print struct pointer c

Question

I'm trying to print the outcome of the function below

void add_votes(struct votes* v1, struct votes* v2) {
    v1->men += v2->men;
    v1->women += v2->women;
}

With this set of code

add_votes(all, michigan);
    printf("all->men: %d\n", &(all->men);

Here are all needed variables and structure

struct votes {
    int men;
    int women;
};
struct votes *michigan = {200, 300};
struct votes *all = {0, 0};

I tried placing an & before the cast in the printf but after a second printf it displays the address of the pointer and when I print it without a & it doesn't display nothing

If it’s supposed to be a pointer how should I do it And if I remove the & from the print it says a series of number I guess the address but I need the value It says something like 1303720 when I need it to be 200 or something like that

Should just be printf("all->men: %d\n", all->men); You don't care about the address at all. Note, though, that you are not correctly initializing your pointer variables; they have to point to something. — Nate Eldredge
– Nate Eldredge, Commented Nov 7, 2020 at 19:08

Barmar · Accepted Answer · 2020-11-07 19:11:41Z

4

Your variable declarations are wrong. They should be structures, not pointers.

struct votes michigan = {200, 300};
struct votes all = {0, 0};

Then you pass the addresses of the structures to the function:

add_votes(&all, &michigan);
printf("all.men: %d\n", all.men);

answered Nov 7, 2020 at 19:11

Barmar

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2 Comments

TechX Over a year ago

Ok but how is the function add_votes going to look if I make it all normal structures it's not assigning values

Barmar Over a year ago

add_votes doesn't change. It's receiving pointers to these structures, so it updates the contents.

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