Problems changing a void pointer value in C

问题

Basically my problem is that i'm trying to change the value inside the valor variable, so that after calling of the cambiar_valor function it chenges to 25. But my problem is that it doesn't chang at all. What am i doing wrong here?. I'm trying to make a really generic function so that depending of the data type i pass to the function it changes dinamically. In this case is an integer type but what i'm trying to do here is to check if i could change the value of the valor variable inside the function

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

void cambiar_valor(void* valor,int* valor_dos) {//assign valor_dos to valor
    valor = valor_dos;
}

int main() {
    void *valor;
    int *valor_dos = 25;
    cambiar_valor(valor,valor_dos);
    printf("%d \n",(int*)valor);//this should show 25
    return 0;
}

回答1:

In your function

void cambiar_valor(void* valor,int* valor_dos) {//assign valor_dos to valor
    valor = valor_dos;
}

You are passing in the pointers by value, meaning that valor and valor_dos are copies of the parameters you pass in. Reassigning valor inside the function has no effect on the calling function.

To fix this, take the parameters in by pointer:

void cambiar_valor(void** valor, int* valor_dos) {//assign valor_dos to valor
    *valor = valor_dos;
}

Then call

cambiar_valor(&valor, valor_dos);

Also, as @Levon has mentioned, your initialization of valor_dos in main is incorrect and will probably cause a segfault at runtime. You might want to change that as well.

Hope this helps!

回答2:

Here

 int *valor_dos = 25;

you are initializing a pointer to an int with the value 25 .. I.e.,it points at memory location 25, that can only lead to trouble. I'm surprised you didn't get a seg fault.

回答3:

int *valor_dos = 25

This statement is incorrect. You are declaring a pointer here, so you cannot assign a value (25) to it.

来源：https://stackoverflow.com/questions/11044555/problems-changing-a-void-pointer-value-in-c

标签

pointers

void-pointers