Thread: integer to pointer without cast

Code:

int criar_contacto()
{
	char name[33], bday[11];
	int cell_phone;

	printf("Introduza o Nome do contacto:\n");

	get_string(name, 33);

	printf("%s", name); // This is just for test purpose


	return 1;
}

void get_string(char *string_to_get, size_t string_size)
{
	int i;

	fgets(*string_to_get, string_size, stdin);

	for(i=0; i<string_size; i++)
	{
		if(string_to_get[i]=='\n')
			{
				string_to_get[i]='\0';
				break;
			}
	}
}

So the first obvious problem here is in "get_string", Eclipse(A linux C SDK) is telling me that "passing argument 1 of ‘fgets’ makes pointer from integer without a cast".

I've got no idea of what it means, I've googled it though, found many things about it but I didn't managed to figure it out by myself.

The code compiles flawlessly, but when I use criar_contacto() which calls get_string() my app just terminates. My main() is just a switch case to create a menu where each case calls a fuction.

My get_string is suposed to get a string with fgets and then cut the "\n" from it, so I don't always have to write the loop to have it done.

I'm trying to make a "notebook for cell phone numbers" (I'm portuguese so I guess that's not what you guys call it, but i hope you get it).

Thanks in advance,

Hugo Ribeira

You want:

Code:

	fgets(string_to_get, string_size, stdin);

The error is because you are passing the first char in string_to_get instead of the whole string. Char is an integer type, the string is a pointer type. Ergo conversion from integer to pointer without a cast.

It works now, but I don't understand why.

what goes in that argument is name[], right? which acts as a pointer, so shouldn't I call the argument type a pointer as well?

EDIT: Understood, thank you.

What goes in that argument is a pointer to a string - not a char pointer to a pointer to a string.

Use string_to_get, just like you would use name.