String input in C issue

Question

I'm trying to assign two strings in a string array but the compiler won't let me do it

typedef char string[20];
    string dizionario[2];

    dizionario[0] = "ergbeciao";
    dizionario[1] = "ciaozio";

Answer 1

try

string dictionary[2] = {"Item1", "Item2"}

Answer 2

Initialize the array variable while declaring. In your program you are using double dimmesion of char array.I hope that u will understand and it will work

#include<stdio.h>
int main()
 {
    typedef char string[20];
    int i;
    string dizionario[2]={ "ergbeciao", "ciaozio"};
    for (i=0;i<2;i++)
      { 
        printf("\nSting i :%s",i,dizionario[i]);
      }
  }

Output

String 0 : ergbeciao
String 1 : ciaozio

Answer 3

Your code could work if you changed the definition of string to something more useful:

typedef const char *string;

int main() {
    string dizionario[2];

    dizionario[0] = "ergbeciao";
    dizionario[1] = "ciaozio";
}

Note that literal strings ("ergbeciao") are immutable.

Answer 4

You cannot use the = operator to copy the contents of one string (or any array type) to another; you must use the strcpy library function instead:

strcpy(dizionario[0], "ergbeciao");
strcpy(dizionario[1], "ciaozio");

This is a function of how C treats array expressions.

First and most importantly, an array expression may not be the target of an assignment (it is not a modifiable lvalue). dizionario[0] and dizionario[1] are array expressions of type char [20]; by this simple virtue, they may not be the target of the = operator.

Secondly, unless it is the operand of the sizeof or unary & operators, or is a string literal used to initialize a character array in a declaration, an expression of type "N-element array of T" will be converted ("decay") to an expression of type "pointer to T", and the value of the expression is the address of the first element of the array. When you write

dizionario[0] = "ergbeciao";

the compiler converts the expression "ergbeciao" from type "10-element array of char" to "pointer to char", and the value of the expression is the address of the first element.

You can use a string literal as an initializer in a declaration:

string dizionario[2] = { "ergbeciao", "ciaozio" };

In this case, the contents of the string literals will be copied to the array elements as you expect. But this only works as part of a declaration.

Answer 5

First of all you can not directly assign value in string by assignment operator you should have to do string operation like this strcpy(dizionario[0] , "ergbeciao"); and strcpy(dizionario[1] , "ciaozio"); otherwise you have to directly assign it while declare that string.

your code should be like this

#include<stdio.h>
#include<string.h>
void main(){

 typedef char string[20];

 string dizionario[2];

 strcpy(dizionario[0] , "ergbeciao");
 strcpy(dizionario[1] , "ciaozio");

 printf("%s\n %s\n",dizionario[0],dizionario[1]);

 }

your output is:

ergbeciao
ciaozio

cheers..........

Answer 6

This depends upon what you want to accomplish.

If you want to declare and allocate two, contiguous strings of 20 characters, fixed width, there are two approaches,

#define STRMAX (20)
typedef char string[STRMAX+1]; //need room for null terminator
string dizionario[2] = { "ergbeciao", "ciaozio" }; //array of char[20+1]

or you can allocate, then assign values using strcpy,

string dizionario[2] = { 0 }; //empty
strcpy( dizionario[0], "ergbeciao" );
strcpy( dizionario[1], "ciaozio" );

However, you may want an array of pointers to strings, so define differently,

#define STRMAX (20)
typedef char string[STRMAX+1]; //need room for null terminator
string* dizionario[2] = { 0 }; //array of null pointers

Then assign the literal strings (pointers) to the pointer array,

dizionario[0] = "ergbeciao";
dizionario[1] = "ciaozio";

Or provide distinct memory using malloc,

dizionario[0] = malloc("ergbeciao");
dizionario[1] = malloc("ciaozio");

Better, provide constructor and destructor for string,

string*
stringNew(char* sp)
{
    char* np = malloc(STRMAX+1);
    if(!np) return np; //malloc failed
    if(sp) strcpy(np,sp); //null source pointer
    return np;
}
void
stringDel(string* sp)
{
    if(!sp) return; //avoid double free
    free(sp);
    return(sp);
}

dizionario[0] = stringNew("ergbeciao");
dizionario[1] = stringNew("ciaozio");