Pointer return pointing to a local character vector?

Question

#include <stdio.h>

char *local(void);

void main(void) {
    char *ponteiro = local();
    printf("%s\n", ponteiro); /*Saida = - (*/
}

char *local(void) {
    char nome[12] = "programador";
    char *ponteiro = nome;
    return ponteiro;
}

In the example above I have the following output commented in the code "- (" I believe it is some garbage in the memory region, however if I access the pointer by means of index it prints me the character, for example ponteiro[0] will print "p", now if I use the storage class type modifier static in the variable nome the output occurs as expected.

Why does this occur on the three occasions cited?

Answer 1

Let’s understand what’s making each line:

char nome[12] = "programador";

Created a space in data stack for a pointer to characters that will be allocated in a static area of the application. There are 12 characters (11 useful because of the terminator). This pointer is called nome.

char *ponteiro = nome;

Created a space in the data stack for a pointer (4 or 8 bytes depending on the architecture) and placed the pointer to the previous space referenced as nome.

return ponteiro;

Ended the execution of the function by returning (copying) the value of this pointer, "destroying" every area allocated in this function (actually it is only unavailable)

Now back to the main function. ponteiro here has the same value of`called function pointer, there was a copy.

printf("%s\n", ponteiro); /*Saida = - (*/

The pointer there is pointing and portando accessing the memory area that was destroyed. Anything can happen. This is called undefined behavior.

When you put one static in the variable is saying that you want to allocate the content in the static area of your application, an area that is never destroyed (and other than the stack), then the die is there as you wish.

This code better demonstrates what you want:

#include <stdio.h>
#include <string.h>

char *local(void) {
    static char nomex[12] = "aaaaaaaaaa";
    char nome[12] = "programador";
    char *ponteiro = nome;
    printf("%s\n", nome);
    printf("%s\n", ponteiro);
    printf("%p\n", (void *)ponteiro);
    printf("%p\n", (void *)nome);
    printf("%p\n", (void *)nomex); 
    printf("%p\n", (void *)&ponteiro); 
    return ponteiro;
}

int main(void) {
    char *ponteiro = local();
    printf("%s\n", ponteiro);
    printf("%c\n", ponteiro[0]);
    printf("%p\n", (void *)ponteiro);
}

Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.

Now look at this:

#include <stdio.h>
#include <string.h>

char *local(void) {
    static char nomex[12] = "aaaaaaaaaa";
    char nome[12] = "programador";
    char *ponteiro = nome;
    return ponteiro;
}

int main(void) {
    char *ponteiro = local();
    printf("%s\n", ponteiro);
    printf("%c\n", ponteiro[0]);
    printf("%p\n", (void *)ponteiro);
}

Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.

Depending on the compiler will give different results, even not changing anything in the variables.