How to assign values to const char* array and print to screen

Question

Although I am using C++, as a requirement I need to use const char* arrays instead of string or char arrays. Since I am new, I am required to learn about how to use const char*. I have declared my const char* array as const char* str[5] . At a later point in the program I need to populate each on of the 5 elements with values.

However, if I try to assign a value like this:

const char* str[5];
char value[5];
value[0] = "hello";
str[0] = value[0];

It will not compile. What is the proper way to add a char array to char to a const char* array and then print the array? Any help would be appreciated. Thanks.

Answer 1

1. The string "hello" is made up of 6 characters, not 5.

    {'h', 'e', 'l', 'l', 'o', '\\0'} 

2. If you assign the string when you declare value , your code can look similar to what your currently have:

    char value[6] = "hello"; 

3. If you want to do it in two separate lines, you should use strncpy() .

    char value[6]; strncpy(value, "hello", sizeof(value)); 

4. To place a pointer to value in the list of strings named str :

    const char * str[5]; char value[6] = "hello"; str[0] = value; 

   Note that this leaves str[1] through str[4] with unspecified values.

Answer 2

Various methods to populate const char* str[5] later point in the program.

int main(void) {
  const char* str[5];

  str[0] = "He" "llo";  // He llo are string literals that concat into 1

  char Planet[] = "Earth";  
  str[1] = Planet;  // OK to assign a char * to const char *, but not visa-versa

  const char Greet[] = "How";
  str[2] = Greet;

  char buf[4];
  buf[0] = 'R'; // For a character array to qualify as a string, need a null character.  
  buf[1] = 0;   // '\0' same _value_ as 0  
  str[3] = buf; // array buf converts to address of first element: char *

  str[4] = NULL; // Do not want to print this.

  for (int i = 0; str[i]; i++)
    puts(str[i]);
  return 0;
}

.

Hello  
Earth  
How  
R  

Answer 3

value[0] = "hello";

How can this hold "hello". It can hold only one character.

Also value[5] is not enough for this . There will be no space for '\\0' and program will exhibit UB .

Thus either use value[6] -

char value[6];
strncpy(value,"hello",sizeof value);

Or declare like this -

char value[]="hello";

And after that just make the pointer point to this array. Something like this-

str[0]=value;