I have been studying C for the past weeks, but I can't fully understand how memory manages pointers.
My question arises from this example obtained from here(page 17 of 19): C-Pointer-Basics
Example code:
#include <stdio.h>
void F(int, int *);
int main()
{
int m = 3;
int n = 5;
F(m, &n);
printf("main: %d, %d\n", m, n); // print 3 7 (Where does the 7 came from?)
return 0;
}
void F(int a, int *b)
{
/*
* `a` being a local variable, gets cleaned after `F()` ends
* BUT how does `b` retain the value of `&a`?
*/
a = 7;
*b = a; // `*b = 7`
b = &a; // `b` points to `a`
*b = 4; // `a = 4` and why `*b` doesn't return 4?
printf("F: %d, %d\n", a, *b); // print 4 4
}
The question here is:
Why when main() prints the values of m and n, It showsm = 3 and n = 7?
My assumptions:
As I know, a pointer goes beyond the scope of the function where it is declared, so in void F(int a, int *b) when the function is no longer needed, it gets destroyed, same with his parameters, BUT the value of int *b remains in memory right(even though int *b no longer exists)? So if this is true, we can 'recover' it from memory and use it in main().
Best,
For the question why m and n print 3,7 is because the first parameter is passed by value hence it is only getting copied, therefore no modification of the original m is happening and in the case of n, you're passing its address so when you do *b=a the value of a gets copied to n. And then when you do b=&a the pointer b now starts pointing to the address of a instead of that of n. Which is why the second time you do *b=4, you are not modifying n but a.
Related
I had to write the following function, which returns two output values.
In order to do so I used a pointer for the second output value of the quotient. However when I wanted to test it with an input it seemed to be crushing. The code is:
#include <stdio.h>
int div( int n, int m, int *quotient)
{
int d = 0;
while (n >= m) {
n = n - m;
d++;
}
*quotient = d;
return n;
}
int main(void)
{
int *p;
int rest;
rest = div(7, 2, p);
printf("n - %i, d - %i", rest, p);
return 0;
}
would be happy to know how to fix it and why it happened at first place
Thanks for your help
Change this:
int *p;
int rest;
rest = div(7, 2, p);
To this:
int p;
int rest;
rest = div(7, 2, &p);
The problem with your code is that p points some random unallocated place (or is a null pointer if you're lucky). The updated version allocates space for the integer and then passes its address to the function. The function then has a pointer to this address and can write the value there. The memory is allocated on the stack (local variable) and so everything is fine.
#include<stdio.h>
int g(int *a, int *b);
int main()
{
int a = 2;
int b = 7;
b = g(&b , &a);
printf("a = %d\n", a);
printf("b = %d\n", b);
return 0;
}
int g(int *a, int *b)
{
(*a) = (*a) + 3;
(*b) = 2*(*a) - (*b)+5;
printf("a = %d, b = %d\n", *a, *b);
return (*a)+(*b);
}
The output is:
a = 10, b = 23
a = 23
b = 33
I'm in an Intro to C programming class and having trouble understanding how this works.
Thanks for the help!
Sequencing the events as presented in question:
int main()
{
Declaration of a and b and value assignment:
int a = 2;
int b = 7;
Here is a trick, the address passed to the parameter int* a is actually of b, and vice-versa on the second parameter:
b = g(&b , &a);
Here just printing values of a and b:
printf("a = %d\n", a);
printf("b = %d\n", b);
return 0;
}
Since the parameters are pointers, the changes made, in the scope of this function, to the variable addresses pointed by them are permanent:
int g(int *a, int *b) {
Here, dereferencing the pointer (*a, the parentheses are not needed in these cases), means you are now working with the value stored in the address pointed by a, so 7 + 3 = 10, now the value stored in the address pointed by a is = 10:
(*a) = (*a) + 3;
Here, the same thing, dereferencing pointers, so 2 * 10 - 2 + 5 = 23, the value stored in the address pointed by b will be 23:
(*b) = 2*(*a) - (*b)+5;
Here printing a = 10 and b = 23, again, dereferencing pointers means you are working with the values stored in the addresses pointed by them:
printf("a = %d, b = %d\n", *a, *b);
The returned value is 10 + 23 = 33, so for b = g(&b, &a), b will be assigned the value of 33, a is already 23 so it stays that way:
return (*a)+(*b);
}
Remember that C passes all function arguments by value - that means that the formal parameter in the function body is a separate object in memory from the actual parameter in the function call, and the value of the actual parameter is copied to the formal parameter.
For any function to modify the value of a parameter, you must pass a pointer to that parameter:
void foo( T *ptr ) // for any type T
{
*ptr = new_T_value(); // write a new value to the thing ptr points to
}
void bar( void )
{
T var;
foo( &var ); // write a new value to var
}
In the code above, all of the following are true:
ptr == &var
*ptr == var
Thus, when you write a new value to the expression *ptr, it's the same as writing a new value to var.
I think part of what's making this confusing for you is that the names of your formal parameters (a and b) and your pointers (a and b) are flipped - g:a points to main:b and vice versa.
g:a == &main:b // I'm using g: and main: here strictly to disambiguate
*g:a == main:b // which a and b I'm talking about - this is not based on
// any real C syntax.
g:b == &main:a
*g:b == main:a
With & you give the address of the variable to the function, instead of the value.
With * you can access the value of an address.
With b = g(&b , &a); you give the address of the variable b and a to the function.
But you can access the address of b with * a because you declare the function that way: int g (int * a, int * b):
*a points to the address of your b variable.
*b points to the address of your a variable.
I think the different variable names are what confuses you.
To make it easier for yourself you could change the declaration to int g (int * b, int * a)
In case you want to change it:
*b would point to the address of your b variable and
*a would point to the address of your a variable.
by using the * you access the object referenced by the pointer. As the pointers are referencing int variables a & b you do the operations on those variables. I think the same variable names are confusing you
int g(int *p1, int *p2)
{
(*p1) = (*p1) + 3;
(*p2) = 2*(*p1) - (*p2)+5;
printf("*p1 = %d, *p2 = %d\n", *p1, *p2);
return (*p1)+(*p2);
}
I would like help understanding this code:
void F (int a, int *b)
{
a = 7 ;
*b = a ;
*b = 4 ;
printf("%d, %d\n", a, *b);
b = &a ;
}
int main()
{
int m = 3, n = 5;
F(m, &n) ;
printf("%d, %d\n", m, n) ;
return 0;
}
I am confused why this does not result in unexpected behavior. At the end of the function F, the value of b is 7. But when I return it is clear that nothing after ' b = &a ' impacts the value of n/b.
I thought that pointing to a local variable would result in garbage/unexpected behavior when the scope changed, but that doesn't appear to be the case.
When you call F, you pass the value of m and the address of n.
In F, b is a pointer to n so that when you change the value of *b, the value of n is changed. However, in "b = &a;" you are changing where b points. After that line, b no longer points to n. Instead, it points to a. That line does nothing at all to the variable n back in main(). After that point, if you change the value of *b, you will change the value of *b and its alias, a. It will not change the value of n in main.
At the end of the function F, the value of b is 7
The value of b is not 7; b is set to point to a, which has the value of 7
But when I return it is clear that nothing after b = &a impacts the value of n
Although b is a pointer, it is passed by value. b inside F behaves as if it were a separate, fully independent, local variable. Any modifications to it made inside F are local to F.
In fact, you cannot trigger undefined behavior in main by actions inside F, because main is not receiving any pointers from F. If you want to make undefined behavior, pass a pointer to pointer into F, and assign it to point to a local variable:
void CauseUB(int a, int **b) {
*b = &a;
}
int main() {
int x = 5, *y = &x;
printf("This is OK: %d\n", *y);
CauseUB(x, &y); // Pointer to pointer
printf("This is UB: %d\n", *y);
return 0;
}
Demo.
I would like to swap two variables containing 2D arrays. I believe this can be simply done by swapping their pointers. I tried this code, but it does not work and I have no idea why, perhaps I am not understanding pointers correctly.
#include <stdio.h>
void swap(int ***a, int ***b) {
int ** temp = *a;
*a = *b;
*b = temp;
}
int main(void) {
int a[10][10];
int b[10][10];
a[1][5] = 4;
b[1][5] = 2;
printf("%d, %d\n", a[1][5], b[1][5]);
swap(&b, &a);
printf("%d, %d\n", a[1][5], b[1][5]);
return 0;
}
This outputs
4, 2
4, 2
I would expect it to output
4, 2
2, 4
So, what am I doing wrong?
a and b in main function are not pointers but arrays.
If you want to use pointers, use pointers.
#include <stdio.h>
#define N 10
void swap(int (**a)[N][N], int (**b)[N][N]) {
int (*temp)[N][N] = *a;
*a = *b;
*b = temp;
}
int main(void) {
int a[N][N];
int b[N][N];
int (*pa)[N][N] = &a;
int (*pb)[N][N] = &b;
(*pa)[1][5] = 4;
(*pb)[1][5] = 2;
printf("%d, %d\n", (*pa)[1][5], (*pb)[1][5]);
swap(&pb, &pa);
printf("%d, %d\n", (*pa)[1][5], (*pb)[1][5]);
return 0;
}
This would not work because what you are swapping is actually what the variables in the swap() function are pointing to, and not what they are pointing at. Its like if a was pointing at 5 and b was pointing at 6 in the swap() function, then it will make a point at 6 and b point at 5, without changing the contents of the memory. This would mean the in the main() function they would be residing at the same place and would be getting pointed by the same variable a and b (different from the variable in swap())
To swap you need to swap the contents of that memory in the swap() function, so that it is reflected in the main() function.
Recall that while doing swapping using pointers, one sends the address and actually swaps the content by dereferencing (*p and *q).
static void swapAddr(int *numOne, int *numTwo)
{
int *tmp;
tmp = numOne;
numOne = numTwo;
numTwo = tmp;
}
int main(void)
{
int a = 15;
int b = 10;
printf("a is: %d\n", a);
printf("Address of a: %p\n", &a);
printf("b is: %d\n", b);
printf("Address of b: %p\n", &b);
swapAddr(&a, &b);
printf("\n");
printf("a is: %d\n", a);
printf("Address of a: %p\n", &a);
printf("b is: %d\n", b);
printf("Address of b: %p\n", &b);
return 0;
}
When I compile and run this piece of code, the output is
a is: 15
Address of a: 0x7fff57f39b98
b is: 10
Address of b: 0x7fff57f39b94
a is: 15
Address of a: 0x7fff57f39b98
b is: 10
Address of b: 0x7fff57f39b94
Clearly the result is not what I intended, since the address does not seem to have been swapped at all.
You generally can't change the address of a variable.
Your 'swapAddr' function changes its parameter values, but these are local to the function - you're not changing anything outside the function. Perhaps the best way of understanding it is that a function parameter always receives a copy of the value that was passed to the function. In this case, you get a copy of the address of a and a copy of the address of b. You can and do change the values of the variables holding those copies (numOne and numTwo), and seeing as they are pointers you could (but don't) change the values that they point at (the values of variables a and b) - but you can't change the addresses of the original variables.
To break it down:
static void swapAddr(int *numOne, int *numTwo)
{
int *tmp;
tmp = numOne;
At this point, tmp and numOne both point to the value of the a variable...
numOne = numTwo;
Now, numOne points instead to the value of the b variable...
numTwo = tmp;
}
And finally, numTwo now points to the value of the a variable. The function returns and numOne and numTwo no longer exist after that point. The addresses of the variables a and b did not change at any stage.
You could however write a function which exchanges the addresses in two pointer variables:
static void swapAddr(int **ptrOne, int **ptrTwo)
{
int *tmp;
tmp = *ptrOne;
*ptrOne = *ptrTwo;
*ptrTwo = tmp;
}
This would allow you to pass the address of two pointer variables, and on return the pointers would be swapped - each one pointing at what the other did previously. But again, this would not change the address of any variable that those pointers happened to point to.
The pointers are passed to the function by value, so changing what they point to isn't going to change the value of the passed parameters in the calling function.
When the function is called, a copy of each pointer is made and saved to the stack. Then the function reads each pointer value off the stack and manipulates them. It never changes the value of the original pointer that was copied onto the stack.
Remember that in C values are passed by value to functions, meaning that the values are copied. When you modify an argument in a function you only modify the local copy inside the function, not the original value that was passed to the function. This goes for pointers as well.
To solve your problem you must pass the arguments by reference, but unfortunately C doesn't have that, it only have pass by value. However, pass by reference can be emulated by passing pointers to the data, just like you do in the function. You must however dereference the pointer to get the values from where the pointers point to, and use those values to do the actual swapping:
int temp = *numOne; // Note: temp is a value not a pointer
*numOne = *numTwo;
*numTwo = temp;
static void swapAddr(int *numOne, int *numTwo)
In this function you are passing 2 pointers by value. This allows you to modify the int pointed to by the pointers but not the pointers themselves.
Use this function definition instead that passes pointers to pointers and allows modifying the pointers themselves
static void swapAddr(int **numOne, int **numTwo) {
int *tmp = *numOne;
numOne = *numTwo;
numTwo = tmp;
}
You could use it like this for example:
int *a = malloc(sizeof(int));
int *b = malloc(sizeof(int));
*a = 15;
*b = 10;
swapAddr(&a, &b);
You canlt change the addresses. The adderss of a is the address of a and that will remain the same until the end of days.
You can do:
static void swapAddr(int **numOne, int **numTwo)
{
int *tmp;
tmp = *numOne;
*numOne = *numTwo;
*numTwo = tmp;
}
int main(void)
{
int a = 15;
int b = 10;
int *pa= &a;
int *pb= &b;
swapAddr(&pa, &pb);
}
What you want to achieve is something like
int *c = &a;
&a = &b;
&b = &a;
This is not possible (you can check: it will not compile). A variable that is created is placed at one place in memory and stays there. So when you create a variable a it will stay variable a and it will not be able to change its identity to that of another variable b.
What you can do is use two pointers int *p1, *p2 to int. These pointers can change their value and point to other objects during lifetime:
p1 = a;
p2 = b;
p1 = b;
p2 = a;
a and b will stay the same, but p1 and p2 can point to different objects over time.
So a thing that would be possible:
static void swapaddr(int **pp1, int **pp2)
{
int *pp;
pp = *pp1;
*pp1 = *pp2;
*pp2 = pp;
}
int main(void)
{
int a = 15, b = 10;
int *pA = &a, *pB = &b;
swapAddr(&pA, &pB);
}
In this example a and b would keep their identity and address, but pA and pB would change their value and pA would point to b and pB would point to pA.
You cannot change the addresses of the variables.however you can change values of pointers,which store addresses as their value,here is an example :
#include <stdio.h>
void swapAddr(int **numOne, int **numTwo)
{
int *tmp;
tmp = *numOne;
*numOne = *numTwo;
*numTwo = tmp;
}
int main(void)
{
int a = 15;
int b = 10;
int *p_a = &a;
int *p_b = &b;
printf("Address of a: %p\n", p_a);
printf("Address of b: %p\n", p_b);
swapAddr(&p_a,&p_b);
printf("\n");
printf("p_a : %p\n",p_a);
printf("p_b : %p\n",p_b);
return 0;
}