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
– Rafaf Tahsin, Commented Sep 4, 2015 at 3:45
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
– M.M, Commented Sep 4, 2015 at 3:49
@RafafTahsin Significance here in this question or in general ? — ameyCU
– ameyCU, Commented Sep 4, 2015 at 3:53
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
– Rafaf Tahsin, Commented Sep 4, 2015 at 4:02
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
– xtofl, Commented Sep 4, 2015 at 4:06

Bill Lynch · Accepted Answer · 2015-09-04 03:07:14Z

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 4, 2015 at 3:07

answered Sep 4, 2015 at 2:59

Bill Lynch

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4 Comments

Captain Obvlious Over a year ago

How about char value[] = "hello";?

Bill Lynch Over a year ago

@CaptainObvlious: Sure. But it's not obvious to revolution how many bytes that occupies.

Fredrick Gauss Over a year ago

@BillLynch, your last example str[0] = value; seems very dangerous. It assigns 4 byte pointer address value to 1 byte char value place.

Bill Lynch Over a year ago

str in this example is a list of strings.

chux · Accepted Answer · 2015-09-04 04:14:03Z

1

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

edited Sep 4, 2015 at 4:14

answered Sep 4, 2015 at 4:00

chux

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Comments

ameyCU · Accepted Answer · 2015-09-04 03:52:30Z

0

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;

answered Sep 4, 2015 at 3:52

ameyCU

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