pointer-to-pointer

I have a question about syntax of pointer to arrays. Well we know arrays are pointers themselves(what our uni professor said) so why when we point to them with another pointer (which would be a pointer to pointer) we use this syntax: int array[10]; int *pointer = array; Instead of this syntax: int array[10]; int **pointer = &array; Although i know this would be correct using malloc but why not in the normal way, is it a compiler or syntax thing or i am wrong somewhere else?? Well we know arrays are pointers themselves No. Arrays are not pointers. Arrays are arrays. Except when it is the

I am new to using pthread and also not that familiar with pointers to pointers. Could somebody perhaps explain why the second argument of pthread_join() is a void ** . Why is it designed like this. int pthread_join(pthread_t thread, void **value_ptr); To return a value via a function's argument you need to pass in the address of the variable to receive the new value. As pthread_join() is designed to receive the pointer value being passed to pthread_exit() , which is a void* , pthread_join() expects the address of a void* which in fact is of type void** . Example: #include <stdlib.h> /* for

I'm am trying to write a program that reads in a series of strings from a text file and stores these in an array of strings, dynamically allocating memory for each element. My plan was to store each string in an array using a pointer and then grow the array size as more were read in. I am having trouble to understand why my test code below is not working. Is this a workable idea? char *aPtr; aPtr =(char*)malloc(sizeof(char)); aPtr[0]="This is a test"; printf("%s",aPtr[0]); In C a string is a char* . A dynamic array of type T is represented as a pointer to T , so for char* that would be char**

I want to change member of structure under double pointer. Do you know how? Example code typedef struct { int member; } Ttype; void changeMember(Ttype **foo) { //I don`t know how to do it //maybe *foo->member = 1; } JaredPar Try (*foo)->member = 1; You need to explicitly use the * first. Otherwise it's an attempt to dereference member. Due to operator precedence, you need to put parentheses around this: (*foo)->member = 1; You can use a temp variable to improve readability. For example: Ttype *temp = *foo; temp->member = 1; If you have control of this and allowed to use C++, the better way is

I've just been helping someone out with some code. He had this: char dataArray[10]; Then wanted to get a pointer to the start of the array. Rather than use: &dataArray[0] or just dataArray He used &dataArray Did he end up with a pointer to a pointer there? I'm not sure what &dataArray would give him. &dataArray[0] is of type char * . That is a pointer to char . dataArray is of type char[10] &dataArray will be of type char (*)[10] . That is a pointer-to-array. Apart from that, the value will be same, i.e., they point to the same address but their types need not be compatible. None of them is a