C-string assignment using for loop - c++

Question

I think this is a relevant topic, while browsing this forum and tutorial I have tweaked my c-string to the proper format(I think) but there's just one topic missing. How can we take an integer, using a for loop, and assign the c-string values from the integer.
I'm just focusing on the integer to binary part right now and I'm sure my number manipulation is solid. However, my prof said we needed to assign the binary values to a c-string. And I'm trying this, it's telling me I'm using a const char and a char* via the compiler. I'm not sure how this is happening, or how to prevent it.
Here's my source code:
//sample integer to binary
#include <iomanip>
#include <iostream>
#include <iomanip>
#include <cstring>
using namespace std;
int main()
{
int num; //the number a user enters
int rem; //the remainder, the 1 and 0 of the binary number
int x; //a variable to store the number after division
char binary[10]; //c-string initialized to 10, perhaps that is too many.
cout << "Enter a number: ";
cin >> num;
for (int i = 0; i < 10; i++)
{
x = num / 2;
cout << x << endl; //this shows that the code is working
rem = num % 2;
cout << num << endl; //this also shows the code is working
char r = (char)rem; //These two lines of code are
strcpy(binary[i], r); //preventing compilation
cout << binary[i] << endl; // this is diagnostic
num = x;
}
cout << "The number " << num << " is " << binary[5] << " in binary.\n";
return 0;
}
Thanks you two, I've been able to make this work (almost).
I'm still getting some unexpected behavior, and I'm not sure how big to initialize the array to, I don't think that it would matter too much, but I don't know exactly how big of numbers the graders use to test.
Anyways, here's the new code:
#include <iomanip>
#include <iostream>
#include <iomanip>
#include <cstring>
using namespace std;
int main()
{
int num; //the number a user enters
int rem; //the remainder, the 1 and 0 of the binary number
int x; //a variable to store the number after division
char binary[5] = {'0', '\0'}; //c-string initialized to 10, perhaps that is too many.
cout << "Enter a number: ";
cin >> num;
for (int i = 0; i < 10; i++)
{
x = num / 2;
//cout << x << endl; //this shows that the code is working
rem = num % 2;
//cout << num << endl; //also shows the code is working
binary[i] = '0' + rem; //not sure what this is doing, but it works.
//cout << binary[i] << endl; // this is diagnostic
num = x;
}
cout << "The number " << num << " is " << binary << " in binary.\n";
return 0;
}
And here's the output:
Enter a number: 5
The number 0 is 1010000000# in binary.
It should show the number, the initial number, and say 101, without the 0's and the # sign.

Answer 1

strcpy is for copying entire NUL-terminated string. If you want to set just one character, you can use =.
binary[i] = r;
However, if you want the line cout << binary[i] to work right, or to treat binary as a C string later, you need to store ASCII digits:
binary[i] = '0' + r;
Don't forget to add a terminating NUL to your string, right now it isn't a C-style string at all.

Answer 2

The arguments to strcpy() are char*, not char. The second argument must point to a null-terminated string, but r is just a single character. To assign a character to an element of an array, do:
binary[i] = r;
But you don't want the binary value of rem, you want the character that represents that binary value. It should be:
char r = '0' + rem;
In order to print binary as a string, you need to give it a null terminator. Since you're putting 10 digits into the string, you need to declare an extra character to hold the terminator, and initialize it with zeroes so it will be terminated properly.
char binary[11] = {0};
And if you want to print the whole string, you shouldn't reference binary[5], you should print the whole array:
cout << "The number " << num << " is " << binary << " in binary.\n";

Answer 3

In addition to problems already pointed out, you get binary digits in the reverse order. You have to count significant digits in your binary representation and reverse characters when you're done.
You didn't initialize the character buffer correctly that's why you are getting garbage in the print out.
As for the buffer size, you need as many characters as there are bits in the integer type you are converting plus one more for the terminating '\0', so for a 32-bit number you need 33 character size buffer. And that's important because if you overrun your buffer, quite nasty things will happen.
And one more note: I assume that numbers you are supposed to convert to a string representation are unsigned, so be explicit about it. Below is a quick and dirty implementation with some minimal error checking:
char *unsigned_to_binstr (unsigned n, char *binary, int buf_len)
{
int i, j;
if (buf_len < 2)
return NULL;
i = 0;
do {
binary[i++] = '0' + n % 2;
n /= 2;
} while (n && i < buf_len);
for (j = 0; j < i / 2; ++j) {
char temp = binary[j];
binary[j] = binary[i-j-1];
binary[i-j-1] = temp;
}
binary[i] = '\0';
return binary;
}

Answer 4

After everyone's comments, and my own research via my textbook I was able to formulate a somewhat working function. However, in the process I began to wonder about the garbage problem, so my solution.. Truncate! I added if clauses (tried to think of a loop representation but didn't get any ideas) and at the end I just use the length provided via if clauses and subtract 1, which should give me the appropriate bit size for the number. Using what basic c++ knowledge we've covered in class this is the solution I came up with, however crude and inefficient it might be!
I want to thank all of you, I couldn't have gotten past the point I was stuck at if it weren't for you!
Here's my rough final revision:
//sample integer to binary
#include <iostream>
using namespace std;
int main()
{
int num; //the number a user enters
int rem; //the remainder, the 1 and 0 of the binary number
int x; //a variable to store the number after division
int l; //length of c-string function
cout << "Enter a number: ";
cin >> num; //user input
if ((num == 1) || (num == 0)) // the following 17 lines of code are to truncate the string size.
l = 2;
if ((num > 1) && (num < 4))
l = 3;
if ((num >= 4) && (num <= 7))
l = 4;
if ((num >= 8) && (num <= 15))
l = 5;
if ((num >= 16) && (num <= 31))
l = 6;
if ((num >= 32) && (num <= 63))
l = 7;
if ((num >= 64) && (num <= 127))
l = 8;
if ((num >= 128) && (num <= 255))
l = 9;
if ((num > 255))
cout << "This number is too large for this string\n"; // I don't think the binary sequence should be larger than 16 bits.
char binary[l]; //c-string initialized to size according to the truncation rules above
for (int i = l - 1; i > 0; i--) //goes in reverse order, as binary counts from bottom to top.
{
x = num / 2;
rem = num % 2;
num = x;
binary[i] = '0' + rem; // added an
}
for (int i = 0; i <= l-1; i++)
{
cout << binary[i];
}
cout << " in binary.\n";
return 0;
}