While this question is a bit easy, I really dont know what is the exact explanation about this.
char *ptr = 'a';
I know a char is not a string but isn't that obvious to store 'a' in the first index of the pointer?
The assignment
char *ptr = 'a';
is equivalent to
char *ptr = 97; // This assumes ASCII encoding
While the assignment itself is valid, dereferencing this pointer is not valid.
If you would like to assign a pointer to point to something that has character 'a' in it, you could use a string literal, an array initializer, or take a pointer of a single-character:
char *ptr = "a"; // String literal
char a[] = {'a'}; // Character array
char *ptr = a;
char a = 'a'; // Single character
char *ptr = &a;
You can specify the address of a pointer. Basically it is a pointer to a pointer.
char* ptr = "a";
char** addrOfPtr = &ptr;
Note that storing the 'a' in the first index of the pointer isn't really what is going on here.
The 'a' is located somewhere in the computer's ram. The value stored in ptr is a number you can use to look up the ram location. It is known as an "address" because when people described these locations they made an analogy to home street addresses.
So 'a' is not stored in the first index of the address, any more than you are stored in the first index of your street address. You reside within the home at your street address, and the value for 'a' resides within the memory specified by the address value stored in ptr.
Note that this means the "pointer to a pointer", addrOfPtr, contains the address where the value within ptr resides.
---- Edit to drive the point home ----
#include <stdlib.h>
#include <stdio.h>
int main(int argc, char** argv) {
char* ptr = "a";
char** ptr_ptr = &ptr;
printf("the value of ptr is '%s', and it is located at %p\n", ptr, &ptr);
printf("the value of ptr_ptr is '%p', and it is located at %p\n", ptr_ptr, &ptr_ptr);
return 0;
}
Will print out a pointer's value and address.
the value of ptr is 'a', and it is located at 0x7ffff79642d8
the value of ptr_ptr is '0x7ffff79642d8', and it is located at 0x7ffff79642d0
You can see that the value of ptr_ptr is the address of ptr, showing that a pointer stores the address of a value, in such a way that the syntax is designed to make it easier to get the value at that address than the actual stored address.
A pointer must have an address assigned to it. 'a' is not an address. It is a integer type with a value of 97, assuming ASCII. If you assign ptr the value 97 and try to dereference it you'll most likely get a segmentation fault because the address 97 does not exist.
char *ptr = "a"; is correct because the compiler will store the characters 'a' and \0' in contiguous memory, with the address of 'a' being stored into ptr.
Related
I have a variable
char **data;
and I'm trying to print the content of the variable. I understand it to be array of arrays. How would I print it?
Thanks
It's a pointer to a pointer not array of arrays
You need use **data to get value
data will give the memory address which data variable holds
*data will give the memory address of the pointer whose address data is holding
**data will give you the actual value
Say you have a variable like
char c = 'X';
char *cp = &c; //cp stores memory address of c
char **cpp = &cp; //now cpp is pointer which points to a existing pointer cp
To get the value stored at the location which pointer points to - we need to dereference the pointer.
printf("%c", *cp); //prints 'X'
printf("%p", *cpp); //prints memory address of cp
printf("%c", **cpp); //prints 'X' (double stars because address->address->value)
char mening[] = "tjena pa dig hog";
This string contains 16 characters. I am then using a function adresss() to find the memory address of a random character in that array. The function adresss() returns a pointer containing the address. The address is at this moment 0x7ffeefbff5f9.
I now need to know what positions that address is pointing to, example is it pointing to the "t" at position 0 in the array, or maybe it is pointing to "d" at position 9. How do I do this?
Edit:
char* adresss(char mening[]){
//Lots of code going on here
return &mening[i];
}
int main(void){
char mening[] = "tjena pa dig hog";
char* ptr;
ptr = adresss(mening);
printf("%p\n", ptr);
That is basically how I get the memory adress. I want to know what "i" was, inside the main function only knowing the memory adress.
If you have two pointers, both pointing to the same array (or to one beyond the end of the array), then you can subtract them from each other.
For example:
char mening[] = "tjena pa dig hog";
char *pointer_to_mening = &mening[10]; // Pointer to the eleventh character
// Should print 10 (which is the index of the eleventh character)
printf("The distance is %zu\n", pointer_to_mening - mening);
Experts I've some doubts in this programme.
#include<stdio.h>
void main() {
char str1[] = "Hello";
char * p = "Hello", * s, * q;
p = "Bye";
printf("%s", p);
s = p;
printf("%s", s);
q = str1;
printf("\n%s", q);
}
Here p,s and q are character pointers. As we know pointers store the address of a variable and suppose if we used p=str1 then it should store the base address of array str1. Then here how p is acting like an array itself which is storing a string? I've not really understood what's character pointer. Is it not really a pointer? Because we are not using & and neither are we getting the address as an output. Please help. Thank you.
You are very much correct here:
As we know pointers store the address of a variable and suppose if we used p=str1 then it should store the base address of array str1.
Your first question:
Then here how p is acting like an array itself which is storing a string?
You can say p is acting like an array but in reality, it is just a pointer which is pointing to the base address of string str1 which is a null-terminated string.
Your second question:
I've not really understood what's character pointer. Is it not really a pointer?
A character pointer is again a pointer like the pointers to other types in C.
But there is catch here. when you do:
char a = 'A';
char *ptr = &a; // ptr points to character 'A'
Here ptr is pointer to a character.
But when you do:
char *str = "Hello";
char *ptr = str; // ptr points to first character of string str
Here ptr is pointer to a string
A point to note here is - pointer to a character is different from the pointer to a string.
In C, strings are defined as an array of characters. The difference between a character array and a string is the string is terminated with a special character ‘\0’.
So,
char str[6] = {'H', 'e', 'l', 'l', 'o', '\0'};
char str[] = "Hello";
Both are same but there is a difference in the way they have been initialized.
Wherever in the program, we use str, this will give the base address of string "Hello". Similarly, string literals are also an array of characters with an exception that they can not be changed because the compiler may put them read-only data section.
So, if we have char *ptr = str then
ptr[0] == str[0]
ptr[1] == str[1]
.....
..... and so on
Because,
ptr[0] is *(ptr + 0) which is character at 0th location of string str and can also be written as *ptr and
ptr[1] is *(ptr + 1) which is a character at the 1st location of string str.
When we increment a pointer, it gets incremented in steps of the object size that the pointer points to. Here, ptr is pointer to char so, ptr+1 will give address of next character and *(ptr + 1) give the character at that location. That's why to the user it looks like its acting like an array.
Your third question:
Because we are not using & and neither are we getting the address as an output.
If I am getting it correctly you want to print the base address of the string, the pointer is pointing to.
In order to get the base address of string a pointer is pointing to you need to use %p. Like this:
#include<stdio.h>
int main() {
char str1[] = "Hello";
char * ptr = str1;
printf ("%s\n", str1);
printf ("%s\n", ptr);
printf ("%p\n", ptr);
printf ("%p\n", str1);
printf ("%p\n", &str1[0]);
return 0;
}
Output on my system:
Hello
Hello
0x7fff5e997b46
0x7fff5e997b46
0x7fff5e997b46
Here you can see - ptr, str and &str[0] giving same address.
In the next code:
char i,*p;
i = 65;
p = &i;
p = (char *) 66;
(*p)++;
printf("%d",p);
I got segmentation fault. I didn't understand why. I have a pointer to a char (in this case char 66=C), and then I change it value, which is also 66 - to 67. Are the values of char "protected" from this change? Is it happen also with others, except char?
I tried to understand the idea that stand behind this thing (and not only fix it). Thanks.
Here is the problem:
p = (char *) 66;
It should be:
*p = 66;
p is a pointer to a char, so you cannot assign values like 66 to it. You can derefernce p in order to assign values to where the pointer "looks".
If you want to print the value where p points to, you must use again the dereference operator (*) like this:
printf("%d", *p); // prints the value where p points to
If you want to print the pointer address you can do this:
printf("%p", p); // prints the address where p points
A character pointer doesn't store a character, it stores an address where a character can be found. So
p = (char *)66;
says that p points to address number 66, where a character can be found. Odds are that address isn't even accessible by your program, much less that it stores a character.
I am speaking in Standard, K&R C.
Given:
const char a[] = {1, 2, 3};
const char *p = NULL;
Are these two statements equivalent:
*p = a;
p = a;
Each of them would be on the third line of the snippet.
1 and 2 certainly don't look the same.
What's the difference between the two then?
No.
p = a initializes the pointer to point to something else (usually it copies another pointer or you will point to a reference, ala p = &a.
*p = a initializes what p refers to. You are "dereferencing" (looking at) what p points to. If p points to NULL as in your example, you will crash (this is good! you do not want to accidentally access something and mess your program up).
In this case, p = a will point to the first of the array a[], and *p = a will attempt to change the first of the array (it won't work; you have it declared const).
Here is a small example program in C++, with almost identical syntax to C.
#include <iostream>
int main()
{
char arr[5] { 'a', 'b', 'c' }; // arr[3] and arr[4] are set to 0
char *ptr = arr; //point to 'a'
for (int i = 0; i != 5; i++)
{
*ptr = 'f'; //this changes the array
ptr++; //this changes what the pointer points to; moves it to next in array
}
for (int i = 0; i != 5; i++)
{
std::cout << *ptr << " ";
}
//outputs f f f f f
}
The * operator is what we call the dereference operator. To understand what it does, you must understand exactly what a pointer is.
When you do
char *p;
the "variable" p does not use the same amount of memory as a normal char, it uses more memory: it uses the amount of memory needed to correctly identify a memory position in your computer. So, let's say you use a 32-bit architecture, the variable p occupies 4 bytes (not the 1 byte you would expect from a char).
So, when you do
p = a;
you see clearly that you are changing the contents of the variable p, that is, you are putting another 32-bit number inside it: you are changing the address it is pointing to.
After that line executes, the value of p is the memory address of the character array a.
Now for the dereference operator. When you do
*p = 'Z';
you are telling the compiler that you want to store the value 'Z' ON THE ADDRESS pointed by p. So, the value of p remains the same after this line: it continues to point to the same address. It's the value of this address that has changed, and now contains 'Z'.
So, the final effect of
char a[] = {'a', 'b', 'c'};
char p = a;
*p = 'Z';
is the same as changing the first position of the array a to 'Z', that is:
char a[] = {'a', 'b', 'c'};
a[0] = 'Z';
NOTE: there is a difference when making a pointer point to an array: the variable that contains the array contains only the address of the first element, so a is the same as "the starting address of the array".
Usually you will see the & operator. It is an operator used to obtain the memory address of a variable. For example:
int number = 42;
int pointer = &number;
printf("%d", *pointer);
Here we have them all. The first line creates an integer variable and stores 42 inside it.
The second line creates a pointer to an integer, and stores the address of the variable number inside it.
The third line reades the value on the address pointed by the pointer.
So, the trick is to read *x as on the address pointed by x and &x as the address of x.
The first dereferences a null pointer, and tries to assign it the address of the array. This will be a compiler error, because char != char []. If it weren't, it would likely crash.
The second sets p to point to the the array.
I think you are mistaking:
char a[8];
char *p=a;
which is legal and does the same as:
char a[8];
char *p=NULL;
p=a;
with:
char a[8];
char *p=NULL;
*p=a;
which as others said would generate a compile error or a segmentation fault.
In the left side of declarations you should read *x as pointer(x) while in
statements it must be read as value_pointed_by(x). &x on the other hand
would be pointer_to(x)
Here's a trick I used when I learned C (and still use today).
Whenever you see the * in front of a variable in your code, automatically read it as "what is pointed to by".
So you should be able to easily see that setting "p" to "a" is very different from setting "what is pointed to by p" to "a".
Also, since p is supposed to be pointing at a char, setting that char p is pointing at (currently the "char" at memory location 0 assuming null is 0) to a char pointer (a) is probably going to fail at compile time if you are lucky (depending on your compiler and lint settings it may actually succeed.)
from comment:In a function declaration like f(char c), I usually try to separate out the variable name from the rest of it--so it would be f( (char) c). so c is a char*. Exactly like a variable definition.
Also & usually reads as "The address of", but that gets even more iffy. A few examples of how I read things to myself. May or may not help you.
int a[] = {1,2,3}; // I mentally parse this as (int[]) a, so a is an int array.
int *p; // p is a pointer to "integers"
int i;
p=a; // p acts exactly as a does now.
i=*p; // i is "What is pointed to by" p (1)
i=p; // i is some memory address
i=*a; // i is what is pointed to by a (1)
i=p[1]; // Don't forget that * and [] syntax are generally interchangable.
i=a+1; // Same as above (2).
p=&i; // p is the address of i (it can because it's a pointer)
// remember from hs algebra that = generally reads as "is", still works!
*p=7; // what is pointed to by p (i) is 7;
a=*i; // whoops, can't assign an array. This is the only difference between
// arrays and pointers that you will have to deal with often, so feel
// free to use which ever one you are more comfortable with.
char c='a';
char * d = &c;// d is a char pointer, and it is the address of c
char ** e ; // e is a pointer to a memory location containing
// a pointer to a char!
e=&d; // gets d's address. a pointer to a pointer gets
// the address of a pointer. Messy but gets the job done
**e=5; // what is pointed to by what is pointed to by e is 5.
*e=&'f'; // what is pointed to by e (which is a char * itself, and is still d!)
// is set to the address of the memory location holding the value 'f'.
// does not change c or e, just d!
I haven't touched c in 10 years, so some of this may be a bit wrong, but it helps me to read it out loud that way.
No, they are not equivalent
If p = NULL, then doing *p = a will give you a segmentation fault.
Because "*p" dereferences the pointer wouldnt this make "p" a "char**" ?
This would point "p" to the first array as expected.
I guess they are not the same.