pointer-to-pointer

问题 Suppose there is a pointer f declared to some function say int foo(int) as : int (*f)(int)=foo; While mentioning about calling foo() function by using this ponter which is passed as argument to some other function. I have come across a statement saying that both y=(*f)(x) and y=f(x) are same in C and calls function foo() ....(x and y are of int type). For arrays I know that if p is pointer to any array a. p[i]=*(p+i)=*(&a[0]+i)=*(a+i)=a[i] . So writing p[i] and *(p+i) are same thing. But I

问题 Suppose there is a pointer f declared to some function say int foo(int) as : int (*f)(int)=foo; While mentioning about calling foo() function by using this ponter which is passed as argument to some other function. I have come across a statement saying that both y=(*f)(x) and y=f(x) are same in C and calls function foo() ....(x and y are of int type). For arrays I know that if p is pointer to any array a. p[i]=*(p+i)=*(&a[0]+i)=*(a+i)=a[i] . So writing p[i] and *(p+i) are same thing. But I

问题 char string1[3][4]={"koo","kid","kav"}; //This is a 2D array char * string[3]={"koo","kid","kav"}; //This is an array of 3 pointers pointing to 1D array as strings are stored as arrays in memory char (*string1Ptr)[4]=string1; //This is a pointer to a 1D array of 4 characters //I want to know differences between string1Ptr(pointer to array mentioned in question) and string(array of pointers mentioned in question). I only typed string1 here to give string1Ptr an address to strings Besides the

问题 I have this code: #include<stdio.h> #include<stdlib.h> struct node { int data; struct node *next; }; void pointerOfPointer(struct node **reference) { struct node *temporary = malloc(sizeof(struct node)); temporary->data = 100; temporary->next = 0; printf("before: temporary->data %d\n", temporary->data); temporary = *reference; printf("after: temporary->data %d\n", temporary->data); } int main() { struct node *linkedlist = malloc(sizeof(struct node)); linkedlist->data = 15; linkedlist->next =

问题 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?? 回答1: Well we know

问题 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?? 回答1: Well we know

问题 I have a problem with the pointers. I know what this does: *name I understand that this is a pointer. I've been searching but I do neither understand what this one does nor I've found helpful information **name The context is int **name, not multiplication Could someone help me? 回答1: NOTE: Without the proper context, the usage of *name and **name is ambiguous. it may portrait (a). dereference operator (b) multiplication operator Considering you're talking about a scenario like char * name;

问题 When you create the multi-dimensional array char a[10][10] , according to my book it says you must use a parameter similar to char a[][10] to pass the array to a function. Why must you specify the length as such? Aren't you just passing a double pointer to being with, and doesn't that double pointer already point to allocated memory? So why couldn't the parameter be char **a ? Are you reallocating any new memory by supplying the second 10. 回答1: Pointers are not arrays A dereferenced char **

问题 When you create the multi-dimensional array char a[10][10] , according to my book it says you must use a parameter similar to char a[][10] to pass the array to a function. Why must you specify the length as such? Aren't you just passing a double pointer to being with, and doesn't that double pointer already point to allocated memory? So why couldn't the parameter be char **a ? Are you reallocating any new memory by supplying the second 10. 回答1: Pointers are not arrays A dereferenced char **

问题 I have a function like this: void myfunc(int** arr, int n) { int i, j; for(i=0; i<n; ++i) { for(j=0; j<n; ++j) { printf("%d,", *(arr + i*n + j) ); // Print numbers with commas } printf("\n"); // Just breakline } } in other function I have an two-dimensional array like this: int main() { int seqs[8][8] = { {0, 32, 36, 52, 48, 16, 20, 4}, {0, 16, 20, 52, 48, 32, 36, 4}, {0, 32, 36, 44, 40, 8, 12, 4}, {0, 8, 12, 44, 40, 32, 36, 4}, {0, 32, 36, 38, 34, 2, 6, 4}, {0, 2, 6, 38, 34, 32, 36, 4}, {0,