如何持久化指针增量

Question

I have the following code

   void fun(int * d)
   {
    d+=1;
   }

   int main()
   {
      int * f=malloc(5);
      f[0]=0;f[1]=1;f[2]=2;f[3]=3;f[4]=4;
      fun(f);
      printf("value : %d\n",*f);
      return 0;
   }

So, i pass an integer pointer to function and increment it there. I want that increment to persist when it returns to main. When I print the value, The output is '0'. But I want the output to be '1' as I have incremented the value in function.

So, briefly, my question is, how to persist the changes made to a pointer?

Answer 1

Assuming you want to increment the pointer , not the value, you have two options:

1. recast the function to void fun(int ** d) , use (*d)+=1; in the function body, and call using fun(&f);

2. recast the function to int* fun(int* d) , and return the new value.

If you want to increase the value , then use (*d)+=1; in the function body.

You don't need the parentheses around *d : I've put them in for clarity.

Answer 2

You forgot * in d+=1;

When you pass that pointer you have access to f[0] with that approach: Take a look here:

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

void fun(int *d){
    *d+=1;
}

int main(void){
    int i;
    int *f=malloc(5 * sizeof int);

    f[0]=0;f[1]=1;f[2]=2;f[3]=3;f[4]=4;

    for(i=0;i<5;i++){
        printf("%d ",f[i]);
    }

    printf("\n");

    fun(f);

    for(i=0;i<5;i++){
        printf("%d ",f[i]);
    }
    free(f);
    return 0;
}

Output:

 0 1 2 3 4 1 1 2 3 4

Or if you try to add +1 to all elements you can do something like this:

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

int *fun(int *d, int len){
    int i;
    int *newArray = d;
    int newValue = 1;

    printf("\n");
    for(i = 0; i < len; i++) {
        newArray[i] += newValue;
    }

    return newArray;
}

int main(void){
    int i;
    int f[] = {0,1,2,3,4};
    int *newArray;

    f[0]=0;f[1]=1;f[2]=2;f[3]=3;f[4]=4;

    for(i=0;i<5;i++){
        printf("%d ",f[i]);
    }

    printf("\n");



    newArray = fun(f,5);

    for(i=0;i<5;i++){
        printf("%d ",newArray[i]);
    }
    return 0;
}

Output:

 0 1 2 3 4 1 2 3 4 5

And by the way you forgot to free f .

Answer 3

You have to pass the address of the pointer, so:

void fun(int **val) {
    *val++;
    ....
}
int main() {
    ...
    fun(&f);
    ...
}