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.

Just curiosity, what is the significance of `const char* str[5];` rather than just `char* str[5];`? — Rafaf Tahsin, Sep 04 '15 at 03:45
You seem mixed up. `char value[5]` is an array of 5 characters. `value[0]` is a single character out of those 5. How do you expect a single character to hold the value `"hello"` ? — M.M, Sep 04 '15 at 03:49
@RafafTahsin Significance here in this question or in general ? — ameyCU, Sep 04 '15 at 03:53
mostly here in this question. what is the significance of using `const char* str[5];` here? and for general purpose, so far I know, the significance of `const char* str[5];` is that it's a variable pointer which points to a `const` variable. Does it have any other significances? >>> @ameyCU — Rafaf Tahsin, Sep 04 '15 at 04:02
It's not bad to learn naked pointer logic, but it _is_ a good thing to define your concepts: `typedef const char* MyString; MyString strings[5];` This will for sure help avoiding mistakes. — xtofl, Sep 04 '15 at 04:06
In C, _string_ is not a type, nor a `char *`, nor a pointer. A _string_ is a character array up to and including a terminating null character `'\0'`. — chux - Reinstate Monica, Sep 04 '15 at 04:18

Bill Lynch · Accepted Answer · 2015-09-04T03:07:14.390

8

The string "hello" is made up of 6 characters, not 5.
```
{'h', 'e', 'l', 'l', 'o', '\0'}
```
If you assign the string when you declare value, your code can look similar to what your currently have:
```
char value[6] = "hello";
```
If you want to do it in two separate lines, you should use strncpy().
```
char value[6];
strncpy(value, "hello", sizeof(value));
```
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.

edited Sep 04 '15 at 03:07

answered Sep 04 '15 at 02:59

Bill Lynch

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How about `char value[] = "hello";`? – Captain Obvlious Sep 04 '15 at 03:08
@CaptainObvlious: Sure. But it's not obvious to revolution how many bytes that occupies. – Bill Lynch Sep 04 '15 at 04:12
@BillLynch, your last example `str[0] = value;` seems very dangerous. It assigns _4 byte pointer address_ value to _1 byte char value_ place. – Fredrick Gauss Mar 23 '17 at 05:47
`str` in this example is a list of strings. – Bill Lynch Mar 23 '17 at 05:56

chux - Reinstate Monica · Answer 2 · 2015-09-04T04:14:03.990

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

score 0 · Answer 3 · answered Sep 04 '15 at 03:52

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;

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

3 Answers3