如何在字符数组中使用默认值？

Question

I have the easiest question possible: how to initialise a value in C? My variable is of the type char[20] and it is declared somewhere outside of my unit. It is impossible to change the type.

Now I would like to give it a default value, let's say all empty characters (or spaces, whatever), but this is not working at all. I have already tried :

Method 1.

host = "";

=> cannot convert from 'const char [1]' to 'char [20]'

Method 2.

host = "                   ";

=> cannot convert from 'const char [20]' to 'char [20]

Method 3.

host = '                   ';

=> cannot convert from 'int' to 'char [20]

Method 4.

host = {"", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", };

=> syntax error : '{'

...

I am getting desperate here: how and why have the inventors of C made it so difficult to simply declare a value to a variable :-( ?

Answer 1

memset(&host, 0, sizeof(host));

将用零填充它。

Answer 2

host = "";

Will work only if the it is char* and not char[20] (if you are doing a function just make the parameter char*). The

host = {"", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", };

Is ok but, first use

{'character', 'character','character'...

(substitute the character with any character of ASCII for example \\0).

and secondly (that is the main problem here) you have a , where it shoudln't be (in the end, so delete this) .

Answer 3

1. Do you know, the meaning of initialize ? Usually, it's performed at defining time only.
2. declared somewhere outside of my unit , or do you mean defined ?

If host is really declared elsewhere, while defining it, you may use

char host[20] = {0};   //to fill with 0

or

char host[20] = { [0 ... 19] = 5 };  //to fill with 5 , supported on gcc

At any later point, if you want to re-initialize to some value, memset() is the way to go. Check the man page here .

Answer 4

I am thinking that you have declared the host variable as an integer variable. If you have declared that as char variable,

host = "";
host = "     ";

These two are possible. Because it will take the characters for corresponding position.

host = ' '; It is not possible because in single quotes we have to give the character only. In this you have to mention the array position otherwise it will throw the error. host[0]=' ';

host = {"", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", };

It is not possible because in this it will take the corresponding characters not a string.