Passing by reference in C

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Passing by reference in C

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#1 by (2 votes)
#2 by (2 votes)

2

In the case of C language, when we want to pass a variable by reference we need to pass the address of the function to the function, I have a question in this case.

See the example:

int main(){
   int *a;
   *a = 5;
   foo(&a);
}


void foo(int **a){
 //tenho que passar o endereço para garantir a mudança?
   bar(&a);
}

void bar(int ***a){
 //como atribuir? ***a = 1; ??
}

My question is as follows, if I want to pass a reference to a pointer to a variable, is it always necessary to pass the address of what I have in the scope of the function to ensure that the result is changed?

If yes, how do I assign the value when I have **a or ***a ?

Is there another approach? How to facilitate this control?

c ponteiro

asked by anonymous 03.07.2015 / 19:12

2 answers

2

You do not need to use pointer to pointer, just use a "simple" pointer. To understand more about potters, I suggest these two other questions:

I have refactored your code to work as you wish:

#include <stdio.h>

void bar(int *a){
    *a = 1;
}

void foo(int *a){
    bar(a);
}

int main(){
   int a;
   a = 5;
   foo(&a);
   printf("a = %d\n", a);
   return 0;
}

Both functions expect as a parameter a pointer to an integer ( int *a ). The foo function simply passes it to the other (because its parameters have the same data type). Note however:

In the main function I did not declare a pointer, but a "normal" variable. In calling foo(&a) I use the & operator to indicate that I want to pass the address of the a variable there.

In the bar function I update the contents of a "pointed" area by a pointer by doing the following: *a = 1 . This basically means "assign 1 to the memory area pointed to by a ".

That is, whenever you see &algo you are getting the address of the memory area where algo is allocated. Whenever you see *algo you are actually accessing this address (both to read and to change).

Q.: I've reversed the order of functions so the compiler does not complain about the functions not being declared when used.

03.07.2015 / 19:29

Bootstrap Tabs Do not Come Back Stored Proc X Entity - Is it a good decision?

score 2 · Accepted Answer

You need malloc() before using the pointer.

You do not need to always add a level of indirection to each new function: just send the address the first time and then reuse that address.

Object declaration follows usage: int *a means *a is int ; int *****b means *****b is int .

If you want to put a value in an object of type int *** it uses ***a .

See this code

#include <stdio.h>
#include <stdlib.h>

void foo(int **a, int *b);
void bar(int ***a, int *b);

int main(void) {
    int *a;
    int b;
    a = malloc(sizeof *a);
    if (a) {
        *a = b = 5;
        printf("antes de foo(): %d, %d\n", *a, b);
        foo(&a, &b);
        printf("depois de foo(): %d, %d\n", *a, b);
        free(a);
    }
}


void foo(int **a, int *b) {
    // tenho que passar o endereço para garantir a mudança?
    printf("antes de bar(): %d, %d\n", **a, *b);
    bar(&a, b);
    printf("depois de bar(): %d, %d\n", **a, *b);
}

void bar(int ***a, int *b) {
    ***a = *b = 42;
}