C++ Assigning int into char* array

Question

Sorry if this has been asked before, but I couldn't find anything that helped my situation. I'm trying to assign an integer to an index in a char array. In this case, assigning 33 into an index of the token array.

char token[100];
int num = 33;
//Assigning num to this place in array
token[1] = num; //How do I make this work?

I want 33 to be an index in the token array, yet when I assign it and print it out it gives me '!', which is the ASCII value for 33. I want to convert the num to a string and then assign it to the index. So how would I convert num to a string and then assign it to the token array?

Related: https://stackoverflow.com/questions/332030/when-should-static-cast-dynamic-cast-const-cast-and-reinterpret-cast-be-used — Sebastian Hoffmann, Feb 17 '20 at 21:16
"_How do I make this work?_" Please explain, what you want to achieve with this? — Algirdas Preidžius, Feb 17 '20 at 21:16
I think `*` should not be here. Did you mean `char token[100]`? — Tarek Dakhran, Feb 17 '20 at 21:19
Instead of `char * token[100]`, you probably want to write `char token[100]`, That will declare an array of type `char`, which is guaranteed to at least be able to represent numbers in the range 0 to 127. Assigning the number 33 to a pointer does not make sense. Or do you maybe want to convert the number 33 to a string? — Andreas Wenzel, Feb 17 '20 at 21:21
please clarify, do you want an array of characters or array of pointers to characters? — 463035818_is_not_a_number, Feb 17 '20 at 21:21
I am not sure why you need such an assigment but for some reason you want it then use char num=33 and then token[1]=&num or char n=num and then token[1]=&n. The point is you have to assign the address of a char, which is not happening in your case thats why you get error. Simply use char token[100] and then your assignment should work. — rootkonda, Feb 17 '20 at 21:23
`char * token[100];` is an array of 100 pointers to `char`, maybe it's not what you want. — anastaciu, Feb 17 '20 at 21:29

Onyrew · Answer 1 · 2020-02-17T21:46:46.883

If you meant, that you have a pointer, that points to a char array, you can do this.

char* token = new char[100];
    token[0] = 33; // You actually get '!' since char 33 is indeed '!'
std::cout << (int)token[0] << std::endl; // cast it to an int, 33 is back

If you want to use an array of char pointers, you can do this:

char* token[100];
char* temp = new char; // You need a char* to store it since char is 1byte and int is 4byte
    *temp = 33; // or *temp = num; in your case
    token[0] = temp; // You still get '!' since char 33 is indeed '!'
std::cout << (int)*token[0] << std::endl; // cast it to an int

C++ Assigning int into char* array

1 Answers1