Modify an array of strings from a struct

Question

I usually deal with array of strings this way, because it allow me to not specify a character limit :

char *names[2] = {"John","Doe"};
printf("%s\n",*((names)));
printf("%s\n",*((names)+1));

I am unable to reproduce this while using a struct.

I tried with both john.names = {"John","Doe"}; and john.*names = {"John","Doe"}. But I am getting an expected expression error.

However, I am able to do so during the initialization with Person john = {{"John","Doe"}};. So I'm not sure if it's allowed to proceed like this only during initialization.

main.h

typedef struct Person Person;

struct Person
{
    char *names[2];
};

main.c

#include <stdio.h>
#include <stdlib.h>
#include "main.h"
int main()
{
    Person john = {{"John","Doe"}};
    john.names = {"John","Doe"}; // Expected expression error
    printf("%s\n",john.names[0]);
    printf("%s\n",john.names[1]);
    return 0;
}

What would be the "expected expression", am I allowed to do this ?

@xing yes of course, I am using this to printf just before my return 0. But what I would like is being able to use brackets. — shellwhale
– shellwhale, Commented Jan 14, 2019 at 17:03

Serge Ballesta · Accepted Answer · 2019-01-14 17:18:01Z

4

Arrays are not first class elements in C. You can initialize a full array, but you can only assign to non array elements, meaning scalars, pointers or structs.

When you write

char *names[2] = {"John","Doe"};

, it is an initialization, not an assignment. And the following assignment would also choke with a syntax error:

char *names[2];
names = {"John","Doe"};    // syntax error here

answered Jan 14, 2019 at 17:18

Serge Ballesta

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static_cast · Accepted Answer · 2019-01-14 17:02:46Z

3

For initialization of a specific field/attribute, you can do something like the following during declaration:

Person john = {.names = {"John","Doe"}};

To initialize additional fields, let's say address and names, this can be done like the following:

Person john = {.names = {"John","Doe"}, .address="foo"};

After declaration, you will have to specify the array index.

john.names[0] = "John"

answered Jan 14, 2019 at 17:02

static_cast

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3 Comments

shellwhale Over a year ago

"After declaration, you will have to specify the array index." So from what I understand, there is no way to use brackets after initialization ?

static_cast Over a year ago

To populate multiple indices? i.e. 0,1 at the same time? No.

Jonathan Leffler Over a year ago

@shellwhale: C does not support array assignment — and john.names = … is attempting to assign to an array.

fratardi · Accepted Answer · 2019-01-24 13:25:10Z

2

Well , i guess that there are may ways to access the content of a struct , the way i learned it on my side is for a structure looking like that //

typedef struct struc_person { char *name[2]; } type_person;

type_person->name[0] = value_of_the_first_name;

type_person->name[1] = value_of_the_surname;

answered Jan 24, 2019 at 13:25

fratardi

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1 Comment

Rup Over a year ago

That depends what type_person represents in your assignment example: if it's a pointer to the type_person that you've typedeffed then yes you would use -> but if it's an instance of the structure then it's .. As written this wouldn't compile, but if you meant type_person to be an instance of type_person then you need to use . not -> here.

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