Char array in a struct - incompatible assignment? [duplicate]

Answer 1

Or you could just use dynamic allocation, e.g.:

struct name {
  char *first;
  char *last;
};

struct name sara;
sara.first = "Sara";
sara.last = "Black";
printf("first: %s, last: %s\n", sara.first, sara.last);

Answer 2

You can try in this way. I had applied this in my case.

#include<stdio.h>

struct name
{

  char first[20];

  char last[30];

};

//globally

// struct name sara={"Sara","Black"};

int main()

 {

//locally

struct name sara={"Sara","Black"};


printf("%s",sara.first);

printf("%s",sara.last);

}

Answer 3

This has nothing to do with structs - arrays in C are not assignable:

char a[20];
a = "foo";   // error

you need to use strcpy:

strcpy( a, "foo" );

or in your code:

strcpy( sara.first, "Sara" );

Answer 4

sara is the struct itself, not a pointer (i.e. the variable representing location on the stack where actual struct data is stored). Therefore, *sara is meaningless and won't compile.

Answer 5

use strncpy to make sure you have no buffer overflow.

char name[]= "whatever_you_want";
strncpy( sara.first, name, sizeof(sara.first)-1 );
sara.first[sizeof(sara.first)-1] = 0;

Answer 6

The Sara structure is a memory block containing the variables inside. There is nearly no difference between a classic declarations :

char first[20];
int age;

and a structure :

struct Person{
char first[20];
int age;
};

In both case, you are just allocating some memory to store variables, and in both case there will be 20+4 bytes reserved. In your case, Sara is just a memory block of 2x20 bytes.

The only difference is that with a structure, the memory is allocated as a single block, so if you take the starting address of Sara and jump 20 bytes, you'll find the "last" variable. This can be useful sometimes.

check http://publications.gbdirect.co.uk/c_book/chapter6/structures.html for more :) .