Write your own trim() or squeeze() function to remove the spaces from a string.

#include < stdio.h >

char *trim(char *s);

int main(int argc, char *argv[])
{
char str1[]=" Hello I am Good ";
printf("\n\nBefore trimming : [%s]", str1);
printf("\n\nAfter trimming : [%s]", trim(str1));

getch();
}


// The trim() function...

char *trim(char *s)
{
char *p, *ps;

for (ps = p = s; *s != '\0'; s++)
{
if (!isspace(*s))
{
*p++ = *s;
}
}

*p = '\0';

return(ps);
}


And here is the output...


Before trimming : [ Hello I am Good ]
After trimming : [HelloIamGood]



Another version of this question requires one to reduce multiple spaces, tabs etc to single spaces...

How to add two numbers without using the plus operator?

Actually,


SUM = A XOR B
CARRY = A AND B


On a wicked note, you can add two numbers wihtout using the + operator as follows


a - (- b)

How to generate prime numbers? How to generate the next prime after a given prime?

This is a very vast subject. There are numerous methods to generate primes or to find out if a given number is a prime number or not. Here are a few of them. I strongly recommend you to search on the Internet for more elaborate information.

Brute Force
Test each number starting with 2 and continuing up to the number of primes we want to generate. We divide each numbr by all divisors upto the square root of that number. If no factors are found, its a prime.

Using only primes as divisors
Test each candidate only with numbers that have been already proven to be prime. To do so, keep a list of already found primes (probably using an array, a file or bit fields).

Test with odd candidates only
We need not test even candidates at all. We could make 2 a special case and just print it, not include it in the list of primes and start our candidate search with 3 and increment by 2 instead of one everytime.


Table method
Suppose we want to find all the primes between 1 and 64. We write out a table of these numbers, and proceed as follows. 2 is the first integer greater than 1, so its obviously prime. We now cross out all multiples of two. The next number we haven't crossed out is 3. We circle it and cross out all its multiples. The next non-crossed number is 5, sp we circle it and cross all its mutiples. We only have to do this for all numbers less than the square root of our upper limit, since any composite in the table must have atleast one factor less than the square root of the upper limit. Whats left after this process of elimination is all the prime numbers between 1 and 64.

Write a program to print numbers from 1 to 100 without using loops!

Another "Yeah, I am a jerk, kick me! kind of a question. I simply dont know why they ask these questions.

Nevertheless, for the benefit of mankind...



Method1 (Using recursion)


void printUp(int startNumber, int endNumber)
{
if (startNumber > endNumber)
return;

printf("[%d]\n", startNumber++);
printUp(startNumber, endNumber);
}




Method2 (Using goto)


void printUp(int startNumber, int endNumber)
{
start:

if (startNumber > endNumber)
{
goto end;
}
else
{
printf("[%d]\n", startNumber++);
goto start;
}

end:
return;
}

Write a program to check if the stack grows up or down.

Try noting down the address of a local variable. Call another function with a local variable declared in it and check the address of that local variable and compare!.


#include < stdio.h >
#include < stdlib.h >

void stack(int *local1);

int main()
{
int local1;
stack(&local1);
exit(0);
}

void stack(int *local1)
{
int local2;
printf("\nAddress of first local : [%u]", local1);
printf("\nAddress of second local : [%u]", &local2);
if(local1 < &local2)
{
printf("\nStack is growing downwards.\n");
}
else
{
printf("\nStack is growing upwards.\n");
}
printf("\n\n");
}

Given two strings A and B, how would you find out if the characters in B were a subset of the characters in A?

Here is a simple, yet efficient C program to accomplish the same...


#include < stdio.h >
#include < conio.h >

int isSubset(char *a, char *b);

int main()
{
char str1[]="defabc";
char str2[]="abcfed";

if(isSubset(str1, str2)==0)
{
printf("\nYes, characters in B=[%s] are a subset of characters in A=[%s]\n",str2,str1);
}
else
{
printf("\nNo, characters in B=[%s] are not a subset of characters in A=[%s]\n",str2,str1);
}

getch();
return(0);
}


// Function to check if characters in "b" are a subset
// of the characters in "a"

int isSubset(char *a, char *b)
{
int letterPresent[256];
int i;

for(i=0; i < 256; i++)
letterPresent[i]=0;

for(i=0; a[i]!='\0'; i++)
letterPresent[a[i]]++;

for(i=0; b[i]!='\0'; i++)
if(!letterPresent[b[i]])
return(1);

return(0);
}

How can we sum the digits of a given number in single statement?

Try something like this


# include < stdio.h >

void main()
{
int num=123456;
int sum=0;

for(;num > 0;sum+=num%10,num/=10); // This is the "single line".

printf("\nsum = [%d]\n", sum);
}


If there is a simpler way to do this, let me know!