Storing integers in char* - c++

Question

I am a beginner is c trying something out. How can i pass an integer array to a function expecting char* in c/c++, since the range of char is only -128 to 127 but i want to store numbers from whole range of integers.
Here is what I was trying:
#include<stdlib.h>
#include<stdio.h>
int size = 12;
void print(char* array){
int i,j;
for(i=0;i<size;i++)
for(j=0;j<size;j++)
printf("%d ",*(array+i*size+j));
printf("\n");
}
int main(){
int array[size][size];
int i=0,j=0,k=0;
for(i=0;i<size;i++)
for(j=0;j<size;j++)
array[i][j]=k++;
for(i=0;i<size;i++)
for(j=0;j<size;j++)
printf("%d ",array[i][j]);
printf("\n");
print((char *) array);
return 0;
}
When array is printed inside the main function, the output is correct (numbers 0-143) but inside the print() function, the output is
0 0 0 0 1 0 0 0 2 0 0 0 3 0 0 0 4 0 0 0 5 0 0 0 6 0 0 0 7 0 0 0 8 0 0
0 9 0 0 0 10 0 0 0 11 0 0 0 12 0 0 0 13 0 0 0 14 0 0 0 15 0 0 0 16 0 0
0 17 0 0 0 18 0 0 0 19 0 0 0 20 0 0 0 21 0 0 0 22 0 0 0 23 0 0 0 24 0
0 0 25 0 0 0 26 0 0 0 27 0 0 0 28 0 0 0 29 0 0 0 30 0 0 0 31 0 0 0 32
0 0 0 33 0 0 0 34 0 0 0 35 0 0 0
I don't know what am I doing wrong.
P.S. I need this since I am using apache Zookeeper and the set and get functions it has require arguments of type char* for your data.

Answer 1

The same way you can use any array or pointer when using e.g. send for sockets: You cast the pointer.
Like
int data[DATA_SIZE];
...
send(s, (const char *) data, sizeof(data), 0);
This is very common, as many functions takes pointer to char even for generic data. The big question is what the function does with the data.

Answer 2

C and C++ are different languages.
"Integer array" is not a C type. Please be specific, such as:
int my_array[42];
You don't pass arrays, in C. Unless it is the operand of the sizeof, unary & or _Alignof operators, or is being used as an initializer, an expression with an array type evaluates to the value of a pointer to the first element of the array. Please see "Array Decay" at http://www.iso-9899.info/wiki/Code_snippets
In C, you can always cast a pointer to any complete or incomplete object type to a pointer to char *. int * is a pointer to a complete object type. Knowing these facts, the following code is valid, inside a function body:
int my_array[42];
char * what_i_want;
what_i_want = (char *) my_array;
Then you can pass the 'what_i_want' value to any function accepting a char * argument.
Having said this, you need to understand what the function is expecting. Share the function and some actual code, then perhaps someone will be able to offer greater insight as to how to achieve your goal.

Answer 3

char * can also be used to point to a string, so you can try using the appropriate conversion (int -> string, string -> int) functions to pass these.
If you post code, I can try and provide a more detailed answer.

Answer 4

Char* - it's the only type of pointer, but you can provide this function everything you want.
You can do something like that:
void func(char* aPointer)
{
int* arr = (int*)aPointer;
int firstInt = *arr;
...
}
Of course, this is very brutal way, so in this case you should be very careful, do some checks and so on.

Answer 5

void func(char* c)
{
...
}
int x[5] = {1,2,3,4,5};
func((char*)x);