C编程：声明字符串和赋值

Question

int main(){
   char inputType[5] = "hello you"; //a string 
   char *word[5]; //a string that will contain 4 character + the null
   word = get_word(inputType);
   return 0;
}
char *get_word(char inputType[]){ //inputType is a string
    char *array[2]; //an array that will store ["hello", "you"]
    // got correct code to make it ["hello", "you"]
    return array[1] //want to return "you"
}

I get this error message:

Error: word = get_word (inputType) assingment to expression with array Type------------------------------ 

Questions:

1. Does char inputType[5] declare a string?
2. Difference between char *word[5] vs inputType[5]
3. The reason of the error message.

Answer 1

This is wrong

char inputType[5] = "hello you"; 

hello you is 9 characters long and you also need the null terminator so you need...

char inputType[10] = "hello you";

This is wrong because the comment is wrong

char *word[5]; //a string that will contain 4 character + the null

char *word[5] is not a C string, it's an array of 5 pointers to C strings that have not yet been allocated.

Answer 2

1. Yes it declares a char array of size 5 (including the NUL character)where you are storing more than 5 characters.

2. First one is array of char* and second one array char s.

3. Yes char*word[5] is not what you are thinking..unless you allocate memory it can just hold addresses of memory locations where characters are stored.

   char* arr[5]

     | ---------------------------------- ~~ | | | | | char* | | | ---------------------------------- ~~ arr[0] arr[1] arr[2] 

Few things that you can do

1. you are basically initializing the char array. So do this char inputType[]= "Hello You";

2. The error message is basically shown due to this line word = get_word(inputType); well what you are doing? You are assigning char* to an array variable which is supposed to contain char* -s

So that will be word[0]=get_word()....

3. Don't forget to keep a place for NUL terminator at the end of char array.