Pretty much what the title says. When I declare a function
void foo(int *x[])
x is considered as a parameter of type int**, what about the second case?
EDIT
The part that I didn't understand was why couldn't I pass a 2D massive using a function with parameter type of int * [], but managed to do it using int ( * )[]. I thought that if the name of an array was converted to a pointer to its first element, then the name of a 2D array would be converted to a pointer of the pointer of its first element, that being said a 2D array is a massive of pointers. And int (*)[] means I am passing a pointer to an integer array. So I'm confused.
Related
How they have use the pointers in the function compare.
I could not understand.
Can somebody explain me this.
Function:-
int compare (const void *a, const void * b)
{
return ( (*(Box *)b).d * (*(Box *)b).w ) -
( (*(Box *)a).d * (*(Box *)a).w );
}
Link: https://www.geeksforgeeks.org/box-stacking-problem-dp-22/
(Box *)b it's a casting type from void to Box struct.
(*(Box *)b).d it's a dereferencing from Box struct's pointer. So, you can handle the structure as an instance, and then get struct's fields values as b.d.
Another option could be to use the pointer and avoid dereferencing with: b->d.
At the end, the compare function does: (dw) - (dw)'. If result were 0, then compare indicates both structures has same values.
PD. At (2): don't forget you are receiving the parameter as a pointer, and this implementation dereferences it as (*b).d
I have got this code:
import std.stdio;
import std.string;
void main()
{
char [] str = "aaa".dup;
char [] *str_ptr;
writeln(str_ptr);
str_ptr = &str;
*(str_ptr[0].ptr) = 'f';
writeln(*str_ptr);
writeln(str_ptr[0][1]);
}
I thought that I am creating an array of pointers char [] *str_ptr so every single pointer will point to a single char. But it looks like str_ptr points to the start of the string str. I have to make a decision because if I am trying to give access to (for example) writeln(str_ptr[1]); I am getting a lot of information on console output. That means that I am linking to an element outside the boundary.
Could anybody explain if it's an array of pointers and if yes, how an array of pointers works in this case?
What you're trying to achieve is far more easily done: just index the char array itself. No need to go through explicit pointers.
import std.stdio;
import std.string;
void main()
{
char [] str = "aaa".dup;
str[0] = 'f';
writeln(str[0]); // str[x] points to individual char
writeln(str); // faa
}
An array in D already is a pointer on the inside - it consists of a pointer to its elements, and indexing it gets you to those individual elements. str[1] leads to the second char (remember, it starts at zero), exactly the same as *(str.ptr + 1). Indeed, the compiler generates that very code (though plus range bounds checking in D by default, so it aborts instead of giving you gibberish). The only note is that the array must access sequential elements in memory. This is T[] in D.
An array of pointers might be used if they all the pointers go to various places, that are not necessarily in sequence. Maybe you want the first pointer to go to the last element, and the second pointer to to the first element. Or perhaps they are all allocated elements, like pointers to objects. The correct syntax for this in D is T*[] - read from right to left, "an array of pointers to T".
A pointer to an array is pretty rare in D, it is T[]*, but you might use it when you need to update the length of some other array held by another function. For example
int[] arr;
int[]* ptr = &arr;
(*ptr) ~= 1;
assert(arr.length == 1);
If ptr wasn't a pointer, the arr length would not be updated:
int[] arr;
int[] ptr = arr;
ptr ~= 1;
assert(arr.length == 1); // NOPE! fails, arr is still empty
But pointers to arrays are about modifying the length of the array, or maybe pointing it to something entirely new and updating the original. It isn't necessary to share individual elements inside it.
How can I convert a int 2d array initialized liked this:
int 2darray[9][9];
Into a void * then back to a 2d array again. This one gives me an incompatible pointer type error
int **sub = *((int **)2darray);
The cast to int** should be like this:
int **sub = ((int **)2darray);
You can not cast the pointer back to an array, it just doesn't make sense. You can, however, cast the pointer back to an array-pointer as shown here https://stackoverflow.com/a/20046703/6024122. As noted in that answer, this is quite uncommon, I've never personally witnessed it.
ETA: pthread_create requires its parameter to be a void *arg, thus you should do something like this:
pthread_create(..., (void *) 2darray);
and then
int **ptr = (int**) arg;
I am trying to translate c code into MATLAB, and I have come across some code that I don't understand. Specifically, there is a variable defined as:
static float *lpfdata;
This gets assigned during a function call to:
envelope_old(&fdata[0], lpfdata, winlength, samprate, BW);
Which accepts input as:
void envelope_old (float *fdata, float *lpfdata, int nsamps, int samprate,
float cutoff)
Within envelope_old, lpfdata is referenced as a vector, being assigned values in a loop in the format "lpfdata[i] = ..." where i is the index variable in the loop.
Later, a function call in the format:
downsample( lpfdata+1, dwndata, winlength, downby);
is called. What does the +1 mean in this instance?
When dealing with a pointer, lpfdata[n] and lpfdata+n are the same - they both add n * sizeof(*lpfdata) to the raw pointer and access the memory at that address.
In this case, lpfdata points to elements of type float, so sizeof(*lpfdata) == sizeof(float)
I know what pointers are but when it comes to strings/arrays I get really confused. If someone has an answer or a website that explains it that would be great. For example:
char * strncopy (char*dest, char * source, size_t);
Why the pointer? what is it pointing to? Does it a pointer usually store an address?
It is sayed in my textbook that each string building function is of type pointer char*.
Also I was trying to see if I could write a program that would clear things up, but it didn't work. Can someone tell me how to fix it, or what I'm doing wrong.
#include <stdio.h>
#include <string.h>
char * getname ()
{
char name [10];
scanf ("%s", name);
return (name);
}
int main (void)
{
char name[10];
printf ("Enter your name\n");
name[] = getname();
printf ("Hi %s", name);
return (0);
}
Inside of your getname function, when you return a pointer to the name array because it's allocated on the stack it gets destroyed leaving you with an invalid pointer. Dereferencing such a pointer causes many, many problems.
You should allocate the name array inside of getname on the heap, with malloc/calloc so that when you return the pointer the data won't be destroyed.
With regards to functions like strncpy, they tend to return a pointer to the resulting string; e.g.: strncpy returns a pointer to the destination.
Pointer itself represents an address, e.g. if you have a pointer typed char *pstr, you can always check the underlying address with printf("address of my pointer %p\n", pstr);
In C programming language, a string is an array of char. If you have a good knowledge of array and its memory layout, it's not too hard for you to understand c-styled string. Generally speaking, an array in C is a continuous chunk of memory with name of array represent address of the first element in the array. So is string who is a chunk of memory with name of the char array address of the first character. In addition, c-styled string terminates with character \0, so if you want to manage memory for string yourself, remember one extra byte for the tailing \0.
As to your second problem, your name in function getname is a local variable whose life time ends when function returns. However, you still want to access name outside the function which is inappropriate. You can solve this be dynamically allocated memory like in dasblinkenlight's and others' post.
Good luck.