give input 511, it will print 00000001 11111111 ;
#include<stdio.h>
int fun(int **ptr);
int main()
{
int i=10;
const int *ptr = &i;
fun(&ptr);
return 0;
}
int fun(int **ptr)
{
int j = 223;
int *temp = &j;
printf("Before changing ptr = %5x\n", *ptr);
const *ptr = temp;
printf("After changing ptr = %5x\n", *ptr);
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>
#include<stdlib.h>
int main()
{
char *i = "55.555";
int result1 = 10;
float result2 = 11.111;
result1 = result1+atoi(i);
result2 = result2+atof(i);
printf("%d, %f", result1, result2);
return 0;
}
Function atoi() converts the string to integer.
Function atof() converts the string to float.
result1 = result1+atoi(i);
Here result1 = 10 + atoi(55.555);
result1 = 10 + 55;
result1 = 65;
result2 = result2+atof(i);
Here result2 = 11.111 + atof(55.555);
result2 = 11.111 + 55.555000;
result2 = 66.666000;
So the output is "65, 66.666000" .