初始化在C中具有确切长度的字符数组？

Question

char ch[2];
ch[0] = 'h';
ch[1] = '\0';

or can i just do

char ch[1];
ch[0] = 'h';

in which '\\0' would already be implied?

The reason why i'm doing it this way is because my program has a loop in it in which i would constantly be changing this character and concatenating it to a longer string.

Answer 1

Assigning just ch[0] will not assign zero to the ch[1] . You can initialize both characters in the array with a string literal in a single line, like this:

char ch[2] = "h";

This will put 'h' into ch[0] , and a terminating zero into ch[1] .

Answer 2

如果愿意，您可以省略终止的空字符，但是您将无法使用需要终止的空字符的函数，例如strlen 。

Answer 3

Well, one option if you plan to be using this single-character string over and over is to just initialise the null once. Then you only ever have to change the first character:

char ch[2] = {0};
ch[0] = 'h';

In your second case, you don't need to declare an array:

char ch = 'h';

But you can't use that as a string. There is no implied null character in your single-character 'array'.

Why can't you just append characters to your string instead of strings? You don't need strcat , if that's what this is about... You just keep appending single characters until you are done and then you append a null.

char somestring[100];
char pos = 0;
while( ... ) {
    somestring[pos] = 'x';  /* or whatever */
    pos = pos + 1;
} 
somestring[pos] = 0;

Answer 4

ch is an char array and not a string.

The compiler can recognize it as a string only if you add the '\\0' ; however, it cannot recognize it as an array of char.

So, you must add the '\\0' , you can do like this:

int ind=0;
while(...)
{
    ch[ind++]='h';
}
ch[ind]='\0';