CODEWITHSUNDEEP
c
iBaer
iBaer

Reputation: 37

Struct function doesn't work with my array

So I basically have a struct with a name, which has to be dinamically, and an ID. I think it can look like this.

typedef struct {
char *name;
unsigned int id;
} person;

Now I shall write a function with the given start:

person *readData();

Both the struct and the name have to be dinamically, I want to do this with the malloc function. For all persons there should also be an array, let's name it "people[1000]".

Here are is my try on the said function with the main function:

int count = 0;

person *readData() {
    int i, len;
    char puffer[1000];

    printf("Name: ");
    scanf_s("%999s", &puffer);

    len = strlen(puffer);
    people[count].name = (char *)malloc((len + 1)*sizeof(char));

    for (i = 0; i < len; i++)
        people[count].name[i] = puffer[i];

    people[count].name[len] = '\0';
}

void main(void)
{
    person *people[1000];
    readData();
    printf("\n%s\n", people[count].name);
}

Well, it doesn't seem to work that way. Visual Studio says in the function, that "people" has to be of type union or struct. Any quick input? It's just basic C since I am at the start of learning it.

EDIT: Full code:

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

typedef struct {
char *name;
unsigned int id;

} person;

person people[1000];

int count = 0;

person *readData() {
int i, len;
char puffer[1000];
printf("Name: ");
scanf_s("%999s", &puffer);
len = strlen(puffer);
people[count].name = (char *)malloc((len + 1)*sizeof(char));
for (i = 0; i < len; i++)
    people[count].name[i] = puffer[i];
people[count].name[len] = '\0';


}

void main(void){
readData();
printf("\n%s\n", people[count].name);

}

Upvotes: 1

Views: 99

Answers (2)

zavg
zavg

Reputation: 11061

people variable is undefined in the body of your readData function. That is why the complier fails with error.

  1. You can pass variable to the function person *readData(person *people). Do not forget to change readData(); to readData(people); in the main function.

  2. Do not mix pointer notation with array notation if you need just 1-dimensional array. Use person people[1000]; instead of person *people[1000];

Upvotes: 2

dutt
dutt

Reputation: 8209

Your function readData doesn't know about your people array. If you rewrite it along these lines:

void readData(person** people) {
    your code
}

void main() {
   person* people[100];
   readData(people);
}

It should go better.

Please see zavgs answer for more things to improve.

Upvotes: 0

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