Let’s re-write your program to make it a little clearer: #include<stdio.h> int main(void) { int arr[4] = {4,3,2,1}; int *p = arr; printf(“\n%d”, *(p+2)); return 0; } Now, *(p+2) is by definition the same as p[2]. Since p points to the first element of arr, then p[2] is the same as arr[2] which is equal … Read more
Time for a crash course on arrays in C. First of all, let’s fix the initializer for the array: int a[3][4] = { { 1, 2, 3, 4}, { 5, 6, 7, 8}, { 9, 10, 11, 12} }; This defines a 3-element array of 4-element arrays of int. The type of the expression a … Read more
void *ptr=10; This is not valid C, you need an explicit conversion with a cast like (void *) 10. printf(“%u”,ptr); This is not valid C, you need an explicit conversion with a cast like (unsigned int) ptr. Regarding the use of void pointers, they are generic object pointers, so you can assign a pointer value … Read more
#include <stdio.h> #include <string.h> double myfunction(const void *data){ double v; memcpy(&v, data, sizeof(v)); return v; } int main (int argc, char *argv[]) { unsigned char data[] = {0x40,0x20,0,0,0,0,0,0}; int i, len = sizeof(data); //reverse data If necessary for(i=0;i<len/2;++i){ unsigned char c = data[i]; data[i] = data[len -1 -i]; data[len -1 -i] = c; } double … Read more
According to the C++ Standard (4.2 Array-to-pointer conversion) 1 An lvalue or rvalue of type “array of N T” or “array of unknown bound of T” can be converted to a prvalue of type “pointer to T”. The result is a pointer to the first element of the array. So for example if you have … Read more
I won’t go into too much detail, but “client.dll”+0065F38 won’t help you too much. Luckily for you, I wrote a tutorial of how to write a trainer in C#. 2 solved Find address using pointer and offset C# [closed]
There’s no reason for them to be, so the compiler does what’s best for itself. Complex compiler optimizations and efficient and linear memory layout don’t all go together, so linear memory layout was sacrificed. When you get to hashtables, you will learn how the most efficient algorithms lead to repeatable pseudo-random output order. The symbol … Read more
From section 7.20.3.2 The free function of the C99 standard: The free function causes the space pointed to by ptr to be deallocated, that is, made available for further allocation. If ptr is a null pointer, no action occurs. Otherwise, if the argument does not match a pointer earlier returned by the calloc, malloc, or … Read more
I end up with this solution It may help someone who runs into the same problem #include<iostream> class Node{ public: int data; Node * next , * prev; }; template<class T> void DeletePtr (T*** ptr) { T** auxiliary = &(**ptr); delete *auxiliary; **ptr = nullptr; *ptr = nullptr; } // Driver Code int main () … Read more
The errors in the code is: M == NULL is not a valid way to check if malloc is succeeded. You should use (*M) == NULL because the value is assigned to (*M), not M. The loop condition in ReservarMemoriaMatriz i<=m seems wrong. It should be i<=n. You forgot to free (*M)[n] in LiberarMemoriaMatriz. Corrected … Read more
It should definitely initialize readIndex to 0. Just put one print statement and check whether your constructor called properly or not. Also try with con::con():readIndex(0) { } 3 solved How to access a member of a pointer to object [closed]
You have your main problem here. double *tabT1; double *tabT2; tabT1[0]=1; tabT1[1]=tabX[j]*(-1); tabT2[0]=1; tabT2[1]=tabX[i]*(-1); You haven’t allocated the memory, instead, you have just declared the double ptrs tabT1 and tabT2 and accessing them by pretending you have allocated. double *tabT1 = new double[2]; double *tabT2 = new double[2]; will fix this, however, I strongly suggest … Read more
A pointer is an address of a variable. Imagine ram as consecutive boxes. Each box has an address. Now in your example. int a = 8; //declares a variable of type int and initializes it with value 8 int *p1; //p1 is a pointer meaning that the variable p1 holds an address instead of a … Read more
These both statements are wrong char[] name = { “Nitish prajapati” }; char* namePointer = &name ; In C++ valid declaration of an array looks like char name[] = { “Nitish prajapati” }; As for the second statement then there is no implicit conversion from type char ( * )[17] to char *. The initializer … Read more
To know how all these three code blocks works, you got to start reading good C book specially array and pointers concept. Case 1 :- In the below code block int main(void) { char x[] = “gate2011”; char *ptr = x; printf (“%s”, ptr+ptr[3]-ptr[1]); return 0; } It looks like x[0] x[1] x[2] x[3] x[4] … Read more