Question

I have array in which i want to store values like this 1203

   char* arr= new char[10];
    arr[0] = 1;
    arr[1] = 2;
    arr[2] = 0;
    arr[3] = 3;

but after storing 0 i am not able to see any further data bcs it’ll consider that as end of the data. is there any way to manage this??

Answer 1

0 means end of string in the char*.

Because char* is a string in C and it exists to be used as char array not as integer array. All function related on string are intended to be used with a string.

You can use int* to store integers.

You can do the binary operations manually like this :

int value = 1245; 

char* temp = new char[4]; 

temp[0] = (char)(value >> 24);
temp[1] = (char)(value >> 16);
temp[2] = (char)(value >> 8); 
temp[3] = (char)value;

return temp;

you can use uint8_t or int instead of char.

the (char*)((int)value) is not good because you cast the int to char*, but the char* is the variable temp not the number to be stored in temp.

If pointers is a new thing for you, it will be helpful to do some exercises in pointers : manipulate arrays, lists,…