char *p=NULL; p = (char *) malloc(10); p++; free(p);What happens, is there a leak?? are 10 bytes freed??
{ int Count; auto int Month; }The example above defines two variables with the same storage class. auto can only be used within functions, i.e. local variables.
{ register int Miles; }Register should only be used for variables that require quick access - such as counters. It should also be noted that defining 'register' goes not mean that the variable will be stored in a register. It means that it MIGHT be stored in a register - depending on hardware and implementation restrictions.
Given an image represented by an NxN matrix, where each pixel in the image is 4 bytes, write a method to rotate the image by 90 degrees. Can you do this in place?
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| Example 1 Example 2INPUT >> OUTPUT | INPUT >> OUTPUT |4 | 4 |1 2 3 4 1 1 1 1 | 11 12 13 14 41 31 21 11 |1 2 3 4 2 2 2 2 | 21 22 23 24 42 32 22 12 |1 2 3 4 3 3 3 3 | 31 32 33 34 43 33 23 13 |1 2 3 4 4 4 4 4 | 41 42 43 44 44 34 24 14 |
const means that something is not modifiable, so a data object that is declared with const as a part of its type specification must not be assigned to in any way during the run of a program.const but not initialized.const and putting it into a pointer to the const-qualified version of the same type is both safe and explicitly permitted; you will be able to use the pointer to inspect the object, but not modify it. Putting the address of a const type into a pointer to the unqualified type is much more dangerous and consequently prohibited (although you can get around this by using a cast). Here is an example:#include <stdio.h>
#include <stdlib.h>
main(){
int i;
const int ci = 123;
/* declare a pointer to a const.. */
const int *cpi;
/* ordinary pointer to a non-const */
int *ncpi;
cpi = &ci;
ncpi = &i;
/*
* this is allowed
*/
cpi = ncpi;
/*
* this needs a cast
* because it is usually a big mistake,
* see what it permits below.
*/
ncpi = (int *)cpi;
/*
* now to get undefined behaviour...
* modify a const through a pointer
*/
*ncpi = 0;
exit(EXIT_SUCCESS);
}
const objects are not modified explicitly by the user.char *const vs const char *
char c; char *const cp = &c;It's simple really;
cp is a pointer to a char, which is exactly what it would be if the const weren't there. The const means that cp is not to be modified, although whatever it points to can be—the pointer is constant, not the thing that it points to. The other way round isconst char *cp;which means that now cp is an ordinary, modifiable pointer, but the thing that it points to must not be modified. So, depending on what you choose to do, both the pointer and the thing it points to may be modifiable or not; just choose the appropriate declaration.
const int x = 10;
int *ptr = &x;
*ptr = 20;
printf ("Value of x is %d\n", x);
Even though the variable x is const the value gets changed with
following warning "initialization discards qualifiers from pointer target type.And its one of the drawback in c;
But the above gives error in Line 2
saying invalid conversion from const int to int
int lcs( char *X, char *Y, int m, int n ){ if (m == 0 || n == 0) return 0; if(X[m-1] == Y[n-1]) return 1 + lcs(X, Y, m-1, n-1); else return max(lcs(X, Y, m, n-1), lcs(X, Y, m-1, n));}Method 2:Dyanamic Programming
int lcs( char *X, char *Y, int m, int n ){ int L[m+1][n+1]; int i, j; /* Following steps build L[m+1][n+1] in bottom up fashion. Note that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */ for (i=0; i<=m; i++) { for (j=0; j<=n; j++) { if (i == 0 || j == 0) L[i][j] = 0; else if(X[i-1] == Y[j-1]) L[i][j] = L[i-1][j-1] + 1; else L[i][j] = max(L[i-1][j], L[i][j-1]); } } return L[m][n];}#include<stdio.h>int fib(int n){ /* Declare an array to store fibonacci numbers. */ int f[n+1]; int i; /* 0th and 1st number of the series are 0 and 1*/ f[0] = 0; f[1] = 1; for (i = 2; i <= n; i++) { /* Add the previous 2 numbers in the series and store it */ f[i] = f[i-1] + f[i-2]; } return f[n];}int main (){ int n = 9; printf("%d", fib(n)); getchar(); return 0;} |
#include<stdio.h>int fib(int n){ int a = 0, b = 1, c, i; if( n == 0) return a; for (i = 2; i <= n; i++) { c = a + b; a = b; b = c; } return b;}int main (){ int n = 9; printf("%d", fib(n)); getchar(); return 0;} |
#include <stdio.h>/* Helper function that multiplies 2 matricies F and M of size 2*2, and puts the multiplication result back to F[][] */void multiply(int F[2][2], int M[2][2]);/* Helper function that calculates F[][] raise to the power n and puts the result in F[][] Note that this function is desinged only for fib() and won't work as general power function */void power(int F[2][2], int n);int fib(int n){ int F[2][2] = {{1,1},{1,0}}; if(n == 0) return 0; power(F, n-1); return F[0][0];}void multiply(int F[2][2], int M[2][2]){ int x = F[0][0]*M[0][0] + F[0][1]*M[1][0]; int y = F[0][0]*M[0][1] + F[0][1]*M[1][1]; int z = F[1][0]*M[0][0] + F[1][1]*M[1][0]; int w = F[1][0]*M[0][1] + F[1][1]*M[1][1]; F[0][0] = x; F[0][1] = y; F[1][0] = z; F[1][1] = w;}void power(int F[2][2], int n){ int i; int M[2][2] = {{1,1},{1,0}}; // n - 1 times multiply the matrix to {{1,0},{0,1}} for ( i = 2; i <= n; i++ ) multiply(F, M);}/* Driver program to test above function */int main(){ int n = 9; printf("%d", fib(n)); getchar(); return 0;} |
#include <stdio.h>void multiply(int F[2][2], int M[2][2]);void power(int F[2][2], int n);/* function that returns nth Fibonacci number */int fib(int n){ int F[2][2] = {{1,1},{1,0}}; if(n == 0) return 0; power(F, n-1); return F[0][0];}/* Optimized version of power() in method 4 */void power(int F[2][2], int n){ if( n == 0 || n == 1) return; int M[2][2] = {{1,1},{1,0}}; power(F, n/2); multiply(F, F); if( n%2 != 0 ) multiply(F, M);}void multiply(int F[2][2], int M[2][2]){ int x = F[0][0]*M[0][0] + F[0][1]*M[1][0]; int y = F[0][0]*M[0][1] + F[0][1]*M[1][1]; int z = F[1][0]*M[0][0] + F[1][1]*M[1][0]; int w = F[1][0]*M[0][1] + F[1][1]*M[1][1]; F[0][0] = x; F[0][1] = y; F[1][0] = z; F[1][1] = w;}/* Driver program to test above function */int main(){ int n = 9; printf("%d", fib(9)); getchar(); return 0;} |
A* [ F(1) F(0) ] = [ F(2) F(1) ]A* [ F(2) F(1) ] = [ F(3) F(2) ] = A^2 * [ F(1) F(0) ]A* [ F(1) F(0) ] = [ F(4) F(3) ] = A^3 * [ F(1) F(0) ]........A* [ F(n) F(n-1) ] = [ F(n+1) F(n) ] = A^n * [ F(1) F(0) ]
Matrix findNthPower( Matrix M , power n ){if( n == 1 ) return M;Matrix R = findNthPower ( M , n/2 );R = RxR; // matrix multiplicationif( n%2 == 1 ) R = RxM; // matrix multiplicationreturn R;}