typedef

[siperi]

int main()
{
	typedef char arr2[5];
	arr2 name="name";
	printf("%s",name);
	return 0;
}

In above code, if I declare arr2[5] name="name"; then it throws error. why?

[Bennie98]

#include <stdio.h>

int main()
{
	typedef char arr2[5];
	arr2 name="name";
	printf("%s\n",name);
	return 0;
}

doesn't give me errors... spells out: "name"...

gave me an error because there wasn't any I/O examples being included.... so I put in "#include <stdio.h>" hopefully this was the case

[laserlight]

In above code, if I declare arr2[5] name="name"; then it throws error. why?

For starters, the syntax is wrong, i.e., it would be arr2 name[5] = "name"; But if you did that, then it would be tantamount to writing char name[5][5] = "name";. That does not make sense since "name" is a const char[5]. Basically, the code snippet you posted is okay (assuming you #include <stdio.h> earlier), changing it in the way you described is not okay.

[siperi]

Hi I used (arr2[5] name="name" instead of (arr2 name="name" in the above program. That time I got error.

[laserlight]

Right. So do you now understand why that is wrong?

[siperi]

I'm not quite clear. Please explain briefly.

[laserlight]

You used the typedef to make arr2 an alias for the type array of 5 char. Therefore, arr2[5] is the type array of 5 array of 5 char. As such, you cannot assign "name" to the array.

But besides this, if you actually wrote:

arr2[5] name = "name;

it would be syntactically wrong regardless of the typedef. The [5] cannot appear at that place. That is, this is also wrong:

char[5] name = "name";