Weird symbols in copied/truncated character array

Question

#include <iostream>
using namespace std;

void truncate(char* s1, char* s2, int n) {
    for (int i = 0; i < n; i++) {
        s2[i] = s1[i];
    }
}

int main() {
    char s1[15] = "Hello World";
    char s2[10];
    int n{ 5 };
    truncate(s1, s2, n);
    cout << s2; // this should display: Hello
}

When I run this in Visual Studio, I get this output:

Hello╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠Hello World

But when I run it using an online compiler (Online GDB), the output is normal:

Hello

I know it has to do with the \0 terminator but I just can't figure it out.

Note: The main function was provided as it is in the post above by my teacher, so I can't change the arrays sizes'. — PixelatedCosmos, Apr 12 '22 at 13:53
And in fact you don't need to change the array sizes. But do make sure that `s2` ends with a null character at the end of the `truncate()` function. — Fred Larson, Apr 12 '22 at 13:57

Adrian Mole · Accepted Answer · 2022-04-12T14:21:54.593

Your truncate function doesn't add the required nul terminator character to the destination (s2) string. Thus, when you call cout in main, that will keep printing characters until it finds its expected nul (zero) character at some unspecified location following the garbage-initialized data that pre-exists in the parts of the s2 local array that weren't explicitly set by your function. (This is undefined behaviour, and almost anything could actually be displayed; the program may also crash, if the cout call tries to read memory it doesn't have the required access to).

To correct this, simply add that nul (zero, or '\0') character to the end of the string in the truncate function. Assuming you don't go beyond the array's bounds, you can use the "left over" value of the loop index, i, to access the required element; but, to do so, you will need to declare the int i outside of the for loop's scope:

void truncate(char* s1, char* s2, int n)
{
    int i; // Must be declared outside the FOR loop ...
    for (i = 0; i < n; i++) {
        s2[i] = s1[i];
    }
    s2[i] = '\0'; // ... so we can use it here.
}

Note: Another way (but apparently prohibited by your teacher) would be to set all elements of the s2 array in main to zero, before calling your truncate function:

char s2[10] = ""; // Set ALL elements to zero

Some compilers (seemingly including the online one you use) will "implicitly" set the elements of uninitialized local arrays to zero; but never rely on this: it is not part of the C++ Standard and compilers are not required to enforce such behaviour.

Note:, as @Fred Larson says, One could also keep the `for` loop as in the original code and just add `s2[n] = 0;` somewhere (anywhere, really) in the `truncate` function. But what I have shown is, IMHO, the more 'idiomatic' C++ or C way of doing such things. — Adrian Mole, Apr 12 '22 at 14:17

Weird symbols in copied/truncated character array

1 Answers1