character array please Help!!!

void main()
{

char name[20];

name = "omshanti";

cout << name;

}


In the above program , if we won't initialize name as the type of its definition , why there's compile time error :
cannot convert from 'char [9]' to 'char [20]'.???????

while using other data type such as int , float , double , we can initialize them after defining them .But same is not the case with character string. WHY???????

Quote:

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But same is not the case with character string.

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Because a string is an array of characters. Use memcpy to put characters in a character array.

Edit : That needs to be strcpy.

You can't assign a string literal to a char array like that. Use strcpy.

Quote:

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Originally Posted by Skizmo

Because a string is an array of characters. Use memcpy to put characters in a character array.

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Why memcpy instead of strcpy?

Quote:

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Originally Posted by GCDEF

Why memcpy instead of strcpy?

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Doh... sorry. Not having my day today. Of course it has to be strcpy.

Code:

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strcpy(name,"omshanti");

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Quote:

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Originally Posted by Tanushreeagr

Code:

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void main()
{
        ...
}

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You are using incorrect signature for main(): it must be int instead of void and it must return an integer value.

Why not std::string?

Code:

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#include <string>
// ...
std::string name = "omshanti";

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or

Code:

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#include <string>
// ...
std::string name;
name = "omshanti";

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Quote:

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Originally Posted by Tanushreeagr

while using other data type such as int , float , double , we can initialize them after defining them .But same is not the case with character string. WHY???????

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You can't do that with other built-in types either.

Code:

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int main()
{
    int a[3];
    a = { 1, 2, 3 }; // error
}

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You cannot assign to an array.

Also, in the case of char name[20] both name and "omshanti" are interpreted by the compiler as pointers. name is essentially the same thing as &name[0].

So you have a pointer to a locally allocated array of chars and you want to assign it the address of a string literal. Doesn't really make conceptual sense.