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!

Is there a way to multiply matrices in lesser than o(n^3) time complexity?

Yes. Divide and conquer method suggests Strassen's matrix multiplication method to be used. If we follow this method, the time complexity is O(n^2.81) times rather O(n^3) times.

Here are some more details about this method.

Suppose we want to multiply two matrices of size N x N: for example A x B = C


[C11 C12] [A11 A12] [B11 B12]
[C21 C22] = [A21 A22] [B21 B22]


Now, this guy called Strassen's somehow :) came up with a bunch of equations to calculate the 4 elements of the resultant matrix


C11 = a11*b11 + a12*b21
C12 = a11*b12 + a12*b22
C21 = a21*b11 + a22*b21
C22 = a21*b12 + a22*b22



If you are aware, the rudimentary matrix multiplication goes something like this


void matrix_mult()
{
for (i = 1; i <= N; i++)
{
for (j = 1; j <= N; j++)
{
compute Ci,j;
}
}
}


So, essentially, a 2x2 matrix multiplication can be accomplished using 8 multiplications. And the complexity becomes


2^log 8 =2^3


Strassen showed that 2x2 matrix multiplication can be accomplished in 7 multiplications and 18 additions or subtractions. So now the complexity becomes


2^log7 =2^2.807


This is how he did it


P1 = (A11+ A22)(B11+B22)
P2 = (A21 + A22) * B11
P3 = A11 * (B12 - B22)
P4 = A22 * (B21 - B11)
P5 = (A11 + A12) * B22
P6 = (A21 - A11) * (B11 + B12)
P7 = (A12 - A22) * (B21 + B22)

C11 = P1 + P4 - P5 + P7
C12 = P3 + P5
C21 = P2 + P4
C22 = P1 + P3 - P2 + P6


Now, there is no need to memorize this stuff!

Write a simple piece of code to split a string at equal intervals.

Suppose you have a big string


This is a big string which I want to split at equal intervals, without caring about the words.


Now, to split this string say into smaller strings of 20 characters each, try this


#define maxLineSize 20

split(char *string)
{
int i, length;
char dest[maxLineSize + 1];

i = 0;
length = strlen(string);

while((i+maxLineSize) <= length)
{
strncpy(dest, (string+i), maxLineSize);
dest[maxLineSize - 1] = '\0';
i = i + strlen(dest) - 1;
printf("\nChunk : [%s]\n", dest);
}

strcpy(dest, (string + i));
printf("\nChunk : [%s]\n", dest);
}

How to fast multiply a number by 7?

Try


(num << 3 - num)


This is same as


num*8 - num = num * (8-1) = num * 7

How do you get the line numbers in C?

Use the following Macros


__FILE__ Source file name (string constant) format "patx.c"
__LINE__ Current source line number (integer)
__DATE__ Date compiled (string constant)format "Dec 14 1985"
__TIME__ Time compiled (string constant) format "15:24:26"
__TIMESTAMP__ Compile date/time (string constant)format "Tue Nov 19 11:39:12 1997"


Usage example


static char stamp[] = "***\nmodule " __FILE__ "\ncompiled " __TIMESTAMP__ "\n***";

...

int main()
{
...

if ( (fp = fopen(fl,"r")) == NULL )
{
printf( "open failed, line %d\n%s\n",__LINE__, stamp );
exit( 4 );
}

...
}


And the output is something like


*** open failed, line 67
******
module myfile.c
compiled Mon Jan 15 11:15:56 1999
***