将结构指针传递给C中的函数时发生奇怪的错误

Question

I've written a simple C test program to practice using structs. However I can't understand what's wrong with the code; it seems that the first printout of either the integer or the double field clears both fields. What's my mistake here?

the code:

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

struct Dummy {
    int i;
    double d;
    char s[8];
};

void print(struct Dummy *dummy) {
    printf("%lf ", dummy->d);
    printf("%lf ", dummy->d);
    printf("%d ", dummy->i);
    printf("%s\n", dummy->s);
}

void set(struct Dummy **dummy) {
    struct Dummy duh;
    duh.i = 1;
    duh.d = 3.14;
    strcpy(duh.s, "Hello!");
    *dummy = &duh;
}

int main(int argc, char **argv) {
    struct Dummy *dummy;
    set(&dummy);
    print(dummy);
    return 0;
}

its output:

3.140000 0.000000 0 Hello!

Thanks a lot!

Answer 1

You've invoked undefined behavior by assigning the address of a local variable in set and returning it.

*dummy = &duh;

duh is no longer valid after set returns.

Answer 2

In set you assign **dummy to variable duh which is defined in the scope of the function. This means that duh will be lost at the end of the function, even if you keep a pointer to it. To correct assign directly to the function's parameter

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

struct Dummy {
    int i;
    double d;
    char s[8];
};

void print(struct Dummy *dummy) {
    printf("%lf ", dummy->d);
    printf("%lf ", dummy->d);
    printf("%d ", dummy->i);
    printf("%s\n", dummy->s);
}

void set(struct Dummy *dummy) {
    dummy->i = 1;
    dummy->d = 3.14;
    strcpy(dummy->s, "Hello!");
}

int main(int argc, char **argv) {
    struct Dummy dummy;
    set(&dummy);
    print(dummy);
    return 0;
}